MATHS TEACHER SUPPORT:
FET Euclidean Geometry: The Power of Proof
25 January 2022
Hosted by Gretel Lampe
Presented by Anne Eadie
Q & A by Jenny Campbell &
Susan Carletti
2022
Maths Teacher
Support
Offerings
Webinars & Videos
Free e-books for teachers
TAS Maths teachers
WhatsApp group
Mathematics
Euclidean Geometry:
Power of Proof
The Major Issues
Language
Knowledge
Logic
and then
Strategies
1
SUMMARY OF
CONTENTS OF BOOKLETS
BOOKLET 1: Teaching Documents
BOOKLET 2: Lines, Angles, Triangles & Quadrilaterals
BOOKLET 3: Circle Geometry
BOOKLET 4: Proportionality, Similarity and
The Theorem of Pythagoras
These Geometry materials (Booklets 1 to 4) were created and produced
by The Answer Series Educational Publishers (Pty) (Ltd)
to support the teaching and learning of Geometry
in high schools in South Africa.
They are freely available to anyone who wishes to use them.
This material may not be sold (via any channel) or used for
profit-making of any kind.
TAS FET EUCLIDEAN
GEOMETRY COURSE
BOOKLET 1
Teaching
Documents
by The TAS Maths Team
Mastering FET
Geometry
Geometry FET
Course Booklets Set
WWW.THEANSWER.CO.ZA
Booklet 1: TEACHING DOCUMENTS
 Exam Mark Distribution (FET Maths Paper 2)
 The 2022 ATP (Proposed)
 Research: Results & Diagnostic Reports (2020)
 Exemplar FET Geometry Questions and Detailed Solutions
 Problem-solving approach: A C T
 The CAPS Curriculum Overview (FET Geometry)
 The FET Geometry CAPS Curriculum per Grade (10 – 12)
 Acceptable reasons for statements in Geometry
 Important Advice for Learners
FET EXAM: Mark distribution
Proofs: maximum 12 marks
PAPER 2
Description
GR 10
GR 11
GR 12
Statistics
15
20
20
Analytical Geometry
15
30
40
Trigonometry
40
50
50
Euclidean Geometry & Measurement
30
50
40
100
150
150
TOTAL
NOTE:
• Questions will not necessarily be compartmentalised in sections, as this table indicates.
Various topics can be integrated in the same question.
• A formula sheet will be provided for the final examinations in Grades 10, 11 and 12.
January
2022
CAPS Curriculum
page
17
A PROPOSED 2022 ATP FOR FET MATHS
Grade 10
Grade 11
Term 4
Term 3
Term 2
Term 1
No. of
weeks
Algebraic Expressions,
Numbers & Surds
4
Exponents, Equations & Inequalities
2
Equations & Inequalities
1
Euclidean Geometry (#1)
3
Grade 12
No.
of weeks
Exponents & Surds
Equations
Equations & Inequalities
1
1
2
Euclidean Geometry
4
Trig functions &
Revision of Grade 10 Trig
Trig identities &
Reduction formulae
1
No.
of weeks
Patterns, Sequences and Series
3
Euclidean Geometry
3
Trigonometry
(Algebra)
4
1
Trigonometry (#1)
3
Trig equation & General solutions
1
Analytical Geometry
2
Number Patterns
1
Quadrilaterals
1
2
Functions (including
Trig Functions (#2))
6
5
2
2
2
5
1
Calculus, including Polynomials
Measurement
Analytical Geometry
Number Patterns
Functions
Trig – sin/cos/area rules
Functions & Inverse Functions
& Exp & Log Functions
Finance
3
Statistics
Probability
Finance (Growth)
Analytical Geometry
2
2
2
2
Trig – sin/cos/area rules
1
Finance
1
Measurement
2
Statistics
3
Statistics (regression & correlation)
3
Probability
4
Counting and Probability
3
Euclidean Geometry (#2)
3
Finance (Growth & Decay)
1
INTERNAL EXAMS
4
Revision
4.
Finance (Growth & Decay)
3
Revision (Paper 1)
1
FINAL EXAMS
3
Revision
1
Revision (Paper 2)
1
FINAL EXAMS
4
Revision (Exam Techniques?)
1
Reporting
1½
Reporting
1½
EXTERNAL EXAMS
6½
RESEARCH:
Results & Diagnostic Reports (2020)
Some interesting statistics ...
ave. over 7 years
2014 – 2020
2020
Statistics
60,1%
74,4%
Analytical Geometry
54,3%
52,3%
Trigonometry
41,6%
46,7%
Euclidean Geometry
46,2%
42,9%
Paper 2
50,6%
50,6%
vs Paper 1
52,2%
50,6%
PAPER 2:
2021?
Note: This information is based on a random sample of candidates and may not reflect
the national averages accurately.
However, it is useful in assessing the RELATIVE degrees of each question
AS EXPERIENCED BY CANDIDATES.
2020: Paper 2
Average % performance per question
Q1 Data Handling
Q2 Data Handling
Q3 Analytical Geometry
Q4 Analytical Geometry
Q5 Trigonometry
Q6 Trigonometry
Q7 Trigonometry
Q8 Euclidean Geometry
Q9 Euclidean Geometry
Q10 Euclidean Geometry
Average performance (%)
and, per sub-question
Sub-question
DIAGNOSTIC REPORT: DBE NOV 2020
QUESTION
8.1
Common Errors and Misconceptions
O is the centre of the circle.
(a)
ˆ = 26º.
KOM bisects chord LN and MNO
ˆ = Kˆ because these
In Q8.1.1 (a) some candidates assumed that Q
2
1
angles appeared to be in the same segment. It was incorrect to assume
K and P are points on the circle with
ˆ = 32º. OP is drawn.
NKP
that KPNO was a cyclic quadrilateral.
L
(b)
K
1
O
2
1
32º
2 3
ˆ = 90°.
were unable to provide the correct reason for the statement M
2
1
M
2
They wrote ‘line from centre to midpoint’ or just ‘perpendicular lines’.
Neither of these was accepted.
26º
P
8.1.1
(c)
N
the size of:
8.1.2
(b) Ô 1
Prove, giving reasons, that
ˆ
KN bisects OKP.
Copyright © The Answer Series
Some candidates made the following incorrect assumptions when
answering Q8.1.2:
Determine, giving reasons,
(a) Ô 2
ˆ = 90°. Other candidates
Some candidates incorrectly assumed that M
2
(2)(4)
Kˆ 2 = Kˆ 1 (as if it was given information), ON || KP and OP ⊥. KN.
ˆ . These candidates
Some candidates incorrectly stated that P̂ = O
1
(3)
assumed that these angles were opposite sides of equal length.
1
QUESTION
MEMO
8.1
8.1
O is the centre of the circle. KOM bisects chord LN and
L
ˆ = 26º. K and P are points on the circle with NKP
ˆ = 32º.
MNO
OP is drawn.
K
L
1
O
2
1
32º
2 3
1
M
2
26º
K
1
O
2
1
32º
2
3
1
2
N
P
M
8.1.1
26º
ˆ
O
2 = 2(32º)
(a)
. . . ø at centre = 2 % ø at circumference
= 64º P
N
ˆ + O
ˆ = 26º + M
ˆ
O
1
2
2
(b)
ˆ = 90º
M
2
&
8.1.1 Determine, giving reasons, the size of :
. . . line from centre to midpoint of chord
ˆ + 64º = 26º + 90º
∴ O
1
(a) Ô 2
(2)
(b) Ô1
(4)
ˆ = 52º ∴ O
1
ˆ = 90º
M
2
OR :
ˆ
8.1.2 Prove, giving reasons, that KN bisects OKP.
ˆ = 64º
∴O
3
(3)
ˆ = 52º ∴ O
1
8.1.2
ˆ
Kˆ 2 = KNO
=
1ˆ
O3
2
. . . ø sum in Δ
s
. . . ø on a straight line
s
. . . exterior ø of Δ
ˆ  KN bisects OKP
2
. . . line from centre to midpoint of chord
. . . ø opp equal radii
= 32º Copyright © The Answer Series
. . . exterior ø of Δ
QUESTION
8.2
Common Errors and Misconceptions
In ΔABC, F and G are points on sides AB
and AC respectively.
(d)
In Q8.2.1 many candidates made the incorrect assumption that FG || BC.
Using this information, they would conclude that B̂ = Fˆ1 because the
D is a point on GC such that D̂1 = B̂ .
A
corresponding angles were equal. Other candidates made unnecessary
constructions and wanted to prove the proportionality theorem.
F
1
G
Some candidates could not provide a valid reason as to why FG was
2
1
2
parallel to BC.
D
C
(e)
B
8.2.1
Those who could identify the correct proportion were unable to
If AF is a tangent to the circle
passing through points F, G and D,
then prove, giving reasons, that
FG || BC.
8.2.2
Candidates were unable to state the correct proportion in Q8.2.2.
simplify correctly.
(4)
7x + 63 = 10x – 30
If it is further given that AF = 2 ,
FB
5
–7x + 10x = –30 + 63
Algebra!
AC = 2x – 6 and GC = x + 9,
then calculate the value of x.
(4)
[17]
Copyright © The Answer Series
Candidates failed to provide a reason for the proportion that they wrote.
3
QUESTION
MEMO
8.2
8.2.1
In ΔABC, F and G are points on sides AB and AC respectively.
ˆ
Fˆ 1 = D
1
ˆ
D is a point on GC such that D̂1 = B.
= B̂
∴ FG | | BC
A
. . . tan chord theorem
. . . given
s
A converse theorem
1
2
Fˆ1 = Bˆ
D
C
B
G
2
8.2.2
D
AG = (2x – 6) – (x + 9)
A
= x – 15
C
AC
= AB
GC
FB
B
x+9
If AF is a tangent to the circle passing through
. . . proportion theorem FG | | BC
that FG || BC.
G
F
x+9
5
5
C
∴ 5x – 75 = 2x + 18
points F, G and D, then prove, giving reasons,
x – 15
2
∴ x – 15 = 2
8.2.2
1
2
1
8.2.1
G
2
. . . corresp. ø :
A
F
1
F
B
∴ 3x = 93
(4)
∴ x = 31 If it is further given that AF = 2 , AC = 2x – 6
FB
and GC =
5
x + 9, then calculate the value of x.
(4)
OR :
[17]
AB : FB = 7 : 5
AC
= AB
GC
FB
. . . proportion theorem FG | | BC
A
∴ 2x – 6 = 7
x+9
2
5
∴ 10x – 30 = 7x + 63
∴ 3x = 93
7
F
5
∴ x = 31 4
x+9
C
B
Copyright © The Answer Series
2x – 6
G
QUESTION 8: Suggestions for Improvement
(a)
Learners should be encouraged to scrutinise the given information and the diagram for clues about which theorems
could be used in answering the question.
(b)
Teachers must cover the basic work thoroughly.
An explanation of the theorem should be accompanied by
showing the relationship in a diagram.
(c)
Learners should be told not to make assumptions based on what they see in the diagram. They should be reminded
that the diagrams are not drawn to scale.
(d)
Learners should be taught that all statements must be accompanied by reasons. It is essential that the parallel lines
be mentioned when stating that corresponding angles are equal, alternate angles are equal, the sum of the
co-interior angles is 180° or when stating the proportional intercept theorem.
(e)
Learners should know that writing a correct statement and reason does not guarantee marks.
They will only get marks if that statement and reason leads to the solution.
Copyright © The Answer Series
5
QUESTION
9.1
Common Errors and Misconceptions
In the diagram, O is the centre of the circle.
(a)
Points S, T and R lie on the circle.
Chords ST, SR and TR are drawn in the circle.
In answering Q9.1, some candidates either did not do
the construction correctly or they failed to do the construction
QS is a tangent to the circle at S.
altogether. Other candidates attempted to use more than
one method to prove this theorem but did not reach a
conclusion through any method.
R
T
O
Q
S
Use the diagram to prove the theorem which
ˆ = R.
ˆ
states that QST
Copyright © The Answer Series
(5)
6
QUESTION
MEMO
9.1
9.1
In the diagram, O is the centre of the circle.
Points S, T and R lie on the circle.
The proof of the tan chord theorem :
ˆ = x
Let QST
Chords ST, SR and TR are drawn in the circle.
QS is a tangent to the circle at S.
ˆ
OSQ
= 90º
. . . radius ⊥ tangent
∴ Sˆ 1 = 90º – x
∴ Tˆ 1 = 90º – x
R
s
. . . ø opp equal radii
T
ˆ
∴ TOS
= 180º – 2(90º – x)
O
= 2x
∴ R̂ = x
Q
Use the diagram to prove the theorem which states that
Copyright © The Answer Series
. . . ø at centre = 2 % ø at circumference
ˆ = R̂ ∴ QST
S
ˆ = R.
ˆ
QST
s
. . . sum of ø of Δ
(5)
7
QUESTION
9.2
Common Errors and Misconceptions
ˆ and intersects chord MP at S.
Chord QN bisects MNP
The tangent at P meets MN produced at R such that
R
N
1 2
1
In Q9.2.1 many candidates did not provide a correct or
complete reason for their statements.
QN || PR. Let P̂1 = x.
M
(b)
3
(c)
2
Instead of using the proportionality theorem to answer Q9.2.2,
candidates attempted to use similar triangles to find the ratios.
S
x
3
Some incorrectly assumed that S was the centre of the circle
2 1
and that MNPQ was a square. From these assumptions, they
P
1 2
then stated that there were a number of sides that were equal
Q
in length.
9.2.1
Determine the following angles in terms of x.
Give reasons.
9.2.2
(a)
N̂ 2
(2)
(b)
Q̂ 2
(2)
Prove, giving reasons, that MN = MS
Copyright © The Answer Series
NR
SQ
(6)
[15]
8
QUESTION
9.2
MEMO
ˆ and intersects chord MP at S.
Chord QN bisects MNP
9.2.1
The tangent at P meets MN produced at R such that QN || PR.
Let P̂1 = x.
1
1
)
. . . alternate ø ; QN | | PR
(
)
. . . tan chord theorem
ˆ = Pˆ = x N
2
1
(b)
ˆ = Pˆ = x Q
2
1
s
R
R
N
M
(
(a)
N
x
3
M
2
3
1 2
x
2
1
2
x
S
S
x
x
3
2 1
1 2
P
3
2 1
P
x
Q
1 2
Q
9.2.2
9.2.1
Mark the sides MN, NR, MS & SQ on the sketch.
Three of these lengths and the | | lines point towards proportion theorem !
Determine the following angles in terms of x.
Give reasons.
In ΔMPR :
9.2.2
(a)
N̂ 2
(2)
(b)
Q̂ 2
(2)
Prove, giving reasons, that MN = MS
NR
SQ
MN
= MS
NR
SP
. . . proportion theorem ; SN | | PR
We need to prove that
SP = SQ
(6)
ˆ
Pˆ 3 = N
1
[15]
ˆ
= N
2
s
. . . ø in same segment
ˆ
. . . Given : QN bisects MNP
= x
&
ˆ = x
Q
2
. . . in Q9.2.1(b)
ˆ (= x)
∴ Pˆ 3 = Q
2
∴ SP = SQ
s
. . . sides opposite equal ø
∴ MN = MS NR
Copyright © The Answer Series
9
SQ
QUESTION 9: Suggestions for Improvement
(a)
Learners should be taught that a construction is required in order to prove a theorem. If the construction is
not shown, then the proof is regarded as a breakdown and they get no marks. Teachers should test theory
in short tests and assignments.
(b)
Learners should be discouraged from writing correct statements that are not related to the solution.
No marks are awarded for statements that do not lead to solving the problem.
(c)
Learners should be forced to use acceptable reasons in Euclidean Geometry. Teachers should explain
the difference between a theorem and its converse.
They should also explain the conditions for which
theorems are applicable and when the converse will apply.
(d)
Learners need to be told that success in answering Euclidean Geometry comes from regular practice, starting off
with the easy and progressing to the difficult.
Copyright © The Answer Series
10
QUESTION
Common Errors and Misconceptions
QUESTION 10
(a)
 In the diagram, a circle passes through
D, B and E.
Some candidates assumed that FBDM was a cyclic quadrilateral instead
of proving that it was. Many candidates did not provide the correct reason
when concluding why FBDM was a cyclic quadrilateral.
 Diameter ED of the circle is produced to C
and AC is a tangent to the circle at B.
A common error was ‘opp øs of a cyclic quad’.
 M is a point on DE such that AM ⊥ DE.
 AM and chord BE intersect at F.
(b)
In Q10.1.2 some candidates did not use the knowledge that FBDM was
a cyclic quadrilateral. They incorrectly assumed that AM was a tangent
to the circle FBDM. Some candidates referred to the incorrect angle at
C
D
1 2
1 2
B 3
M
1 2
41
3 F2
point B. They considered the entire B̂ instead of Bˆ 1 ; Bˆ 2 or Bˆ 3 .
E
(c)
When answering Q10.1.3, some candidates attempted to prove that the
triangles were congruent instead of trying to prove them similar.
Some candidates were able to identify the equal pairs of angles but
A
could not provide the correct reasons for them being equal. Other
10.1 Prove, giving reasons, that:
10.1.1 FBDM is a cyclic quadrilateral (3)
candidates could not name the angles correctly. They would state that
ˆ = F
ˆ
10.1.2 B
3
1
(4)
ˆ = CBE
ˆ .
B̂ = Ê instead of Bˆ 1 = Eˆ 1 and that D̂ = B̂ instead of D
1
10.1.3 ΔCDB ||| ΔCBE
(3)
Copyright © The Answer Series
11
QUESTION
MEMO
QUESTION 10
10.1.1
Shade the quadrilateral
 In the diagram, a circle passes through D, B and E.
ˆ = 90º
M
2
 Diameter ED of the circle is produced to C and AC
&
is a tangent to the circle at B.
Bˆ 2 = 90º
. . . ø in semi-?
 M is a point on DE such that AM ⊥ DE.
ˆ = Bˆ
∴ M
2
2
 AM and chord BE intersect at F.
∴ FBDM is a cyclic quad. 10.1.2
ˆ
Bˆ 3 = D
2
...
= Fˆ 1 . . .
C
D
1
M
2
1
. . . CONVERSE of ext. ø of cyclic quad.
tan chord theorem
ext.ø of c.q. FBDM (10.1.1)
E
2
C
4 1
1
3 F2
1
2
B
10.1.3
1
3 F2
1
(1)
Ĉ is common
(2)
Bˆ 1 = Ê
A
Prove, giving reasons, that :
10.1.1
FBDM is a cyclic quadrilateral
(3)
10.1.2
ˆ = F
ˆ
B
3
1
(4)
10.1.3
ΔCDB ||| ΔCBE
(3)
...
2
B
tan chord theorem
∴ ΔCDB | | | ΔCBE 12
2
4 1
3
Copyright © The Answer Series
M
2
Mark the Δs clearly
In Δs CBD & CBE
10.1
D
. . . øøø
3
A
E
QUESTION
Misconceptions
10.2* If it is further given that CD = 2 units and DE = 6 units,
(d)
calculate the length of:
In Q10.2.1 many candidates could not establish the correct
proportion from Q10.1.3. They incorrectly assumed that
10.2.1 BC
(3)
10.2.2 DB
(4)
[17]
EC = DE and some candidates substituted BC = DE – CD
(e)
Many candidates did not attempt Q10.2.2. Of those who did,
some were able to establish the correct proportion but could
not proceed any further.
C
D
1 2
M
1 2
1 2
B 3
E
41
3 F2
A
Copyright © The Answer Series
13
QUESTION
MEMO
10.2*
10.2.1 ∴
If it is further given that CD = 2 units and DE = 6 units,
calculate the length of :
∴
10.2.1
BC
(3)
10.2.2
DB
(4)
CD BC ⎛ DB ⎞
=
⎜=
⎟
BC
CE ⎝ BE ⎠
. . . | | | Δs
2
BC
=
BC
2+6
∴ BC2 = 16
∴ BC = 4 units [17]
10.2.2
DB
CD
2
1
=
=
=
BE
BC
4
2
∴ BE = 2DB
C
D
1
M
2
1
2
4 1
3 F2
1
2
B
E
Let DB = x, then BE = 2x
In ΔDBE :
Bˆ 2 = 90º
∴ DB2 + BE2 = DE2
3
. . . ø in semi-?
. . . Theorem of Pythagoras
∴ x2 + (2x)2 = 62
∴ x2 + 4x2 = 36
A
∴ 5x2 = 36
∴ x2 =
36
5
∴ x  2,68 units Copyright © The Answer Series
14
QUESTION 10: Suggestions for Improvement
(a)
More time needs to be spent on the teaching of Euclidean Geometry in all grades.
More practice on Grade 11
and 12 Euclidean geometry will help learners to learn theorems and diagram analysis. They should carefully read
the given information without making any assumptions. This work covered in class must include different activities
and all levels of taxonomy.
(b)
Teachers should require learners to make use of the diagrams in the answer book to indicate angles and
sides that are equal and record information that has been calculated.
(c)
Learners need to be made aware that writing correct but irrelevant statements will not earn them any marks
in an examination.
(d)
Learners need to be exposed to questions in Euclidean Geometry that include the
theorems and the converses.
When proving that a quadrilateral is cyclic, no circle terminology may be used when referring to the quadrilateral.
Copyright © The Answer Series
15
GR 10 – 12 EXEMPLAR GEOMETRY
GRADE 10: QUESTIONS
GRADE 10: MEMOS
1. PQRS is a kite such that the diagonals intersect in O.
ˆ = 20º.
OS = 2 cm and OPS
Q
P
O
R
20º
2.2
the longer diagonal of a kite
bisects the shorter diagonal
OQ = 2 cm ...
1.2
ˆ = 90º POQ
the diagonals of a kite
...
intersect at right angles
1.3
ˆ = 20º
QPO
...
2 cm
. . . opp. sides of ||m AODE
â AE || BO
1.1
and OF || AB
. . . proven above
â OE || AB
â ABOE is a ||m
the longer diagonal of a
kite bisects the (opposite)
angles of a kite
ˆ = 40º â QPS
AE || OD
both pairs of opposite
sides are parallel
...
OR: In ||m AODE : AE = and || OD . . .
opp. sides
of ||m
But OD = BO . . . O proved midpt of BD
S
1.1 Write down the length of OQ.
(2)
ˆ
1.2 Write down the size of POQ.
(2)
ˆ
1.3 Write down the size of QPS.
(2) [6]
2.
â ABOE is a ||m A
B
A
The highlighted s
(and their sides)
refer to Question 2.3.
O
E
C
D
F
2.1
C
(4)
2.2 Prove that ABOE is a parallelogram.
(4)
Copyright © The Answer Series
m
diagonals of || BCDE
bisect each other
m
& F is the midpt of AD . . . diagonals of || AODE
2.1 Prove that OF || AB.
2.3 Prove that ABO ≡ EOD.
In DBA :
O is the midpt of BD . . .
D
(5) [13]
. . . 1 pr of opp. sides
= and ||
E
F
O
â AE = and || BO
Use highlighters to mark the various ||ms and s
2. In the diagram, BCDE and AODE are parallelograms.
B
of BD in 2.1
Hint :
bisect each other
â OF || AB . . .
the line joining the
midpoints of two sides
of a  is || to the 3rd side
1
2.3
In s ABO and EOD
1) AB = EO
. . . opposite sides of ||m ABOE
2) BO = OD
. . . proved in 2.1
3) AO = ED
. . . opposite sides of ||m AODE
â ABO ≡ EOD . . . SSS
2.2
GRADE 11: QUESTIONS
In the diagram, M is the centre of the circle.
A, B, C, K and T lie on the circle.
3.1
The angle between a chord and a tangent
at the point of contact is . . .
AT produced and CK produced meet in N.
1.1 Complete the statement so that it is valid :
The line drawn from the centre of the circle
perpendicular to the chord . . .
Also NA = NC and B̂ = 38º.
(1)
B
Q
M
C
D
DE = 20 cm and CE = 2 cm.
B
A
1 2
3
K
Calculate the length of the following with reasons :
1.2.2 PQ
In the diagram, EA is a tangent to circle ABCD
at A.
CE and AG intersect at D.
C
O
4
1
2
(2)(4) [7]
T
C 1
In the diagram, O is the
centre of the circle and
A, B and D are points on
the circle.
2
D
2.2.1
O
(b) Tˆ 2
(d) Kˆ 4
(c) Ĉ
B
(2)(2)
G
2
x
4 5
3 2 1
3
1
A
D
1
2
F
(2)(2)
1
2 1
y
E
Use Euclidean geometry methods to prove the
(5)
2.2.2
Show that NK = NT.
(2)
If Aˆ 1 = x and Eˆ 1 = y, prove the following with
reasons :
2.2.3
Copyright © The Answer Series
Calculate, with reasons, the size of the
following angles :
ˆ
(a) KMA
A
ˆ = 2ADB.
ˆ
theorem which states that AOB
2
3
N
2.1
(1)
AC is a tangent to circle CDFG at C.
E
The diameter DE is
perpendicular to the
chord PQ at C.
1.2.1 OC
3.2
38º
P
1.2 In the diagram, O is the
centre of the circle.
Complete the following statement so that it
is valid :
Prove that AMKN is a cyclic
quadrilateral.
2
(3) [18]
3.2.1
BCG || AE
(5)
3.2.2
AE is a tangent to circle FED
(5)
3.2.3
AB = AC
(4) [15]
GRADE 11: MEMOS
3.1
B
2.2
. . . equal to the angle subtended by the chord
in the alternate segment. 38º
1.1
1.2.1
OE = OD = 1 (20) = 10 cm
2
radii
K
= 1 diameter
A
12
3
2
â OC = 8 cm . . . CE = 2 cm
B
3.2
M
C
. . . bisects the chord 4
1
2
T
C 1
3 2
N
2
1.2.2 In OPC :
PC2 = OP2 - OC2
2.2.1
. . . Pythagoras
= 102 - 82
= 36
â PC = 6 cm
ˆ = 2(38º)
(a) KMA
...
= 76º P
C
(b) Tˆ 2 = 38º Q
O
Proof :
2.2.2
O
A
radii;
øs opp = sides
Similarly: Let Dˆ 2 = y
ˆ = 2y
then, O
2
ˆ
â AOB
= 2x + 2y
= 2(x + y)
ˆ = 2 ADB
Copyright © The Answer Series
. . . ext. ø of c.q. CKTA
C
In NKT : Kˆ 4 = Tˆ 2
ˆ = x
A
1
. . . given
â Cˆ 2 = x
. . . tan chord theorem
ˆ = x
â G
2
. . . tan chord theorem
ˆ
â Aˆ 1 = (alternate) G
2
. . . sides opp equal øs
3.2.2
ˆ
Fˆ1 = C
3
ˆ = 2(38º)
KMA
& N̂ = 180º - 2(38º)
s
. . . (alternate ø equal)
. . . ext. ø of cyclic quad. CGFD
= Eˆ 1 (= y)
B
2.2.3
y
. . . both = 38º in 2.2.1
â NK = NT 12
ˆ = 2x
â O
1
. . . ext. ø of DAO
2 1
â BCG || AE 12
A
D
2 1
D
Construction : Join DO and
produce it to C
...
3.2.1
ˆ
= 38º . . . øs opp = sides
(d) NAC
â Kˆ 4 = 38º then  = x
1
1
E
or, ext. ø of cyclic quad. CKTA
â PQ = 12 cm . . . line from centre  chord
Let Dˆ 1 = x
G
4
3 2 1
. . . ext. ø of cyclic quad. BKTA
s
(c) Ĉ = 38º . . . ø in the same segment
D
2.1
2
F
E
x
5
ø at centre =
2 % ø at circumference
3
. . . see 2.2.1(a)
. . . alternate øs ; BCG || AE
â AE is a tangent to ?FED . . . converse of tan chord theorem
. . . sum of øs in NKT
(see 2.2.2)
ˆ + N̂ = 180º
â KMA
3.2.3
â AMKN is a cyclic quadrilateral . . . opposite øs supplementary
3
ˆ = CAE
ˆ
C
1
= B̂
â AB = AC . . . alternate øs ; BCG || AE
. . . tan chord theorem
. . . sides opposite equal øs
2.
GRADE 12: QUESTIONS
In the diagram, M is the centre of the circle and diameter
AB is produced to C. ME is drawn perpendicular to AC
such that CDE is a tangent to the circle at D. ME and
chord AD intersect at F. MB = 2BC.
3.2
In the diagram, ADE is a triangle having BC | | ED
and AE | | GF. It is also given that AB : BE = 1 : 3,
AC = 3 units, EF = 6 units, FD = 3 units and
CG = x units.
1.1 Complete the following statement :
A
The angle between the tangent and the chord at
the point of contact is equal to . . .
3
(1)
B
1.2 In the diagram, A, B, C, D and E are points on the
circumference of the circle such that AE | | BC.
BE and CD produced meet in F. GBH is a
tangent to the circle at B. Bˆ = 68º and F̂ = 20º.
M
A
3
1
1
2
B
1
2
2
F 3
G
2
3
1
A
F
1
68º
1
B
4
E
2
x
C
1
G
x
E
20º
angles each equal to x.
1
2 D
C
H
(3)
2.2 Prove that CM is a tangent at M to the circle
passing through M, E and D.
(4)
2.3 Prove that FMBD is a cyclic quadrilateral.
(3)
2.4 Prove that DC2 = 5BC2.
(3)
2.5 Prove that DBC | | | DFM.
(4)
2.6 Hence, determine the value of
DM
.
FM
(2) [19]
Determine the size of each of the following:
1.2.1
Eˆ 1
(2)
1.2.2
Bˆ 3
(1)
1.2.3
ˆ
D
1
(2)
1.2.4
Eˆ 2
(1)
1.2.5
Ĉ
Copyright © The Answer Series
(2) [9]
3.1
A
In the diagram, points D
and E lie on sides AB
and AC respectively of
ABC such that DE | | BC.
Use Euclidean Geometry
methods to prove the
theorem which states
that
AD AE

.
DB EC
F
3
D
Calculate, giving reasons:
ˆ = x, write down, with reasons, TWO other
2.1 If D
4
3
6
D
E
2
3
4
C
B
D
E
C
(6)
4
3.2.1
the length of CD
(3)
3.2.2
the value of x
(4)
3.2.3
the length of BC
(5)
3.2.4
area ABC
the value of
area GFD
(5) [23]
GRADE 12: MEMOS
1.1
. . . the angle subtended by the chord in the alternate
segment.
1.2.1
Eˆ 1 = Bˆ 1
= 68º . . . tan chord theorem
1.2.2
Bˆ 3 = Eˆ 1
. . . alt. øs ; AE || BC
2.4
Let BC = a ; then MB = 2a
â MD = 2a
. . . radii
ˆ = 90º
In MDC : MDC
â DC2 =
=
=
=
=
ˆ = Bˆ
D
1
3
. . . ext. ø of 
Ĉ = 180º - Eˆ 2
FM
. . . opp. øs of cyclic quad.
= 92º 2.1
 = x
ˆ = x
D
2
. . . tan chord theorem
3.1
. . . øs opp. equal sides
=
5 BC
BC
=
5 Proof :
M
A
3
2
1 2
1
F3
2
2
1
3
4
B
1
x
D
& area of ADE =
area of EDC
. . . ext. ø of 
ˆ = 90º . . . radius MD  tangent CDE
& MDE
. . . sum of øs in MED
â CM is a tangent at M to ?MED 2.3
ˆ
ADB
= 90º
ˆ
& M3 = 90º
. . . converse tan chord
theorem
. . . ø in semi-?
3.2.1
A
4p
area of ABC
=
area of GFD
h
E
=
B
C
. . . equal heights
h
= AE
EC
3p
CD
=
3
p
% 3) â CD = 9 units . . . equal heights
1
2
ˆ
DG . DF sin D
1
. 3 . 94 . sin Dˆ
2
1
. 4 . 3 . sin Dˆ
2
. . . corr. øs ; BC || ED
9
4
4
9
=
16
1
2
1
2
. p . 3 . sin Aˆ
. 4 p . 12 . sin Aˆ
16
& area of GFD =
area of AED
1
. 4 . 3 . sin Dˆ
2
1
. 12 . 9 . sin Dˆ
2
3
3
â area of GFD = 1 area of AED
A
p
B
area of GFD
C
x
6
F
9
–  — : â area of ABC =
3
3
1
16
1
9
16
16
... –
= 1
9
... —
area of AED
area of AED
= 9 D
= 1
4
â area of ABC = 1 area of AED
i.e. same height
G
5
ˆ
AC . BC sin ACB
area of AED
same base DE &
3p
E
1
2
OR : area of ABC =
. . . proportion thm; BC || ED
ˆ = ADB
ˆ
â M
3
Copyright © The Answer Series
h′
D
h
. . . |||  s
AE
=
. . . ME  AC
â FMBD is a cyclic quad . . . converse ext. ø of cyclic quad
. . . equiangular  s
4
Let AB = p ; then BE = 3p
In AED :
â ABC ||| AED
% 9) â BC = 9 units â area of ADE = area of ADE
area of DBE
area of EDC
â AD = AE DB
EC
. . . ME  AC
. . . corr. øs ; BC || ED
9
But, area of DBE = area of EDC . . . betw. same || lines,
= 2x
ˆ = 90º - 2x
â M
2
1
AE .
2
1
EC .
2
ˆ = Ê
(2) ABC
p
â BC =
3.2.4
= AD
DB
E
ˆ = Aˆ + D
ˆ
M
1
2
area of ADE
=
area of DBE
C
= 3x + 9
= - 45
= 5 In s ABC and AED
ED
. . . see 2.4
1
AD . h
2
1
DB . h
2
. . . prop. thm. ; AE || GF
6
â BC = AB
s
. . . |||  s
BC
= 3
(1) Â is common
. . . equiangular 
Construction :
Join DC and EB
and heights h and h′
2.2
â Ê = 2x
ˆ = Ê
â M
1
3.2.3
. . . both = x
â DM = DC
2.6
= 88º 1.2.5
. . . theorem of Pythagoras
â DBC ||| DFM ˆ + 20º
Eˆ 2 = D
1
9-x
x+3
â 54 - 6x
â - 9x
â x
2
1
ˆ = D
ˆ
(2) D
4
2
. . . ext. ø of cyclic quad.
= 68º 1.2.4
MC2 - MD2
(3a)2 - (2a)2
9a2 - 4a2
5a2
5BC2 CG = x ; so GD = 9 - x
In DAE :
. . . radius  tangent
In s DBC and DFM
(1) Bˆ = Fˆ
. . . ext ø of c.q. FMBD
2.5
= 68º 1.2.3
3.2.2
PROBLEM-SOLVING: An Active Approach
A
Be Active . . .
C
Use all the Clues/Triggers
T
Recall the Theory systematically
ACT!
6
CAPS Curriculum page 14
THE CAPS CURRICULUM: OVERVIEW OF TOPICS
Paper 2
7. EUCLIDEAN GEOMETRY AND MEASUREMENT
Grade 10
(a) Revise basic results established in earlier
grades.
(b) Investigate line segments joining the midpoints of two sides of a triangle.
(c) Properties of special quadrilaterals.
Grade 11
(a) Investigate and prove theorems of the
geometry of circles assuming results from
earlier grades, together with one other result
concerning tangents and radii of circles.
(b) Solve circle geometry problems, providing
reasons for statements when required.
(c) Prove riders.
Grade 12
(a) Revise earlier (Grade 9) work on the
necessary and sufficient conditions for
polygons to be similar.
(b) Prove (accepting results established in earlier
grades):
that a line drawn parallel to one side of a
triangle divides the other two sides
proportionally (and the Mid-point Theorem
as a special case of this theorem);
that equiangular triangles are similar;
that triangles with sides in proportion are
similar;
the Pythagorean Theorem by similar
triangles; and
riders
THE CAPS CURRICULUM TERM BY TERM CONTENT
GRADE 10 - GRADE 12
CAPS
GRADE 10: TERM 1
Weeks
Topic
Curriculum statement
1. Revise basic results established in earlier
grades regarding lines, angles and
triangles, especially the similarity and
congruence of triangles.
Clarification
Comments:
•
ˆ = Dˆ ,
equal: Triangles ABC and DEF are similar if A
2. Investigate line segments joining the midpoints of two sides of a triangle.
3
Euclidean
Geometry
3. Define the following special quadrilaterals:
the kite, parallelogram, rectangle, rhombus,
square and trapezium. Investigate and
make conjectures about the properties of
the sides, angles, diagonals and areas
of these quadrilaterals. Prove these
conjectures.
Triangles are similar if their corresponding angles are equal, or if the ratios of their sides are
similar if
Bˆ = Eˆ and Cˆ = Fˆ . They are also
.
•
We could define a parallelogram as a quadrilateral with two pairs of opposite sides parallel.
Then we investigate and prove that the opposite sides of the parallelogram are equal, opposite
angles of a parallelogram are equal, and diagonals of a parallelogram bisect each other.
•
It must be explained that a single counter example can disprove a Conjecture, but numerous
specific examples supporting a conjecture do not constitute a general proof.
Example:
In quadrilateral KITE, KI = KE and IT = ET. The diagonals intersect at M. Prove that:
1. IM = ME and
(R)
2. KT is perpendicular to IE.
(P)
As it is not obvious, first prove that
.
MATHEMATICS GRADES 10-12
25
MATHEMATICS GRADES 10-12
GRADE 10: TERM 4
28
Weeks
Topic
Curriculum statement
Solve problems and prove riders using the
properties of parallel lines, triangles and
quadrilaterals.
Clarification
Comment:
Use congruency and properties of quads, esp. parallelograms.
CURRICULUM AND ASSESSMENT POLICY STATEMENT (CAPS)
Example:
EFGH is a parallelogram. Prove that MFNH is a parallelogram.
E
2
F
Euclidean
Geometry
M
G
N
H
M
F
G
(C)
N
H
E
34
No. of
weeks
Topic
Curriculum Statement
CURRICULUM AND ASSESSMENT POLICY STATEMENT (CAPS)
Accept results established in earlier grades
Comments:
as axioms and also that a tangent to a circle
Proofs of theorems can be asked in examinations, but their converses
is perpendicular to the radius, drawn to the
(wherever they hold) cannot be asked.
point of contact.
Example:
Then investigate and prove the theorems of
1. AB and CD are two chords of a circle with centre O. M is on AB and N is on CD such that
OM ⊥ AB and ON ⊥ CD. Also, AB = 50mm, OM = 40mm and ON = 20mm. Determine the
radius of the circle and the length of CD.
(C)
the geometry of circles:
3
Euclidean
Geometry
Clarification
•
The line drawn from the centre of a circle
perpendicular to a chord bisects the chord;
•
The perpendicular bisector of a chord
passes through the centre of the circle;
•
The angle subtended by an arc at the
centre of a circle is double the size of the
angle subtended by the same arc at the
circle (on the same side of the chord as the
centre);
•
Angles subtended by a chord of the circle,
on the same side of the chord, are equal;
•
The opposite angles of a cyclic quadrilateral
are supplementary;
•
Two tangents drawn to a circle from the
same point outside the circle are equal in
length;
•
The angle between the tangent to a circle
and the chord drawn from the point of
contact is equal to the angle in the alternate
segment.
Use the above theorems and their
converses, where they exist, to solve riders.
2. O is the centre of the circle below and Oˆ1 = 2x .
2.1. Determine Ô2 and M̂ in terms of x .
(R)
2.2. Determine K̂1 and K̂ 2 in terms of x .
(R)
2.3. Determine Kˆ1 + Mˆ . What do you notice?
(R)
2.4. Write down your observation regarding the measures of K̂ 2 and M̂ .
(R)
MATHEMATICS GRADES 10-12
GRADE 11: TERM 1
48
No. of
Weeks
Topic
Curriculum statement
CURRICULUM AND ASSESSMENT POLICY STATEMENT (CAPS)
1. Revise earlier work on the necessary and
sufficient conditions for polygons to be
similar.
Example:
2. Prove (accepting results established in
earlier grades):
Let D be on
•
3
Euclidean
Geometry
Clarification
that a line drawn parallel to one side
of a triangle divides the other two
sides proportionally (and the Mid-point
Theorem as a special case of this
theorem) ;
• that equiangular triangles are similar;
• that triangles with sides in proportion are
similar; and
• the Pythagorean Theorem by similar
triangles.
Consider a right triangle ABC with
such that
. Let
and
. Determine the length of
.
in terms of a and c .
(P)
MATHEMATICS GRADES 10-12
GRADE 12 TERM 1
Euclidean Geometry : Theorem Statements & Acceptable Reasons
LINES
The adjacent angles on a straight line are supplementary.
øs on a str line
If the adjacent angles are supplementary, the outer arms
of these angles form a straight line.
adj øs supp
The adjacent angles in a revolution add up to 360º.
øs round a pt OR øs in a rev
Vertically opposite angles are equal.
vert opp øs
If AB || CD, then the alternate angles are equal.
alt øs ; AB || CD
If AB || CD, then the corresponding angles are equal.
corresp øs ; AB || CD
s
If AB || CD, then the co-interior angles are supplementary. co-int ø ; AB || CD
If the alternate angles between two lines are equal, then
the lines are parallel.
alt øs =
If the corresponding angles between two lines are equal,
then the lines are parallel.
corresp øs =
If the co-interior angles between two lines are
supplementary, then the lines are parallel.
co-int øs supp
TRIANGLES
The interior angles of a triangle are supplementary.
ø sum in Δ OR sum of øs OR
int øs in Δ
The exterior angle of a triangle is equal to the sum of the
interior opposite angles.
ext øs of Δ
The angles opposite the equal sides in an isosceles
triangle are equal.
øs opp equal sides
The sides opposite the equal angles in an isosceles
triangle are equal.
sides opp equal øs
In a right-angled triangle, the square of the hypotenuse is
equal to the sum of the squares of the other two sides.
Pythagoras OR
Theorem of Pythagoras
If the square of the longest side in a triangle is equal to
the sum of the squares of the other two sides then the
triangle is right-angled.
Converse Pythagoras
OR Converse Theorem of
Pythagoras
Copyright © The Answer Series
If three sides of one triangle are respectively equal to
three sides of another triangle, the triangles are
congruent.
SSS
If two sides and an included angle of one triangle are
respectively equal to two sides and an included angle of
another triangle, the triangles are congruent.
SAS OR SøS
If two angles and one side of one triangle are
respectively equal to two angles and the corresponding
side in another triangle, the triangles are congruent.
AAS OR øøS
If in two right angled triangles, the hypotenuse and one side
of one triangle are respectively equal to the hypotenuse
and one side of the other, the triangles are congruent.
RHS OR 90ºHS
The line segment joining the midpoints of two sides of a
triangle is parallel to the third side and equal to half the
length of the third side.
Midpt Theorem
The line drawn from the midpoint of one side of a
triangle, parallel to another side, bisects the third side.
line through midpt || to
2nd side
A line drawn parallel to one side of a triangle divides the
other two sides proportionally.
line || one side of Δ OR
prop theorem; name || lines
If a line divides two sides of a triangle in the same
proportion, then the line is parallel to the third side.
line divides two sides of Δ in
prop
If two triangles are equiangular, then the corresponding
sides are in proportion (and consequently the triangles
are similar).
||| Δs OR equiangular Δs
If the corresponding sides of two triangles are
proportional, then the triangles are equiangular (and
consequently the triangles are similar).
Sides of Δ in prop
If triangles (or parallelograms) are on the same base (or on
bases of equal length) and between the same parallel lines,
then the triangles (or parallelograms) have equal areas.
same base; same height OR
equal bases; equal height
CIRCLES
QUADRILATERALS
GROUP I
s
The interior angles of a quadrilateral add up to 360º.
sum of ø in quad
The opposite sides of a parallelogram are parallel.
opp sides of ||m
If the opposite sides of a quadrilateral are parallel, then
the quadrilateral is a parallelogram.
opp sides of quad are || OR
converse opp sides of ||m
The opposite sides of a parallelogram are equal in length.
opp sides of ||m
If the opposite sides of a quadrilateral are equal, then the
quadrilateral is a parallelogram.
opp sides of quad are = OR
converse opp sides of a parm
The opposite angles of a parallelogram are equal.
opp øs of ||m
If the opposite angles of a quadrilateral are equal then
the quadrilateral is a parallelogram.
opp øs of quad are = OR
converse opp angles of a
parm
The diagonals of a parallelogram bisect each other.
diag of ||m
If the diagonals of a quadrilateral bisect each other, then
the quadrilateral is a parallelogram.
diags of quad bisect each
other OR
converse diags of a parm
If one pair of opposite sides of a quadrilateral are equal
and parallel, then the quadrilateral is a parallelogram.
pair of opp sides = and ||
The diagonals of a parallelogram bisect its area.
diag bisect area of ||m
The diagonals of a rhombus bisect at right angles.
diags of rhombus
The diagonals of a rhombus bisect the interior angles.
diags of rhombus
All four sides of a rhombus are equal in length.
sides of rhombus
All four sides of a square are equal in length.
sides of square
The diagonals of a rectangle are equal in length.
diags of rect
The diagonals of a kite intersect at right-angles.
diags of kite
A diagonal of a kite bisects the other diagonal.
diag of kite
A diagonal of a kite bisects the opposite angles.
diag of kite
Copyright © The Answer Series
O
O
O
O
O
x
O
2x
O
O
xi
The tangent to a circle is perpendicular
to the radius/diameter of the circle at
the point of contact.
tan ⊥ radius
tan ⊥ diameter
If a line is drawn perpendicular to a
radius/diameter at the point where the
radius/diameter meets the circle, then
the line is a tangent to the circle.
converse tan ⊥ radius OR
converse tan ⊥ diameter
The line drawn from the centre of a
circle to the midpoint of a chord is
perpendicular to the chord.
The line drawn from the centre of a
circle perpendicular to a chord bisects
the chord.
The perpendicular bisector of a
chord passes through the centre of
the circle.
The angle subtended by an arc at the
centre of a circle is double the size of
the angle subtended by the same arc
at the circle (on the same side of the
chord as the centre)
The angle subtended by the diameter
at the circumference of the circle
is 90º.
If the angle subtended by a chord at
the circumference of the circle is 90°,
then the chord is a diameter.
line ⊥ radius OR
line from centre to midpt of chord
line from centre ⊥ to chord
perp bisector of chord
øat centre
= 2 % ø at circumference
øs in semi circle OR
diameter subtends right angle
OR ø in ½ ?
chord subtends 90º OR
converse øs in semi circle
GROUP II
Angles subtended by a chord of the
circle, on the same side of the
chord, are equal
x
x
y
y
x
If a line segment joining two points
subtends equal angles at two
points on the same side of the line
segment, then the four points are
concyclic.
x
(This can be used to prove that the
four points are concyclic).
x
GROUP III
x
s
y
The opposite angles of a cyclic
quadrilateral are supplementary
(i.e. x and y are supplementary)
180º – x
If the opposite angles of a
quadrilateral are supplementary
then the quadrilateral is cyclic.
ø in the same seg
x
s
line subtends equal ø
OR
Equal chords subtend equal angles
at the circumference of the circle.
The exterior angle of a cyclic
quadrilateral is equal to the
interior opposite angle.
x
equal chords; equal øs
If the exterior angle of a quadrilateral
is equal to the interior opposite
angle of the quadrilateral, then the
quadrilateral is cyclic.
x
x
Equal chords subtend equal angles
at the centre of the circle.
x
xO
equal chords; equal øs
A
x
x
B
x
Equal chords in equal circles subtend
equal angles at the circumference of
the circles.
Equal chords in equal circles
subtend equal angles at the centre
of the circles.
(A and B indicate the centres of
the circles)
Two tangents drawn to a circle
from the same point outside the
circle are equal in length (AB = AC)
A
equal circles; equal chords;
equal øs
C
equal circles; equal chords;
x
equal øs
x
Copyright © The Answer Series
converse opp øs of cyclic quad
ext ø of cyclic quad
ext ø = int opp ø
OR
converse ext ø of cyclic quad
GROUP IV
B
x
opp øs quad sup OR
converse øs in the same seg
x
x
opp øs of cyclic quad
xii
Tans from common pt
OR
Tans from same pt
tan chord theorem
y
The angle between the tangent to
a circle and the chord drawn from
the point of contact is equal to the
angle in the alternate segment.
converse tan chord theorem
y
If a line is drawn through the endpoint of a chord, making with the
chord an angle equal to an angle
in the alternate segment, then the
line is a tangent to the circle.
y
x
a
b
(If x = b or if y = a then the
line is a tangent to the circle)
OR
ø between line and chord
IMPORTANT ADVICE FOR MASTERING MATHS
™
Don't focus on what you haven't done in the past.
Put that behind you and start today! Give it your all – it is well worth it!
–
Don't worry about marks! Just focus on the work and the marks will take care
of themselves.
Worrying is tiring and time-wasting and gets in the way of your progress!
Your marks will gradually improve if you work consistently.
TIMETABLE / PLANNING
■
Draw up a timetable of study times.
■
Revise your schedule from time to time to ensure optimum focus and
š
awareness of time.
Motivation will not be a problem once you've done this,
YOUR APPROACH
The most important thing of all is to remain positive. Some times will be
tough, some exams WILL BE TOUGH, but in the end, your results will reflect
all the effort that you have put in.
because you will see that you need to use every minute!
—
WORK FOCUS
ROUTINE
■ Despair can destroy your Mathematics
Routine is really important. Start early in the morning, at the same time every
Mathematics should be taken on as a continual challenge (or not at all!).
Teach your ego to suffer the 'knocks' which it may receive – like a poor
test result. Instead of being negative about your mistakes (e.g. 'I'll never
be able to do these sums'), learn from them. As you address each one,
they will help you to understand the work and do better next time!
day, and don't work beyond 11 at night.
Arrange some 1 hour and some 2 hour sessions on
particular subjects.
Schedule more difficult pieces of work for early in the day
and easier bits for later when you're tired.
■ Work with a friend occasionally.
Discussing Mathematics makes it alive and enjoyable.
Reward yourself with an early night now and again!
Allow some time for physical exercise – at least ½ hour a
day. Any sport or walking, jogging (or skipping when it rains)
will improve your concentration.
˜
›
A GREAT GENERAL STUDY TIP
■
Don't just read through work!
■
Study a section and then, on your own, write down all you can remember.
Knowing that you're going to do this makes you study in a logical, alert
way.
■ at the television
■
You're then only left to learn the few things which you left out.
■ on Facebook and any other social networks
■
This applies to all subjects. In this text, read the explanations very
carefully and actively, trying all worked examples yourself first as you
master each topic.
■
A subject like Maths also requires you to practise and apply the concepts
regularly.
'NO-NO'S'
Limit the time you spend
■ on your phone
■ in the sun
All these activities break down your commitment,
focus and energy.
Copyright © The Answer Series
8
œ
ž
ABOUT THE MATHS
■ Try each problem on your own first – no matter how inadequately –
before consulting the answer. It is only by encountering the difficulties
which you personally have that you will be able, firstly, to pin-point them,
and then, secondly, to understand and rectify them and then make sure
you don't make them again!!
THE EXAMS
Finally, for the exams themselves, make sure you have all you need (e.g. your
calculator, ruler, etc.) and don't allow yourself to be upset by panicking friends.
Plan your time in the exam well – allowing some time to check at the end.
Whatever you do, don't allow yourself to get stuck on any difficult issues in the
exam. Move on, and rather come back to problem questions if you have time
left. If you're finding an exam difficult, just continue to do your absolute best
right until the end! Partial answers can earn marks.
■ Learn to keep asking yourself 'why'?
It is when you learn to REASON that you really start enjoying Maths and,
quite coincidentally, start doing well at it!!
Answers are by no means the most important thing in Mathematics.
When you've done a problem, don't be satisfied only to check the answer.
We wish you the best of luck in your studies and hope that this book
will be the key to your success – enjoy it!!
Check also on your layout and reasoning (logic). Systematic, to-the-point,
logical and neat presentation is very important.
The Answer Series Maths Team
■ Revise earlier work as often as possible. Set up a revision plan with at
least one session a week for this purpose. Familiarity is the key to success
in Maths!

EXAM PREPARATION
The best way to prepare for your exam is to start early – in fact, on the
first day of the year!!! Working past papers is excellent preparation for any
exam – and The Answer Series provides these – but, only when you're ready.
Focus on WORKING ON ONE TOPIC AT A TIME first. It is the most effective
way to improve, particularly as you will build up your confidence this way.
The Answer Series provides thorough topic treatment for all subjects,
enabling you to cover all aspects of each topic, from the basics to the top
level questions. Thereafter, working past papers is a worthwhile and
rewarding exercise.
PLEASE NOTE:
These Geometry materials (Booklets 1 to 4) were created
and produced by The Answer Series Educational Publishers
(Pty) (Ltd) to support the teaching and learning of Geometry
in high schools in South Africa.
They are freely available to anyone who wishes to use them.
This material may not be sold (via any channel) or used for
profit-making of any kind.
Copyright © The Answer Series
9
GRADES 8 - 12
ALL MAJOR SUBJECTS IN
ENGLISH & AFRIKAANS
WWW.THEANSWER.CO. ZA
Philosopher, Immanuel Kant (18th century philosopher)
Theory without practice
is empty
Practice without theory
is blind
1
CONTENT FRAMEWORK
 LINES
(Gr 8
 TRIANGLES
 QUADRILATERALS
 CIRCLES (Gr 11)
2
Gr 12?
10)
Gr 12:
Theorem of Pythagoras (Gr 8)
Similar Δs (Gr 9)
Midpoint Theorem (Gr 10)
&
The Proportion Theorem
Ratio
Proportion
3
Area
LINES
2 Situations
2
1
1
2
4
3
5 6
8 7
NB:
Converse Statements
Logic !
1 2
4 3
Vocabulary first,
then facts
4
TRIANGLES
Sum of Interior øs
Isosceles
Δ
Exterior ø of Δ
Equilateral Δ
NB:
Vocabulary first,
then facts
Right-ød Δ
–
Theorem
of Pythagoras
Area of a Δ and related facts
Similar
Δs
Congruent Δs
Midpoint
Theorem
5
These involve
converse theorems
QUADRILATERALS - definitions, areas & properties
A Rectangle
All you need to know
The Square
'Any' Quadrilateral
f
e
a
A Parallelogram
A Trapezium
b
DEFINITION :
b
d
A ||m with one right ø
2
c
h
Sum of the ø of
any quadrilateral = 360º
Sum of the interior angles
= (a + b + c) + (d + e + f)
= 2 % 180º
. . . ( 2 Δs )
= 360º
DEFINITION :
Quadrilateral with 2 pairs opposite sides ||
1
s
a
DEFINITION :
Quadrilateral with 1 pair of opposite sides ||
B
= 1 ah + 1 bh
2
2
Q
Area = s 2
D
where ΔQCD ≡ ΔPBA . . . RHS / 90º HS
â ||m ABCD = rect. PBCQ (in area)
= BC % QC
Properties :
2 pairs opposite sides equal
2 pairs opposite angles equal
& DIAGONALS BISECT ONE ANOTHER
A Kite
a
2
a
2
Properties :
It's all been said 'before' !
Since a square is a rectangle,
a rhombus, a parallelogram,
a kite, . . . ALL the properties
of these quadrilaterals apply.
C
||m ABCD = ABCQ + ΔQCD
rect. PBCQ = ABCQ + ΔPBA
'Half the sum of the || sides
% the distance between them.'
Copyright © The Answer
A
DIAGONALS are EQUAL
Area = Δ 1 + Δ 2
2
See how the properties
accumulate as we
move from left to right,
i.e. the first quad. has
no special properties
and each successive
quadrilateral has all
preceding properties.
P
the 'ultimate' quadrilateral !
A Rhombus
= 1 (a + b) . h
The arrows indicate
various ‘pathways’
from ‘any’
quadrilateral to the
square (the ‘ultimate
quadrilateral’). These
pathways, which combine
logic and fact, are
essential to use when
proving specific types
of quadrilaterals.
Area = base % height
Area = ℓ % b
DEFINITION :
A ||m with one pair of
adjacent sides equal
sides
Area
= 1 product of diagonals (as for a kite)
2
diagonals
or
= base % height (as for a parallelogram)
DEFINITION :
Quadrilateral with 2 pairs of
adjacent sides equal
b
Given diagonals a and b
1 a
Area = 2Δs = 2 ⎛⎜ b . ⎞⎟ = ab
⎝2
2⎠
angles
THE DIAGONALS
2
'Half the product of the diagonals'
• bisect one another PERPENDICULARLY
• bisect the angles of the rhombus
• bisect the area of the rhombus
Note :
y
x
y
x
Quadrilaterals play a
prominent role in both
Euclidean & Analytical
Geometry right through
to Grade 12 !
2
THE DIAGONALS
• cut perpendicularly
• ONE DIAGONAL bisects the other diagonal,
the opposite angles and the area of the kite
x
x
y y
s
ø of Δ or
2x + 2y = 180º . . .
co-int. øs ; || lines
² x + y = 90º
SUMMARY OF CIRCLE GEOMETRY THEOREMS
I
II
III
x
The
'Centre'
group
The
' No Centre '
group
The
' Cyclic Quad.'
group
2x
2x
x
x
Equal
chords !
x
There are ' 3 ways to
prove that a quad. is
a cyclic quad '.
y
x + y = 180º
IV
The
'Tangent'
group
Copyright © The Answer Series
Equal
tangents !
There are ' 2 ways to prove that
a line is a tangent to a ?'.
Equal
radii !
TAS FET EUCLIDEAN GEOMETRY COURSE
BOOKLET 2
Lines, Angles, Δ & Quadrilaterals
s
(Grade 8 to 10 Revision)
by The TAS Maths Team
Lines, Angles, Δ s &
Quadrilaterals
Geometry FET
Course Booklets Set
WWW.THEANSWER.CO. ZA
CONTENTS of Booklet 2
Lines & Angles
 Vocabulary and Facts
Triangles
 Vocabulary, Facts and Proofs
 Area
Quadrilaterals
 Properties
 Definitions
 Theorems
 Area
Grade 10: Midpoint Theorem
 Statement and Converse
 Riders
EXERCISES & FULL SOLUTIONS on all 4 sections
LINES, ANGLES & TRIANGLES
Know the meanings
of all the WORDS.
LINES: 2 SITUATIONS
The Language (Vocabulary)

PARALLEL LINES
PERPENDICULAR LINES
AB || CD
A
B
C
A
A
C
C
SINGLE ANGLES
B
arms
vertex
1ˆ and 4ˆ ;
D
4ˆ and 3̂ , etc.
90º
D
A
D
( adjacent øs
AB ⊥ CD
B
a right angle
90º
acute
angles
a straight angle
180º

The ANGLE is the
amount of rotation
about the vertex.
obtuse
angles
3
1ˆ and 3ˆ
2
or
2ˆ and 4ˆ
WHEN 2 LINES ARE CUT BY A TRANSVERSAL . . .
the
transversal
1 2 3 4
5 6 7 8
• corresponding øs : 1 & 5 ; 2 & 6 ; 3 & 7 ; 4 & 8
reflex
angles
• alternate øs : 3 & 5 and 4 & 6
1 2
4 3
5 6
8 7
• co-interior øs : 4 & 5 and 3 & 6
• Complementary øs add up to 90º
e.g. 40º and 50º; x and 90º – x
• Supplementary øs add up to 180º
e.g. 135º and 45º; x and 180º – x
40º
135º
50º
45º
x
• Adjacent ø have a common vertex
and a common arm and lie on
opposite sides of the common arm.
40º
50º
( Corresponding means their positions correspond.
90º – x
x
( Alternate means:
180º – x
common
vertex
A pair of adjacent
supplementary øs :
1
øs lie on opposite sides of the transversal;
whereas, 'co-' means: 'on the same side of the transversal'.
s
Copyright © The Answer Series
1
PAIRS OF ANGLES , one from each family:
PAIRS OF ANGLES
e.g. A pair of adjacent
complementary øs :
4
2 'families' of 4 angles are formed
a revolution
360º
270º
( vertically opposite øs,
&
B
C
D
NOTE: The plural of vertex is vertices!
0º
WHEN 2 LINES INTERSECT, WE HAVE . . .
common arm
e.g. corresponding
angles
140º
40º
1
alternate
angles
co-interior
angles
2
4
3
5
6
8
7
!
TRIANGLES
The Facts
Classification according to . . .
SIDES
LINES
ANGLES

d
Scalene Δ
Acute ø Δ
(all 3 øs are acute)
(all 3 sides different in length)
Fact 1
d
Isosceles Δ
Right ø Δ
Equilateral Δ
Obtuse ø Δ
(2 sides equal in length)
When two lines intersect, any pair of adjacent angles is
supplementary.
1
2
4
3
e.g. 4̂ + 1̂ = 180º & 4̂ + 3̂ = 180º
(one ø = 90º)
d
(all 3 sides equal in length)
(one ø is obtuse)
Fact 2
the vertical angle
the base angles
In an isosceles triangle:
Vertically opposite angles are equal.
s
We can classify Δ according to sides and ø simultaneously:
Special case:
Perpendicular lines
Examples

This is a/an . . . . -angled Δ

This is a/an . . . . -angled Δ

This is a/an . . . . -angled Δ
 An isosceles right-ød Δ  An isosceles acute-ød Δ  A scalene obtuse-ød Δ
Interior and Exterior angles
&
b
y
x
z
x
a
c
e
or
d
PARALLEL LINES
. . . whether the lines are parallel or not !
Exterior angles:
f
x is not an exterior ø
. . . the side is not 'produced'.
The sizes of the
other angles?
When ANY 2 lines are cut by a transversal, 2 'families' of four øs are formed,
and there are:
corresponding øs ;
1 2
alternate øs ;
4 3
& co-interior øs
Answers
Interior angles:
2 intersecting lines
form 4 angles
e.g. 1̂ = 3̂ and 2̂ = 4̂
BASE
s

INTERSECTING LINES
An exterior angle
is formed between
a side of the triangle
and the produced
(extension) of
an adjacent side of
the triangle.
5 6
8 7
Fact 3
If 2 PARALLEL lines are cut by a transversal,
corresponding angles are equal ;
alternate angles are equal ; and
co-interior angles are supplementary.
2
1 2
4 3
5 6
8 7
Copyright © The Answer Series
The Facts, continued
CONVERSE STATEMENTS
C
FACT 1 says:
1
2
If ABC is a straight line, then 1̂ + 2̂ = 180º.
TRIANGLES
B
A
FACT 1: The Sum of the interior angles of a triangle . . .
The CONVERSE STATEMENT of FACT 1 says
A
The sum of the (interior) angles
of a triangle is 180º.
If 1̂ + 2̂ = 180º, then: ABC is a straight line.
B
R
Example
If, in the sketch alongside, x = 40º and y = 130º,
x
is PQR a straight line?
Answer:
 + B̂ + Ĉ = 180º
y
FACT 2: The Exterior angle of a triangle . . .
Q
P
x + y = 40º + 130º = 170º ≠ 180º
â No, PQR is not a straight line
2
The exterior angle of a triangle equals
the sum of the interior opposite angles.
Whether it looks
...
like it or not!
1̂ = 2̂ + 3̂
1
3
FACT 3: An Isosceles triangle . . .
A
Now, refer to FACT 3 on the previous page . . .
If a pair of corresponding angles is equal
B
If a pair of alternate angles is equal
A
OR
C
1
2
then:
C
If
If 2 angles of a triangle are equal,
then: the sides opposite them are equal.
D
If a pair of co-interior angles is supplementary
line AB is PARALLEL to line CD
. . . whether it looks like it or not!
determine the size of the base øs.
3
Answer
A
If in isosceles ΔABC,
ˆ is a right ø,
the vertical angle (A)
B
x
1̂ = 2̂ ,
then: AB = AC
Example
then:
1̂ = 2̂
So, conversely:
The CONVERSE states:
B
OR
If AB = AC,
In an isosceles triangle,
the base angles are equal.
The CONVERSE STATEMENT of FACT 3 says
Copyright © The Answer Series
C
x C
2 x = 90º
∴ x = 45º
The angles of an equilateral triangle
all equal 60º.
60º
60º
The symbol: |||
FACT 7: Similar triangles
FACT 4: An Equilateral triangle . . .
Definition of Similarity
3 x = 180º
60º
Two figures are SIMILAR if:
∴ x = 60º
2 conditions
 they are equiangular, and
 their corresponding sides are in proportion.
FACT 5: THE THEOREM OF PYTHAGORAS
e.g.
...
THE THEOREM OF PYTHAGORAS states:
triangles
quadrilaterals
pentagons
The square on the hypotenuse of a right-angled triangle
equals the sum of the squares on the other two sides (in area).
In the case of TRIANGLES (only):
If triangles are equiangular,
If the square on one side
of a triangle equals the
sum of the squares on the
other two sides (in area),
then the triangle is
A
c
b
a
C
right-angled.
2
their corresponding sides are in proportion
and therefore: they are similar.
i.e. If
Conversely:
2
B
then:
If Ĉ = 90º,
then:
2
c = a2 + b2
The CONVERSE states:
If c = a + b ,
then:
AB
= BC = AC ,
DE
EF
DF
and therefore: ΔABC
Ĉ = 90º
A
ˆ
 = D̂ , B̂ = Ê and Ĉ = F,
then:
2
D
||| ΔDEF.
E
B
F
C
The CONVERSE states:
Note:
• Only one angle can be 90º (a right angle)
If the sides of two triangles are in proportion,
• The side opposite the right-angle is called the hypotenuse.
then: these triangles will also be equiangular
and therefore: the triangles are similar.
FACT 6: The Area of a triangle
The area of a triangle =
i.e. If
AB
= BC = AC ,
DE
EF
DF
then:
base % height
2
h
b
 = D̂ , B̂ = Ê and Ĉ = F̂ ,
and therefore:
4
ΔABC
||| ΔDEF
Copyright © The Answer Series
The possibilities are:
The symbol:
FACT 8: Congruent triangles

Two triangles are congruent if they have
• 3 sides the same length
. . . SSS
• 2 sides & an included angle equal
. . . SøS
• a right angle, the hypotenuse & a side equal
• 2 angles and any side equal
SSø – the case where the angle is NOT INCLUDED
between the sides.
. . . RHS or SS90º
. . . øøS
A
This is the ambiguous case because
there are 2 possible ∆s
which we could draw
D
B
C
If we can prove that ΔABC
F
SSS – only one size (& shape) of triangle could
be constructed.
E
obtuse ød ∆
acute ød ∆

RHS – the angle is not included,
but it is a right angle,
so only 1 option is possible:

SøS – the angle is included.
Only 1 option is possible.

øøS – given 2 angles, we actually have all 3 angles since the sum of
the angles must be 180º.
ΔDEF, then we can conclude that
all the sides and angles are equal.
Congruent Δs have the same SHAPE, and the same SIZE.
or
Similar Δs have the same SHAPE, but not necessarily the same SIZE.
The side restricts the size of the triangle.
Conditions of Congruency

So, only 1 option is possible.
Congruency can be understood best by constructing triangles,
even casually imagining the construction!
However, when comparing triangles,
the given equal sides must correspond
in relation to the angles.
A triangle has 6 parts, 3 sides and 3 angles, which can be measured.
However, we use 3 measurements at a time to construct a triangle.
øøø – A side is
required!
Any number of possible options.
Copyright © The Answer Series
5
or

or

Comparing Triangles – Congruence vs Similarity
Draw line DAE through A, parallel to BC.
We write: ABC
A
D
( If 2 triangles are equal in every respect (i.e. all 3 sides and
all 3 angles), we say they are congruent.
E
1 2 3
PQR.
B
( Enlarging or reducing a triangle, as on a copier, will produce a triangle
similar to the original one.
We know:
i.e. It will have the same shape (all angles will be the same size as before)
but the respective sides will not be the same length.
The sides will, however, be proportional.
C
. . . adjacent øs on a straight line, DAE
1̂ + 2̂ + 3̂ = 180º
But 1̂ = alternate B̂
&
. . . DAE || BC
3̂ = alternate Ĉ
â B̂ + 2̂ + Ĉ = 180º
We write: ABC ||| PQR.
Triangle Fact 2
Proofs of Triangle Facts 1 and 2
Why is the exterior angle of a triangle equal to
the sum of the two interior opposite angles?
A
Triangle Fact 1
Why is the sum of the interior angles of a triangle 180º?
x
B
A
See ΔABC:
x̂ + ŷ = 180º
B
C
but
Can we prove that B̂ + Â + Ĉ = 180º ?
ˆ + B)
ˆ = 180º
x̂ + (A
ˆ + B
ˆ
â ŷ = A
6
y
C
D
. . . see LINES fact 1 (Page 2)
. . . see TRIANGLES fact 1 (Page 3)
Logical !
Copyright © The Answer Series
Investigating quadrilaterals, using diagonals:
QUADRILATERALS
fig. 1 : Use a diagonal to determine the sum of
the interior angles of a quadrilateral.
fig. 2 : Use a diagonal to find the area of
a trapezium.
The Facts
fig. 3 - 6 : Which of these quadrilaterals have their areas bisected by
the diagonal?
Properties of Quadrilaterals
fig. 3 - 6 : Draw in the second diagonal. For each figure, establish whether the
diagonals are :
equal
bisect each other
intersect at right angles
bisect the angles of the quadrilateral
• Recall all the quadrilaterals ... (kite, trapezium, parallelogram, rectangle, rhombus, square).
• What properties do they have?
Equal sides?
kite
How would you find the sum
of the interior angles of
a pentagon? A hexagon?
fig. 6 : Find the area of a kite in terms of its diagonals.
Could this formula apply to a rhombus? A square?
rectangle
parallelogram
rhombus
Defining Quadrilaterals
square
A trapezium
'any' quadrilateral
rhombus
kite
square
A parallelogram
Parallel sides?
'any'
quadrilateral
Definition : A trapezium is a quadrilateral with
ONE PAIR OF OPPOSITE SIDES parallel.
We have observed the properties of a parallelogram :
both pairs of opposite sides parallel
We will, however, define the
both pairs of opposite sides equal
parallelogram in terms of
both pairs of opposite angles equal
its parallel lines.
diagonals which bisect one another.
rectangle
trapezium
square
parallelogram
rhombus
Definition : A parallelogram is a quadrilateral with
TWO PAIRS OF OPPOSITE SIDES parallel.
Angles?
Which are equal? Which are supplementary? Which are right angles?
Diagonals?
1
Observe the progression of quadrilaterals below as we discuss further definitions :
4a
5
3
2
rectangle
'any'
quadrilateral
trapezium
square
parallelogram
rhombus
4b
6a
6b
or
Copyright © The Answer Series
A diagonal of a
quadrilateral is
a line joining
opposite vertices.
7
•
•
•
•
Which property does a parallelogram need to become a rectangle?
Which property does a parallelogram need to become a rhombus?
Which property does a rectangle need to become a square?
Which property does a rhombus need to become a square?
A rectangle
A kite
Definition : A kite is a quadrilateral with 2 pairs of adjacent sides equal.
Definition : A rectangle is a parallelogram with 1 right angle
OR : A rectangle is a quadrilateral with . .?. . right angles?
Is a rhombus a kite?
A rhombus is a kite with .....................................................................................
A rhombus
A square is a kite with ........................................................................................
Definition : A rhombus is a parallelogram with a pair
of adjacent sides equal
Definitions tell us what the minimum (least) is that one needs !
OR : A rhombus is a quadrilateral with . .?. . equal sides?
See the summary of quadrilaterals on the next page:
A square
'Pathways of definitions, areas and properties'
Definition : A square is a rhombus with . . . . . . . . . . . . . .
A square is a rectangle with . . . . . . . . . . . . . .
A square is a parallelogram with . . . . . . . . . . . . . .
A square is a quadrilateral with . . . . . . . . . . . . . .
Observing the progression of quadrilaterals – along routes 1 & 2 – is essential for
understanding definitions, properties (especially diagonals) and area formulae.
None of these facts and proofs should need to be memorised.
Note:
The further back you go on the route,
the more the definition requires !
Proving conjectures
All these definitions are extremely important to know when you're doing sums.
Note : The curriculum says :
• Define the quadrilaterals.
e.g. If asked to prove that a particular quadrilateral is a rectangle, and you already
know it is a parallelogram, all you need to show is that one angle is a right angle.
• Investigate and make conjectures about the properties of the
sides, angles, diagonals and areas of these quadrilaterals.
Observe the following progression of quadrilaterals before the next definitions :
• Prove these conjectures.
e.g. 1 Conjecture : The diagonals of a rhombus bisect the angles of the rhombus.
'any' quadrilateral
kite
rhombus
square
x2
Proof :
x1
x3
x4
x̂1 = x̂ 2
. . . øs opp = sides
= x̂ 3
. . . alt øs ; || lines
= x̂ 4
. . . øs opp = sides
e.g. 2 Conjecture : The diagonals of a rhombus intersect at right angles.
(The proof is on the SUMMARY on the next page)
Copyright © The Answer Series
8
QUADRILATERALS - definitions, areas & properties
A Rectangle
All you need to know
The Square
'Any' Quadrilateral
f
e
a
A Parallelogram
A Trapezium
b
DEFINITION :
b
d
A ||m with one right ø
2
c
h
Sum of the ø of
any quadrilateral = 360º
Sum of the interior angles
= (a + b + c) + (d + e + f)
= 2 % 180º
. . . ( 2 Δs )
= 360º
DEFINITION :
Quadrilateral with 2 pairs opposite sides ||
1
s
a
DEFINITION :
Quadrilateral with 1 pair of opposite sides ||
B
= 1 ah + 1 bh
2
2
Q
Area = s 2
D
where ΔQCD ≡ ΔPBA . . . RHS / 90º HS
â ||m ABCD = rect. PBCQ (in area)
= BC % QC
Properties :
2 pairs opposite sides equal
2 pairs opposite angles equal
& DIAGONALS BISECT ONE ANOTHER
A Kite
a
2
a
2
Properties :
It's all been said 'before' !
Since a square is a rectangle,
a rhombus, a parallelogram,
a kite, . . . ALL the properties
of these quadrilaterals apply.
C
||m ABCD = ABCQ + ΔQCD
rect. PBCQ = ABCQ + ΔPBA
'Half the sum of the || sides
% the distance between them.'
Copyright © The Answer
A
DIAGONALS are EQUAL
Area = Δ 1 + Δ 2
2
See how the properties
accumulate as we
move from left to right,
i.e. the first quad. has
no special properties
and each successive
quadrilateral has all
preceding properties.
P
the 'ultimate' quadrilateral !
A Rhombus
= 1 (a + b) . h
The arrows indicate
various ‘pathways’
from ‘any’
quadrilateral to the
square (the ‘ultimate
quadrilateral’). These
pathways, which combine
logic and fact, are
essential to use when
proving specific types
of quadrilaterals.
Area = base % height
Area = ℓ % b
DEFINITION :
A ||m with one pair of
adjacent sides equal
sides
Area
= 1 product of diagonals (as for a kite)
2
diagonals
or
= base % height (as for a parallelogram)
DEFINITION :
Quadrilateral with 2 pairs of
adjacent sides equal
b
Given diagonals a and b
1 a
Area = 2Δs = 2 ⎛⎜ b . ⎞⎟ = ab
⎝2
2⎠
angles
THE DIAGONALS
2
'Half the product of the diagonals'
• bisect one another PERPENDICULARLY
• bisect the angles of the rhombus
• bisect the area of the rhombus
Note :
y
x
y
x
Quadrilaterals play a
prominent role in both
Euclidean & Analytical
Geometry right through
to Grade 12 !
2
THE DIAGONALS
• cut perpendicularly
• ONE DIAGONAL bisects the other diagonal,
the opposite angles and the area of the kite
x
x
y y
s
ø of Δ or
2x + 2y = 180º . . .
co-int. øs ; || lines
² x + y = 90º
These slides on Quadrilaterals are from
The Answer Series Gr 12 Video series
on Analytical Geometry.
Defining Quadrilaterals
A trapezium
Definition: A trapezium is a quadrilateral with
ONE PAIR OF OPPOSITE SIDES parallel.
A parallelogram
We have observed the properties of a parallelogram:
both pairs of opposite sides parallel
both pairs of opposite sides equal
both pairs of opposite angles equal
We will, however, define the
parallelogram in terms of
its parallel lines.
diagonals which bisect one another.
Definition: A parallelogram is a quadrilateral with
TWO PAIRS OF OPPOSITE SIDES parallel.
2
Observe the progression below as we discuss further definitions . . .
rectangle
'any'
quadrilateral
trapezium
parallelogram
square
rhombus
Which property does . . .
1.
a parallelogram need to become a rectangle?
2.
a parallelogram need to become a rhombus?
3.
a rectangle need to become a square?
4.
a rhombus need to become a square?
5.
And, which property(s) does a parallelogram need to become a square?
3
QUADRILATERALS
A Rectangle
'Any' Quadrilateral
f
e
a
A Trapezium
A Parallelogram
b
b
c
d
h
The Square
2
1
A Rhombus
a
The 'ultimate'
quadrilateral
A Kite
a
2
The arrows indicate various
‘ pathways’ from ‘any’
quadrilateral to the square
(the ‘ultimate quadrilateral’)
a
2
b
Consider
Sides: parallel?
equal?
Angles: equal?
supplementary?
Diagonals . . . . ?
4
right  ?
Investigating diagonals . . .
'Any quadrilateral'
 How do we find the sum of the interior angles of a quadrilateral?
And a pentagon?
And a hexagon?
f
a
e
b
The sum of the interior angles = (a + b + c) + (d + e + f)
d
= 2 % 180º
c
. . . (2 s)
= 360º
Pause now while you continue the pattern
A quadrilateral
No. of Sides
No. of Diagonals
No. of Triangles
Sum of the interior  s
4
1
2
2 % 180º = 360º
A pentagon
A hexagon
A polygon with n sides
5
A Trapezium
 Can you derive a formula for the area of a trapezium?
b
h
The Area =  1 +  2
2
= 1 ah + 1 bh
2
2
= 1 (a + b).h
2
1
a
 The area of a trapezium:
Half the sum of the || sides % the distance between them.
6
A Kite
Can you derive a formula for the area of a kite?
A Kite
a
2
Given diagonals a and b . . .
a
2
b
Area = 2s = 2  1 b . a  = ab
2
2
2
...
the product of the diagonals
2
 The area of a kite: 'Half the product of the diagonals'
Could this formula apply to a rhombus?
And to a square?
7
Now, draw in a second diagonal . . .
Consider each quadrilateral
Determine in which quadrilaterals the diagonals:
A Rectangle
1. bisect each other?
A Parallelogram
A Square
2. intersect at right angles?
A Rhombus
3. bisect the angles of the quadrilateral?
4. are equal?
A Kite
8
A Rectangle
The diagonals . . . .
1. bisect one another?
Parallelogram
Rectangle
Rhombus
A Rhombus
A Kite
Square
2. intersect at right angles?
Kite
Rhombus
Square
3. bisect the  s of the quadrilateral?
Rhombus
Square
4. equal?
Rectangle
The Square
A Parallelogram
Square
9
A SUMMARY: DIAGONALS
Kite
cut perpendicularly, and
the long diagonal bisects:
the short diagonal the opposite angles & the area of the kite
Parallelogram
Rectangle
bisect one another
are equal
Rhombus
bisect one another perpendicularly
y
...
bisect the angles of the rhombus
x
bisect the area of the rhombus
x
x
y
y y
x
2x + 2y = 180º
² x + y = 90º
co-int.  s suppl.
 The  s at the point of intersection
of the diagonals are right angles.
Square
Since a square is a rectangle, a rhombus, a parallelogram, a kite . . .
. . . ALL the properties of these quadrilaterals apply.
10
s
. . .  of , or
SUMMARY: AREAS
A Rectangle
A Parallelogram
The Square
A Trapezium
b
Area = ℓ % b
2
h
1
Area = base % height
a
P
Area =  1 +  2
= 1 ah + 1 bh
2
2
1
= (a + b).h
2
B
a
2
a
2
b
Q
Area = s 2
D
C
A Rhombus
m
|| ABCD = ABCQ + QCD
'Half the sum of the || sides
% the distance between them.'
A Kite
A
Since a square is a rectangle,
a rhombus, a parallelogram,
a kite, . . . ALL the properties
of these quadrilaterals apply.
rect. PBCQ = ABCQ + PBA
where QCD ≡ PBA . . . SS90º
â ||m ABCD = rect. PBCQ (in area)
= BC % QC
Area of a rhombus
= 1 product of diagonals (as for a kite)
2
Given diagonals a and b
Area = 2s = 2 1 b . a = ab
2
2
2
'Half the product of the diagonals' 11
or
= base % height (as for a parallelogram)
K Theorems and Proofs
Theorem 4 : If a QUADRILATERAL has 2 pairs of opposite angles equal,
then the quadrilateral is a parallelogram.
The following section deals with the properties of a parallelogram. We firstly prove
all the properties. Secondly, we prove that a quadrilateral with any of these
properties has to be a parallelogram.
Theorem 5 : If a QUADRILATERAL has 2 pairs of opposite sides equal,
then the quadrilateral is a parallelogram.
Geometry is an exercise in LOGIC. Initially, we observe, we measure, we
record . . . But, finally . . . We decide on how to define something and
then we prove various properties logically, using the definition.
Theorem 6 : If a QUADRILATERAL has 1 pair of opposite sides equal and
parallel, then the quadrilateral is a parallelogram.
Theorem 7 : If a QUADRILATERAL has diagonals which bisect one another,
then the quadrilateral is a parallelogram.
THE DEFINITION OF A PARALLELOGRAM
A parallelogram is a quadrilateral with
2 PAIRS OF OPPOSITE SIDES PARALLEL.
O AN ASSIGNMENT O
Beyond the DEFINITION of a parallelogram, we noticed other facts / properties
regarding the lines, angles and diagonals of a parallelogram. The statement and
proofs of these properties make up our first three THEOREMS !
TASK A: Theorems 1 → 3
Prove each of these properties yourself,
STARTING WITH THE DEFINITION as the 'given'.
The PROPERTIES of a parallelogram
TASK B: Theorems 4 → 7
All the properties are to be deduced from the definition!
Prove these four converse theorems,
Hint
Use your FACTS on II lines
and congruent triangles.
Theorem 1 : The opposite angles of a parallelogram are equal.
WORKING TOWARDS THE DEFINITION,
Theorem 2 : The opposite sides of a parallelogram are equal.
i.e. you need to prove, given any one of these situations, that the quadrilateral
would have 2 pairs of opposite sides parallel, i.e. that, by definition, the
Theorem 3 : The diagonals of a parallelogram bisect one another.
quadrilateral is a parallelogram.
The CONVERSE theorems
Given a property, prove the quadrilateral is a parallelogram,
i.e. prove both pairs of opposite sides are parallel.
There are four converse statements, each claiming that IF a quadrilateral
has a particular property, it must be a parallelogram.
In these cases, we work towards the definition !
Copyright © The Answer Series
10
K The Theorem Proofs
Theorem 3: The diagonals of a parallelogram bisect
one another.
THE PROOFS OF THE PROPERTIES
Given : ||m ABCD with diagonals AC and BD intersecting at O.
DON'T EVER MEMORISE THEOREM PROOFS!
RTP:
Develop the proofs/logic for yourself before checking against the methods shown below.
A
AO = OC and BO = OD
Proof: In Δs AOB and DOC
Theorems :
Definition
Converse theorems :
Property
1) 1ˆ = 2ˆ
2) 3ˆ = 4ˆ
Make sense of
THE LOGIC!
Property
Definition
RTP:
 = Ĉ
Proof:
 + B̂ = 180º
A
and B̂ = D̂
â ΔAOB ≡ ΔCOD
But, Â + D̂ = 180º
C
. . . øøS
Note: We used the result in theorem 2 in the proof of theorem 3 –
but, we could've started from the beginning,
i.e. from the definition of a parallelogram.
We would just have needed to prove an extra pair of Δs
congruent (as in theorem 2).
C
. . . co-interior øs ; AD || BC
. . . co-interior øs ; AB || DC
â B̂ = D̂
Similarly, Â = Ĉ
2
â AO = OC and BO = OD
B
D
4
>
D
øs
3
B
>
. . . vert opp
3) AB = DC . . . opposite sides of ||m – see theorem 2 above
Theorem 1: The opposite angles of a ||m are equal.
Given : ||m ABCD
i.e. AB || DC and AD || BC
. . . alt øs ; AB || DC
1
RTP: Required to prove
THE CONVERSE PROOFS
m
Theorem 2: The opposite sides of a || are equal.
Theorem 4: If a QUADRILATERAL has 2 pairs of opposite
Given : ||m ABCD
i.e. AB || DC and AD || BC
RTP:
Given : Quadrilateral ABCD with  = Ĉ and B̂ = D̂
AB = CD and AD = BC
Construction: Draw diagonal AC
Proof:
angles equal, then the quadrilateral is a ||m
It doesn’t matter which
...
diagonal you draw
In Δs ABC and ADC
1) 1̂ = 2̂ . . . alternate øs ; AB || DC
2) 3̂ = 4̂
A
. . . alternate øs ; AD || BC
3) AC is common
â ΔABC ≡ ΔCDA
. . . øøS
â AB = CD and AD = BC
1
4
D
>
RTP:
B
>
2
A
>
x
ABCD is a parallelogram,
i.e. AB || DC and AD || BC
y
D
>
y
B
x
C
Proof: Let  = Ĉ = x and D̂ = B̂ = y
then  + B̂ + Ĉ + D̂ = 360º . . . sum of the øs of a quadrilateral
â 2x + 2y = 360º
÷ 2)
â x + y = 180º
3
C
i.e.  + D̂ = 180º and  + B̂ = 180º
â AB || DC and AD || BC . . . co-interior øs are supplementary
â ABCD is a parallelogram . . . both pairs of opposite sides ||
We could, of course, also have proved the first theorem this way!
11
Copyright © The Answer Series
Theorem 7: If a QUADRILATERAL has diagonals which
bisect one another, then the quadrilaterals
Theorem 5: If a QUADRILATERAL has 2 pairs of opposite
m
sides equal, then the quadrilateral is a ||
Given : Quadrilateral ABCD with AB = CD
and AD = BC
RTP:
ABCD is a parallelogram,
i.e. AB || DC and AD || BC
A
3
is a ||m.
B
1
2
D
Given : Quadrilateral ABCD with diagonals AC and BD intersecting at O and
AO = OC and BO = OD.
A
B
4
3
C
RTP:
Construction: Draw diagonal AC . . . it doesn’t matter which diag. you draw
Proof:
Proof:
In Δs ACD and CAB
1) AC is common
2) AD = BC
. . . given
3) CD = AB
. . . given
â ΔACD ≡ ΔCAB
. . . SSS
â 1̂ = 2̂ and 3̂ = 4̂
â AB || DC and AD || BC
â ABCD is a parallelogram
A
4
ABCD is a parallelogram,
i.e. AB || DC and AD || BC
D
>
2
C
1ˆ = 2ˆ
In Geometry, we never have
to repeat a 'logic sequence'
(as would’ve been required
here) –
we just say: Similarly, . . . !
. . . 2 pairs of opp. sides are ||
In our sums, we may use ALL properties and theorem statements . . .
To prove that a quadrilateral is a ||m we may choose one of 5 ways :
1) Prove both pairs of opposite sides || (the definition).
3
C
2) Prove both pairs of opposite sides = (a property).
Construction: Draw diagonal AC . . . It doesn’t matter which
. . . THE SIDES
3) Prove 1 pair of opposite sides = and || (a property).
diagonal you draw
Proof:
4
In Δ AOB and COD
â ABCD is a parallelogram
B
>
1
D
Similarly, by proving ΔAOD ≡ ΔCOB it can be shown that AD || BC
sides equal and ||, then the quadrilateral is a ||m
Given : Quadrilateral ABCD with AB = and || DC
2
AO = OC . . . given
. . . vert opp øs
BO = O
. . . given
ΔAOB ≡ ΔCOD
. . . SøS
ˆ
ˆ
â BAD = OCD
â AB || DC . . . alternate øs equal
. . . alternate øs are equal
. . . both pairs of opposite sides ||
1
s
1)
2)
3)
â
Theorem 6: If a QUADRILATERAL has 1 pair of opposite
RTP:
ABCD is a parallelogram,
i.e. AB || DC and AD || BC
In Δs ABC and CDA
4) Prove both pairs of opposite angles = (a property). . . . THE ANGLES
1) AB = DC
. . . given
2) 1̂ = 2̂
. . . alternate øs ; AB || DC
3) AC is common
â ΔABC ≡ ΔCDA . . . SøS
5) Prove that the diagonals bisect one another (a property).
â 3̂ = 4̂
â AD || BC
. . . alternate øs equal
But AB || DC
. . . given
â ABCD is a parallelogram
Copyright © The Answer Series
. . . THE DIAGONALS
Using diagonals . . .
. . . both pairs of opposite sides ||
12
To prove a parallelogram is a rectangle:
prove that the diagonals are equal.
To prove a parallelogram is a rhombus :
prove that the diagonals intersect
at right angles, or
prove that the diagonals bisect
the angles of the rhombus.
K Area of Quadrilaterals and Triangles
Why is the area of a parallelogram = base % height?
A SUMMARY OF FORMULAE FOR
Compare ||m ABCD and rectangle EBCF
AREAS OF QUADRILATERALS
Δ1
m
|| ABCD = ABCF + Δ2
So far, we have established:
The area of a trapezium =
The area of a kite =
1
2
a
>
1
(a
2
+ b) . h
& rectangle EBCF = ABCF + Δ1
h
>
A
E
B
F
D
Δ2
C
b
the product of the diagonals
But, we know that the opposite sides of rectangles and
parallelograms are equal
But also, remember:
â Δ1  Δ2
The area of a square = s2,
s
where s = the length of a side of a square
The area of a rectangle = ℓ % b
where ℓ = the length & b = the breadth
. . . SS90º or RHS
â Δ1 = Δ2 in area
â Area of ||m ABCD = Area of rectangle EBCF
b
= BC . FC
ℓ
...
ℓ%b
= base % height (of the ||m )
The area of a parallelogram = base % height
. . . See the explanation of
this formula below:
The area of a rhombus = . . . . . . . (1) . . . . . . . . . . . . .
NB:
or = . . . . . . . (2) . . . . . . . . . . . . .
Study THE SUMMARY OF QUADRILATERALS for
Answers
'pathways of definitions, areas, properties', etc.
(1) base % height (because a rhombus is a parallelogram), or
(2)
KNOW THIS WELL ! (page 9)
1 product of the diagonals (because a rhombus is a kite)
2
13
Copyright © The Answer Series
4. The diagonal of a parallelogram (or rhombus or
SOME IMPORTANT FURTHER FACTS ON AREAS
rectangle or square!) bisects the area
OF ΔS & QUADRILATERALS
A
B
ΔABC ≡ ΔCDA . . . (SøS)
D
The height of these Δs & ||ms is the distance between the parallel lines . . .
C
â ΔABC = ΔCDA [= 1 ||m ABCD] in area
2
1. Δs on the same base & between the same || lines
are equal in area
P
Area of Δ = 1 base (BC) % height (h)
Q
A
5. If two triangles lie between
h
2
C
â ΔPBC = ΔABC = ΔQBC in area
Q
P
B
1. x. h
1
area of ΔABD
2
=
=
1 . 4x . h
4
area of ΔDAC
h
A
2
â ||m ABCD = ||m PQCD in area
C
D
⎛ base of ΔABD
⎜=
⎝ base of ΔDAC
3. The median of a Δ bisects the area of the Δ
â Area of ΔABD =
A
Area of ΔABD = 1 BD.h &
2
h
Area of ADC = 1 DC.h
2
But BD = DC
...
B
x
D
x
C
median AD bisects
the base
â Area of ABD = Area of ADC [ 1 Area of ABC!]
2
Copyright © The Answer Series
B x
= THE RATIO OF THEIR BASES
2. Parallelograms on the same base and between
the same || lines are equal in area
h
THE RATIO OF THEIR AREAS
B
Area of ||m = base (DC) % height (h)
A
the same parallel lines:
14
D
4x
C
1
. x. h
area of ΔABD
1
Also,
= 2
=
1 . 5 x. h
area of ΔABC
5
2
⎞
1
area of ΔABC
!⎟ â Area of ΔABD =
5
⎠
1
Area of ΔDAC
4
PROOFS
Gr 10: THE MIDPOINT THEOREM
Use the diagrams below to prove facts 1 and 2:
FACT 1
The line segment through the midpoint of one side of a triangle, parallel to a second
side, bisects the third side.
1.
A
Given :
P midpoint AB & PQ || BC
P
Result :
Q
Q midpoint AC & PQ = 1 BC
2
B
C
(See Exercise 4 Q3.2 for the proof)
FACT 2
The line segment joining the midpoints of two sides of a triangle is parallel to the
third side and equal to half of the third side.
A
Given :
P & Q midpoints of AB & AC
P
Result :
2.
Q
PQ || BC & PQ = 1 BC
2
B
C
(See Exercise 4 Q3.1 for the proof)
Regard these Facts 1 & 2 as a special case
of the Proportion Theorem in Gr 12 Geometry.
Copyright © The Answer Series
15
Grade 8 to 10
EUCLIDEAN GEOMETRY
EXERCISES & FULL SOLUTIONS
EXERCISE 1:
Lines and Angles
EXERCISE 2:
Lines, Angles and Triangles
EXERCISE 3:
Quadrilaterals
EXERCISE 4:
Midpoint Theorem
EXERCISE 1: Lines and Angles
6.
ˆ = 200º (see figure below).
Given : reflex AOD
(Answers on page 22)
1.
140º
140º and 40º are adjacent supplementary angles :
x
The angle supplementary to x is :
O
40º
In this figure
â x+y=
x
y
(2.1)
7.
. . . ø on a straight line
(2.2)
and so they are called
A
145º
D
1̂ and 2̂ are
angles.
(7.1)
100º
40º
?
60º
A
(4.1)
D
x
(6.5)
(6.6)
1
2
1̂ and 2̂ are
øs (NAME)
(7.2)
1̂ and 2̂ are
øs (NAME)
RELATIONSHIP :
When two lines intersect,
the vertically opposite øs
are equal. Why?
y
x
z
B
8.
The sum of the
adjacent angles
about a point
is 360º.
(4.2)
?
9.
(7.6)
1
3 5
4 6
1̂ and
(8.1)
are corresponding øs and they are
(8.2)
1̂ and
(8.3)
are alternate øs and they are
(8.4)
1̂ and
(8.5)
are co-interior øs and they are
(8.6)
NB : It is ONLY BECAUSE THE LINES ARE
(9)
, that
the corresponding and alternate ø ARE EQUAL and the
co-interior øs are SUPPLEMENTARY !!!
x + y = 180º . . . øs on a
straight line
and z + y = 180º . . . øs on a
straight line
â
Copyright © The Answer Series
øs (NAME)
RELATIONSHIP :
(7.5)
s
5.
(7.3)
(3.2)
x
2
4.
ˆ
AOC
=
E
90º - x
B
(6.4)
1
2
(7.4)
C
45º
(6.3)
1
Is ADB a straight line? Give a reason for your answer.
(3.1)
ˆ =
BOD
s
: x + 90º + y = 180º
C
(6.2)
(1)
?
RELATIONSHIP :
3.
B
D
2
2.
ˆ
Obtuse AOD
=
200º
C
A
Reasons :
(6.1)
(5)
17
3.
EXERCISE 2:
Lines, Angles and Triangles
3.1
(Answers on page 22)
1.
In the sketch,
AB is a straight line.
If x - y = 10º, find the
value of x and y.
Determine the values of x and y in the following
diagrams. Give reasons for your answers.
A
E
y
120º
x+y
y
T
3.2
F
A
60º
35º
x
3x y
B
G
e
2x
4.
2.2.2
120º
3x 4x
5x
110º
A
D
6.1.2
P
x
R
5.
B
56º
Q
R
(Hint : first prove that ∆PQR ≡ ∆SRQ)
2.4 State whether PQ || RS,
giving reasons.
P
37º
30
A
5.2
V 104º
76º
Q
18
30
53º
E
27º
D
x
C
ˆ ,
ˆ and CED
6.2.1 Write down the sizes of BED
giving reasons.
18
Y
D
A
B
6.2.2 Hence, or otherwise, calculate the
value of x, showing all working and
giving reasons.
87º
C
x + 30º
6.2 In the diagram,
AB || ED and
BE = EC.
ˆ = 27º
Also, ABE
ˆ = 53º.
and BAC
X
E
17
T
B
3x - 10º
Z
A
R
U
S
State whether the following pairs of triangles are
congruent or similar, giving reasons for your choice.
5.1
A
C
C
Q
T
x
S
B
P
= T̂
Prove that P
by first proving that
the 2 triangles
are congruent.
x
2.3 If PQ = SR and
øPQR = øSRQ,
prove that
PR = SQ.
x
3x
60º
f
2.2 Calculate x and give reasons.
Copyright © The Answer Series
D
E
a
2.2.1
6.1.1
C
66º
g
d
R
6.1 Find the value of x, by forming an equation first
and then solving for x.
Show all your working and give reasons.
B
D
P
Q
F
x
K
C
60º
2.1 Find the size of angles a to g (in that order),
giving reasons.
b
c
x
100º
5.3
S
W
F
7.
L
23
25
H
51
Calculate the area
of the kite alongside.
B
E
4 cm
D
12 cm
69
13 cm
M
18
75
N
C
8.
Calculate the value
of x giving
reasons.
x + 10º
B
9.
2x + 36º
C
3x - 50º
D
A
y
D
C
10. In ΔABC, BC is produced to D and CE || BA.
Prove that:
ˆ = Aˆ + Bˆ
10.2 ACD
14.2 What is the relationship
between CE and AD?
Explain.
E
1
B
ˆ = 45º.
11. Prove that DOC
2
3
D
C
6 5
7
D
B
A
O
A
D
A
C
B
35º 2 1
y
1
E
D
x
S
D
Q
E
A
16.3
R
16.4
A
D
A
B
B
E
D
16.5
x
C
E
y
x
D
x
y
B
C
D
17. Given ΔPQR and ΔABC.
P
A
12
x
8
C
15.2 Complete the following statement :
∆ABE ||| ∆ . . . ||| ∆ . . .
15.3 If BC = 18 cm and BE = 12 cm,
calculate the length of
15.3.1 AE
15.3.2 AB correct to one decimal.
15.4 Hence calculate the area of rectangle ABCD.
19
B
C
B is the centre of the circle
A
D
B
140º
O
P
C
y
y
E
3
16.2
B
E
15.1 Make a neat copy of this sketch and fill in all the
other angles in terms of x. Reasons are
not required.
x x
Copyright © The Answer Series
8
ˆ
14.3 If CE bisects ACD
C
what is the relationship
between BE and CD? Explain.
C
C
9
ˆ , ABE
ˆ and
14.1 Express CEB
ˆ in terms of x.
AEB
ˆ = 180º
10.1 Aˆ + Bˆ + C
1
Determine,
stating reasons,
the values of
x and y.
B
13.4 9ˆ + 6ˆ = . . . (the no. of one angle)
13.5 7̂ = . . . . (list only one angle)
= 10
, then GD = . . . .
13.6 If 12
B
ˆ = 35º.
AOC
10
= . . . . (degrees)
13.2 1̂ + 11
13.3 6ˆ + 7ˆ + 2ˆ = . . . . (degrees)
x
12. If AB || CD,
ˆ = 140º and
BOE
16.1 A
13.1 1ˆ + 2ˆ + 3ˆ + 4ˆ = . . . . (degrees)
z
A
16. In each of the following, state whether the given
triangles are congruent or not , and in each case
give a reason for your answer. Do not prove the
triangles congruent, but name each congruent
pair correctly.
F
12 G
11 2
1 3
4
Complete the
following statements
(giving reasons) :
A
If AC || DB, prove with
geometric reasons that
x = y + z.
H
13. In the accompanying
figure BG || CF.
A
9
R
6
Q
C
6
3
B
17.1 Show that ΔPQR is NOT similar to ΔABC.
Clearly show all relevant calculations
and reasons.
17.2 Prove that ΔPQR is not right-angled.
6.
EXERCISE 3: Quadrilaterals
(Answers on page 23)
1.
Which of the following quadrilaterals is definitely a
parallelogram? (Not drawn to scale.)
A.
7.
D.
If the area of
parallelogram PQRS
is equal to 90 cm2
then PT is equal to
. . . cm.
T
R
P
8.
U
5.
Y
A
E
B
x
125º
D
B
9.
R
D
B
F
E
D
B
y
4y - 18º
F
In the accompanying
figure ABCD is a
parallelogram.
D
The diagonal is
produced to E and
AB = BE and
A
AD = CE.
C
x
B
D
F
E C
11.2 Show that BC = CF.
11.3 Prove that DF = EC.
A
1
2
1 2
ˆ and
AP bisects DAB
ˆ .
BP bisects ABC
B
1 2 3
D
P
C
13.1 Prove the theorem which says that if both pairs of
opposite angles of a quadrilateral are equal,
then it is a parallelogram.
P
1
Q
2
E
S
13.3 Write down 2 facts about a rhombus which are
not generally true of any parallelogram.
B
14. PQRS is a parallelogram.
P
M
m
m
x
B
R
A
P
What type of quadrilateral is LMPN? Give reasons.
Find NM and MS.
20
R
Prove that it is a parallelogram.
L
10. LM = 5 cm,
LP = 6 cm and
PS = 1 cm.
N
C
y
y
C
ˆ = x, prove, giving reasons, that FDC
ˆ = 3x.
If CEB
C
A
F
x
P
13.2 PQRS is a quadrilateral
with PS = PR = QR and
PQ || SR.
65º
A
Copyright © The Answer Series
C
E
x
11.1 Find the magnitude
ˆ .
of APB
Prove that AB = 2BC
115º
65º
T
ABCD is a rhombus.
By setting up an equation
and showing all steps and
reasons in the process,
find the value of y.
A
A
AE and BF bisect  and B̂
respectively.
P is the point of intersection
of AE and BF.
Prove that ABCD is a parallelogram.
Q
Prove that TRUP is a parallelogram.
Prove that ECF
is a straight line.
7.2 ABCD is a square.
ˆ = 125º.
BFD
S
11. ABCD is a parallelogram, and
12. ABCD is a parallelogram.
D
In the given diagram,
PQRS is a parallelogram.
SP = PT and
UR = RQ.
In the diagram,
ABCD is a
parallelogram.
AB is produced
to E and AD to F.
3x - 30º
F
S
4.
D
C
10 cm
Q
3.
15 cm
P
E
In each of the following cases, calculate the value
of x giving reasons.
W
X
7.1 WXYZ is a
x + 10º
parallelogram.
Z
2.
A
ˆ .
Calculate the size of EBC
B.
C.
ABCD is a parallelogram.
E is a point on AD such
that AE = AB and
EC = CD.
ˆ = 90º.
BEC
B
n
S
Prove that ABCD
is a rectangle.
S
C
D
n
y
y
R
Q
x
EXERCISE 4: Midpoint Theorem
(Answers on page 25)
1.1 Complete (giving missing words only):
The line segment joining the midpoints of two sides
of a triangle is . . . . . . to the third side and
equal to . . . . . . .
1.2 ΔABC has medians
BN and CM drawn,
cutting each other
M
at O.
P and Q are the
P
midpoints of BO
B
and CO
respectively.
3.2 The diagram shows a
triangle PQR with M the
midpoint of PQ.
MNT is drawn parallel
to QR so that RT || QP.
Use this diagram to
prove the theorem
which states that
MN bisects PR.
A
4.
N
O
Q
C
5.
15
S
2.1 Prove VR = 7
T
1
cm.
2
6.
2.2 Calculate PQ if PQ = 16 VR and hence
ˆ = 90º.
prove that PQR
5
D
Copyright © The Answer Series
8.
H
J
E
A
In ΔPQR, PT = TQ, while PV = VR = RS.
T
V
Q
R
W
60º
T
M
Show that WR = 1 QR.
S
4
80º
B
ABCD is a square.
The diagonals AC and
BD intersect at O.
M is the midpoint of BO
and AM = ME.
6.3.1 Dˆ 1
C
x
C
F
A
M, N and T are the
midpoints of AB,
BC and AC in ∆ABC.
6.3 Determine, with reasons,
the size of :
Q
O
P
6.2 Prove that DOEC is
a parallelogram.
A
P
F
G
N
C
A
D
9.
ABCD is any quadrilateral.
E is a point BC. P, Q, R and S are the midpoints
of AB, AE, DE and DC.
D
1
A
O
P
M
6.1 Prove that MD || EC.
2.3 Write down the length of ST.
3.1 With reference to the diagram,
prove the theorem which
states that PQ is parallel to DC
and half its length.
y
D
Triangle DEF has G
the midpoint of DE,
H the midpoint of DF.
GH is joined.
HJ is parallel to DE.
Calculate the angles
of ∆MNT.
R
9
D
R
y
 = 60º and B̂ = 80º.
V
9
Q
ˆ = JHF
ˆ
4.2 GDH
4.3 JF = GH
P
B
7.2 ΔBCA ≡ ΔAED
x
Prove :
4.1 GHJE is a parallelogram.
In the diagram below QS || TV; PQ || ST;
QT = TR = 9 cm and PS = 15 cm.
Q
With reference to the diagram,
prove that
7.1 FE = EC = CA
T
N
M
E
Prove that MNQP is a parallelogram.
2.
7.
P
B
C
B
R
Q
E
Prove that :
E
9.1 PQ || RS
9.2 PQ + QR + RS = 1 (AD + BC)
ˆ
6.3.2 ECD
2
21
S
C
ANSWERS TO EXERCISES
2.2.3 In ∆s PQR & SRQ
EXERCISE 1: Lines & Angles
1.
180º - x
2.1
3.1
3.2
4.1
90º
2.2 complementary
No
. . . 45º + 145º ≠ 180º
Yes . . . 90º – x + 90º + x = 180º
160º
4.2 360º - x
6.1
s
6.3
6.5
ˆ = SRQ
ˆ
PQR
160º
revolution / ø about a point
6.2
20º
20º
5. â x = z
. . . (given)
corresponding
equal
8.1
5̂
8.2
equal
8.3
4̂
8.4
equal
8.5
3̂
8.6
supplementary
7.2 co-interior
7.5 supplementary
7.3
7.6
alternate
equal
But x - y = 10º
â x = 50º and y = 40º
c = 35º
d = 85º
3.2
4.
. . . øs on a straight line
2x + 2y = 180º
â x + y = 90
b = 35º
. . . because corresponding
x = 100º - 60º = 40º
. . . ext. ø of EFC
. . . ext. ø of ABC
. . . by inspection
. . . vertically opposite angles
. . . alternate øs ; || lines
. . . øs opp = sides
. . . sum of øs in ∆
øs are equal
corresp. øs ; || lines
g = 60º – 35º = 25º . . . ext. ø of ∆
. . . revolution
Copyright © The Answer Series
=
=
=
=
2x + 60º
. . . corr. øs ; BA || DC
60º
30º
180º - 4x = 60º . . . øs on a straight line
In ∆s QRP and SRT
(1)
Q̂ = Ŝ (= x)
(2)
ˆ = SRT
ˆ
QRP
7.
. . . given
â ∆QRP ≡ ∆SRT
. . . øøS
Congruent ; SøS . . . [ Ĉ = 87º
5.2
Similar ; sides in proportion
2
8.
5.3
. . . 2 prs. co-int. øs;
|| lines
6.1.1
10.2
Cˆ 3 + Cˆ 2 + Cˆ 1
= 180º
. . . øs on a straight line
But Cˆ 2 = Â
. . . alternate øs ; CE || BA
& Cˆ 3 = B̂
. . . corresp. øs ; CE || BA
ˆ
= 180º - Cˆ 1
ACD
& Â + B̂ = 180º - Cˆ 1
. . . str. line
. . . ø sum of ∆
ˆ
â ACD
= Â + B̂
. . . [no sides]
6.1.2 (3x - 10º) + (x + 30º)
â 4x + 20º
â 4x
â x
. . . alt. øs ; AC || DB
. . . ext. ø of ∆
Ĉ = z
â B̂ + Â + Cˆ 1 = 180º
Similar ; equiangular (alt. øs & vert. opp. øs )
3x = 66º + x
â 2x = 66º
â x = 33º
. . . why?]
2x + 36º = 3x - 50º + x + 10º . . . ext. ø of ∆
â - 2x = -76º
A VERY IMPORTANT THEOREM !
â x = 38º
â x = y+z
ø sum of ∆]
LOOKS MAY NOT DECEIVE !
... 2%∆
2
= 80 cm
[OR : 1 product of diagonals
10.1
. . . [17 : 23 : 25 = 51 : 69 : 75]
. . . diagonals of a kite
. . . 5 : 12 : 13 ∆ ; Pythag.
2
. . . vertically opposite øs
(3) QR = SR
BE ⊥ AC
BE = 5 cm
Area of kite = 2. 1 (12 + 4).5
9.
5.1
f = 35º + 35º = 70º . . . either ext. ø of ∆ or
2.2.2 120º + 110º + x = 2(180º)
â 230º + x = 360º
â x = 130º
4x
â 2x
â x
y
. . . øs opp = sides
â In ∆ABC : 53º + x + x - 27º = 180º . . . sum of øs in ∆
â 2x = 154º
â x = 77º
â P̂ = T̂
e = 60º - 35º = 25º . . . ext. ø of ∆
12x = 360º
â x = 30º
â Yes, PQ || RS
y = 120º - 40º = 80º
9. parallel
EXERCISE 2:
Lines, Angles and Triangles
a = 60º
3.1
. . . corresp. øs ; AB || ED
ˆ
= x
ABD
ˆ
â EBC
= x - 27º
â Ĉ = x - 27º
ˆ = 180º - 76º = 104º . . . øs on str line
PUV
ˆ = RVW
ˆ
â PUV
ø on a straight line
6.4
6.2.2
. . . SøS
s
. . . corresp. øs ; AB || ED
ˆ
= 53º
CED
. . . (given)
(3) QR is common
2.4
7.1
7.4
2.2.1
(2)
. . . alt. øs ; AB || ED
ˆ
6.2.1 BED
= 27º
â PR = SQ
ˆ
øs on a straight line or vertically opposite to BOD
2.1
PQ = SR
â ∆PQR ≡ ∆SRQ
6.6
1.
(1)
[Note: order of letters !]
11.
ˆ
= x+y
DOC
But 2x + 2y = 90º
â x + y = 45º
. . . ext ø of ∆
. . . ext. ø of ∆
. . . ø sum of rt. ød ∆ABC
ˆ = 45º
â DOC
=
=
=
=
180º
180º
160º
40º
22
. . . co-int. øs; AB || CD
12.
s
Oˆ 1 = 40º & Oˆ 3 = 40º . . . ø on a straight line
â x = 40º
Cˆ 1 = 35º
â y = 145º
. . . alt. or corresp. øs ; AB || CD
. . . alt. øs ; AB || CD
. . . øs on a straight line
13.1
. . . revolution
360º
16.1
s
13.2 180º
. . . co-interior ø ; BG || CF
13.3 180º
. . . vert. opp. øs ; ø sum of ∆
13.4
12
13.5
9̂
16.2 No
The angle is not
included, but
it is a right angle
. . . RHS
16.3 Yes, ∆ABD ≡ ∆ACD
. . . corresp. øs ; BG || CF
= 10
. . . 12
ˆ
CEB
ˆ
ABE
= 9 ; sides opp = øs
11
s
= x
. . . ø opp = sides
= 2x
. . . ext. ø of ∆
s
ˆ = (180º - 2x)
 + AEB
ˆ
â AEB = 90º - x
. . . sum of ø in ∆
A
2x
B
17.1
12
4
=
;
9
3
8
4
=
6
3
2
2
2
â 12 g 6 + 8
ˆ and AEB
ˆ are complementary
Explanation : CEB
. . . x + (90º - x) = ?
14.3 They are parallel; i.e. BE || CD ;
ˆ =x
Explanation : ECD
ˆ
. . . CE bisects ACD
ˆ
ˆ
â ECD
= alternate angle, CEB
A
90º - x
E
x
2
NB: The order of
the letters!
AB2 = 122 - 82 = 80 . . . Theorem of Pythagoras
â AB = 80
j 8,9 cm
15.4 Area of rect. ABCD = 8,9 % 18 j 161 cm2
Copyright © The Answer
5.
P
U
x
x
x
6.
Q
A
ˆ = y . . . diagonals of a
DAC
y
x̂ 1 = x̂ 2
= x̂ 3
T
. . . alt. øs ; AD || BC in ||m
A
Let Ŝ = x
ˆ = x
Then PTS
ˆ
â TPQ
= x
But Q̂ = x
ˆ
= x
â RUQ
But PU || TR
. . . alt. øs ; PQ || SR in ||m PQRS
. . . opp. øs of ||m PQRS
. . . øs opp = sides
s
. . . corresponding ø equal
m
. . . opposite sides of || PQRS
23
y2
y1
y3
C
Similarly, y1 = y2 = y3 (= y, say)
. . . øs opp = sides
â TRUP is a parallelogram
E
B
ˆ
ˆ
= RUQ
â TPQ
â PT || UR
x2
x1
x3
R
C
. . . øs opp = sides
x
S
y
. . . ø sum of 
(= x, say)
x
B
y
rhombus bisect the
øs of the rhombus
ˆ
â DCA = y . . . øs opp = sides,
4y - 18º
D
sides of a
rhombus (or alternate angles)
â 4y - 18º + 2y = 180º
â 6y = 198º
â y = 33º
Area of || PQRS = 15 % PT = 90 . . . base % height
â PT = 6 cm
C
. . . sides in proportion
AE = BE
BE
BC
2
2
%BE) â AE = BE = 12 = 8 cm
BC
18
15.3.2
CONVERSE
of Theorem
of Pythag.
m
x
18
15.2 ∆ABE ||| ∆ECB ||| ∆DEC
15.3.1 ∆ABE ||| ∆ECB
â ECF is a straight line . . . conv. of 'øs on a str. line'
D : diagonals bisect one another
3.
. . . ø sum of 
ˆ + x + y = 180º
â BCD
2
90 º- x
90º - x
B
D
x
12
2.
. . . corr. øs ; BC || AD(F) in ||m
ˆ . . . opposite øs of ||m
But  = BCD
EXERCISE 3: Quadrilaterals
1.
E
In ΔAEF : Â + x + y = 180º
â Q̂ g 90º, i.e. ∆PQR is not rt. ød. . .
D
C
Similarly, let Ê = y
17.2 12 = 144 and 6 + 8 = 36 + 64 = 100
E
y
x
ˆ = y
then DCF
14.2 They are perpendicular to each other; i.e. CE ⊥ AD;
15.1
ˆ = x
then BCE
â ∆PQR ||| ∆ABC
x
C
Let F̂ = x
[Note: equal radii]
6
=2
3
but
F
x
y
â The sides are not in proportion
. . . øs opp = sides
D
B
. . . 2 angles and a side, but they don’t correspond
2
x
. . . SSS
16.4 Yes, ∆ADB ≡ ∆CDB
16.5 No
A
4.
. . . 2 sides and an angle, but the angle isn’t included
. . . ext. ø of ∆
13.6 BD
14.1
. . . øøS
Yes, ∆ABC ≡ ∆EDC
ˆ or øs of BEC
x + y = 90º . . . str. AED
and y = 2x
. . . opp. øs of ||m
â 3x = 90º
ˆ = 30º
â x = 30º, i.e. EBC
7.1 3x - 30º = x + 10º . . . opp. øs of a ||m
â 2x = 40º
W
â x = 20º
. . . both pairs opp. sides ||
x + 10º
3x - 30º
Z
Y
X
D
7.2
ˆ = 45º . . . diagonals of a square bisect
BAE
(right) angles of a square
ˆ
â ABE = x - 45º . . . ext. ø of Δ
A
ˆ = 90º . . . ø of square
FAB
E
F 125º
ˆ = ABE
ˆ + FAB
ˆ
. . . ext. ø of Δ
BFD
â 125º = x - 45º + 90º
â x = 80º
8.
E
115º
65º
65º
M
5
C
P
1
â RM = 4 cm
ˆ = 90º ; 3:4:5∆ ; Pyth.
. . . LRM
â NM = 8 cm
. . . diagonals bisect
MP = 5 cm
. . . sides of rhombus
â MS =
C
. . . co-interior øs supplementary
B̂ = 65º
. . . øs opp = sides
24 j 4,9 cm
ˆ = 65º
& CED
. . . str. line AED
A
11.1
x
. . . øs opp = sides
y
D
â AB || DC
. . . co-interior øs supplementary
m
â ABCD is a || . . . 2 pairs opp. sides ||
x
x
C
2x
F
E C
2x + 2y = 180º . . . co-int. øs ; AD || BC in ||m
â x + y = 90º
11.2
x
ˆ = ABF
ˆ =y
BFC
ˆ
ˆ
â CBF = CFB
â BC = CF
. . . alt. øs ; AB || DC
. . . sides opp = øs
B
s
ˆ = x
EAB
ˆ = x
â DCA
. . . ø opp = sides
ˆ = x
CBE
ˆ = 2x
â ACB
ˆ = 2x
â DAC
. . . øs opp = sides
. . . alt. øs ; DC || AB in ||m
. . . ext. ø of Δ
. . . alt. øs ; AD || BC in ||m
ˆ = DAC
ˆ
ˆ + DCA
FDC
= 2x + x
= 3x
Copyright © The Answer Series
. . . ext. ø of Δ
S
Pˆ1
C
P
= Aˆ 2
. . . alt. øs ; AB || DC in ||m
= Aˆ 1 (= x)
. . . given
. . . øs opp = sides
â DP = DA
Similarly, Pˆ3 = Bˆ1 = Bˆ 2 = y
â CP = BC
But BC = DA
. . . opp. sides of ||m
& AB = DC
. . . opp. sides of ||m
= DP + CP
= DA + BC
= 2BC
13.1 In the notes (Theorem 4)
13.2 PQ || SR
x
E
D
2x
B
ˆ = 90º . . . ø sum of 
â APB
F
A
y
y
2
P
. . . ø sum of Δ
ˆ = 65º + 50º + 65º = 180º
â B̂ + DCB
9.
x
y
x 1 2 3y
. . . diagonals bisect
LR = RP = 3 cm
B
y 1
1 2x
D
â MS2 = 52 - 12 = 24 . . . Pythagoras
â AE(D) || (B)FC
ˆ = 50º
â ECD
A
x
ˆ + FCE
ˆ = 115º + 65º = 180º
AEC
â D̂ = 65º
12.
3 R
N
50º
F
L
3
65º
65º
A rhombus ; 2 prs. opp. sides ||
and diagonals intersect
at right øs
x
D
65º
B
B
D
A
10.
11.3
DF = DE - FE
& EC = CF - FE
ˆ = BAE
ˆ (= x)
AED
â DE =
=
=
â DF =
AD
BC
CF
EC
. . . alt. øs ; AB || DC in ||m
. . . øs opp = sides
. . . opp. sides of ||m
. . . proved in 11.2
24
P
Let Ŝ = x; then
1
ˆ
= x . . . øs opp = sides
PRS
â Pˆ2 = x . . . alt. øs ;
PQ || SR
â Q̂ = x . . . øs opp = sides
â In ∆s PSR and PRQ
Q
2
x
x
x
x
S
R
ˆ = 180º - 2x . . . ø sum of  s
Pˆ1 and PRQ
â PS || QR
. . . alt. øs equal
â PQRS is a ||m
. . . 2 prs. opp. sides ||
OR: Could've proved 2 prs. opp. øs equal
OR : Could've proved 2 prs. opp. sides equal
13.3 The diagonals intersect at right angles.
The diagonals bisect the angles of the rhombus.
(& 2 adjacent sides are equal)
P
14.
m
x
m
Q
2.1
x
â S midpoint PR . . .
B
A
â SR = 15 cm
converse of
midpoint theorem
â VR = 1 (15) = 7 1 cm
2
y
y
n
R
S
2.2
s
2m + 2n = 180º . . . co-int. ø ; PQ || SR in ||
â m + n = 90º
5
HJ || GE
15
ˆ = 90º
â PQR
9
Q
2
T
9
R
4.2
i.e. converse Pythag.
ˆ = 90º
â QCR
2.3
ˆ = 90º
â BCD
ST = 1 (24 cm) = 12 cm . . . midpoint
2
theorem
â B̂ = 90º . . . ø sum of 
. . . all angles = 90º
â ABCD is a rectangle
Extend PQ to R
. . . 2 pairs opp. sides ||
ˆ = JHF
ˆ
GDH
. . . corresp. øs ; HJ || DE
Q
R
A
5.
60º
D
Proof :
. . . parallel . . . half of the third side.
APCR is a ||
m
C
. . . diagonals bisect one another
80º
B
â CR = and || PD
â PDCR is a ||
N
M
O
B
â PQ || and = 1 DC
2
C
3.2.
Join MN and PQ
MQRT is a ||m
...
â TR = and || MQ
In ∆ABC : M & N are midpoints of AB & AC
â MN || BC and MN = 1 BC
2
. . . 1 pr. opp. sides = and ||
â PR || and = DC
Q
P
m
. . . PQ =
. . . midpoint thm
â MT || BC
N
Q
â PQ || BC and PQ = 1 BC
. . . midpoint thm
ˆ
ˆ
(2) PNM
= RNT
. . . vertically opp øs
â MN || PQ and MN = PQ
. . . both || and =
ˆ
ˆ
(3) PMN
= NTR
. . . alternate øs ; PM || TR
â ∆PNM ≡ ∆RNT
. . . øøS
Copyright © The Answer
. . . 1 pair of opp. sides = and ||
â PN = NR
25
. . . sum of øs in 
R
. . . midpoint thm.
. . . corresp. øs ; || lines
Similarly, M & N are midpoints of AB & BC
ˆ
â BMN
= 60º
â In ∆ PNM & RNT
. . . proved
â MNQP is a ||m
T
s
(1) PM = RT
1
BC
2
C
In ΔABC : M & T are midpoints of AB and AC
& In ∆OBC : P & Q are midpoints of OB & OC
2
Ĉ = 180º - (80º + 60º)
ˆ = 80º
â AMT
M
â TR = and || PM
N
= 40º
1
PR
2
P
2 pairs
opp. sides ||
T
M
â CR = and || AP
A
. . . see 4.1
2
(â JF = 1 EF too)
2
Join AR, PC and CR
EXERCISE 4: Midpoint Theorem
. . . opp. sides of ||m
EJ = GH
â JF = GH
P
such that PQ = QR
â GHJE is a ||m
= 1 EF
A
3.1 Construction :
â The 4th angle, D̂ = 90º . . . ø sum of quad.
4.3
OR : Pythag
9 : 12 : 15
= 3:4:5 !
So, too, n + y = 90º
1.2
â GH || EJ(F) [and GH = 1 EF]
. . . ratio of sides = Pythag. 'triple'
Similarly, x + y = 90º
F
. . . given
V
â sides of ∆PQR :
ˆ
â BAD
= 90º . . . vertically opp øs
J
& In ∆DEF : G & H are midpoints of DE & DF
S
2
18 : 24 : 30 = 3 : 4 : 5
H
E
P
2
PQ = 16 % 15 = 24 cm
m
ˆ
â PAS
= 90º . . . ø sum of 
1.1
G
Similarly : In ∆RQS : V midpoint SR
C
D
n
D
4.1
In ∆RQP : T midpoint QR & TS || QP
. . . corresp. øs ; MN || AC
ˆ
= 180º - (80º + 60º) . . . øs on a str. line
â TMN
= 40º
The same method can be followed to determine the
other two angles of ΔMNT.
Answer : 40º ; 80º ; 60º
6.1
. . . diagonals bisect
In ∆AEC : O midpoint AC
A
We have DO || CE
& DO = OB
. . . in 6.1
diagonals of
sq. bisect
. . . midpoint. theorem
P
C
Q
R
W
These Geometry materials (Booklets 1 to 4)
were created and produced by
The Answer Series Educational Publishers (Pty) (Ltd)
to support the teaching and learning of
Geometry in high schools in South Africa.
. . . right øs of sq. bisected by diagonal
ˆ
ˆ
= BOA
ECA
= 90º
. . . diagonals of sq. int. at rt. øs
. . . just as Dˆ above
1
. . . converse of midpoint thm
â FE = EC
& In ∆EDA : O midpoint AD and (B)OC || DE
. . . converse of midpoint thm
â EC = CA
â FE = EC = CA
In ∆ BCA & AED
F
E
O
C
x
This material may not be sold (via any channel)
or used for profit-making of any kind.
â PQ || BE(C)
Similarly, in ΔDEC :
D
A
parallel
to BEC
B
A
R
Q
P
E
s
9.2 In Δ ABE, AED and DEC :
ˆ
ˆ
(1) CBA
= EAD
. . . both = x
ˆ
ˆ
= AED
(2) BCA
. . . corresp. øs ; DE || BC
(3) CA = FE
= ED
. . . in 7.1
. . . øs opp = sides
= 1 (AD + BE + EC)
2
â ∆BCA ≡ AED
. . . øøS
= 1 (AD + BC)
Copyright © The Answer Series
They are freely available to anyone
who wishes to use them.
P & Q are midpoints of AB & AE
â PQ || RS
. . . both are
x
D
y
This drawing looks confusing at first. But, look at each
triangle separately – the 'middle' one is just upside down!
– and apply the facts to each, one at a time.
RS || (B)EC
B
y
9.
9.1 In ΔABE :
In ∆FBC : D midpoint FB & DE || BC
s
S
. . . corresp. øs ; MO || EC in 6.1
ˆ = 90º + 45º = 135º
â ECD
7.2
V
E
. . . 1 pair of opp. sides = and ||
ˆ = 45º
& OCD
7.1
2
= 1 ( 1 QR)
2 2
= 1 QR
4
T
= EC
theorem
â WR = 1 TV
M
B
. . . midpoint
â In ∆STV : R midpoint VS and WR || TV
1
O
. . . M midpt. BO
6.3.1 Dˆ 1 = 45º
2
D
= 2OM
â DOEC is a ||m
6.3.2
...
In ∆PQR : T & V are midpoints of PQ & PR
â TV || QR and = 1 QR
. . . midpoint theorem
â MO(D) || EC
6.2
8.
. . . given
& M midpt. AE
PQ + QR + RS = 1 BE + 1 AD + 1 EC
2
2
2
26
2
S
C
TAS FET EUCLIDEAN GEOMETRY COURSE
BOOKLET 3
Circle Geometry
by The TAS Maths Team
Circle Geometry
Geometry FET
Course Booklets Set
WWW.THEANSWER.CO. ZA
"SUBTEND" . . .
CIRCLE GEOMETRY
P
P
O
O
Understand the word!
The Language (Vocabulary)
GROUP 1 AND
Centre
B
Figure 1
A
A
B
Figure 2
O
A
O
B
180º
Figure 3
Figure 4
radius
diameter
Central and Inscribed angles
B
O
centre
A
or chord AB, subtends:
In all the figures, arc AB (AB),
Circumference
ˆ
a central AOB
at the centre of the circle, and
ˆ
an inscribed APB
at the circumference of the circle.
Chords
B
C
Diameter
Radius
2
A
P
P
B
chord
Consider that
subtend
means support.
major arc AB
A
To ensure that you grasp the meaning of the word 'subtend' :
B
Arcs (major & minor)
Place your index fingers on A & B ;
A
A
sector
Copyright © The Answer Series
move along the radii to meet at O and back ; then,
minor arc AB
move to meet at P on the circumference and back.
major
segment
A
O
O
Turn your book upside down and sideways.
You need to recognise different views of these situations.
Segments (major & minor)
Sectors
P
Take each of the figures:
chord AB
minor
segment
Take note of whether the angles are acute, obtuse, right, straight or reflex.
Redraw figures 1 to 4 leaving out the chord AB completely and observe
the arc subtending the central and inscribed angles in each case.
B
B
1
GROUP
GROUP
3
Cyclic Quadrilaterals
Tangents
A cyclic quadrilateral is a quadrilateral which has all 4 vertices on the circumference
of a circle.
Special lines
D
4
A tangent is a line which touches a circle at a point.
Points A, B, C and D
are concyclic,
i.e. they lie on the same circle.
O
A
tangent
point of contact
C
B
Note: Quadrilateral AOCB is not a cyclic quadrilateral because point O
is not on the circumference ! (A, O, C and B are not concyclic)
A secant is a line which cuts a circle (in two points).
We name quadrilaterals by going around, either way, using
consecutive vertices, i.e. ABCD or ADCB, not ADBC.
secant
Exterior angles of polygons
The exterior angle of any polygon is an angle which is formed between one side of
the polygon and another side produced.
e.g. A triangle
A
E
D
B
A
e.g. A quadrilateral /
cyclic quadrilateral
C
D
B
NB : It is assumed that the
tangent is perpendicular
C
ˆ is an exterior ø of ΔABC.
ACD
[NB : BCD is a straight line ! ]
Copyright © The Answer Series
to the radius (or diameter)
ˆ is an exterior ø of c.q. ABCD.
ADE
at the point of contact.
[NB : CDE is a straight line ! ]
2
SUMMARY OF CIRCLE GEOMETRY THEOREMS
I
II
III
x
The
'Centre'
group
The
' No Centre '
group
The
' Cyclic Quad.'
group
2x
2x
x
x
Equal
chords !
x
There are ' 3 ways to
prove that a quad. is
a cyclic quad '.
y
x + y = 180º
IV
The
'Tangent'
group
Equal
tangents !
There are ' 2 ways to prove that
a line is a tangent to a ?'.
Copyright © The Answer Series
3
Equal
radii !
GROUPING OF CIRCLE GEOMETRY THEOREMS
The grey arrows indicate how various theorems are used to prove subsequent ones
I
II
III
x
The
'Centre'
group
The
' No Centre '
group
The
' Cyclic Quad.'
group
2x
2x
x
Equal
chords !
x
Equal chords subtend equal angles
and, vice versa,
equal angles are subtended
by equal chords.
x
There are ' 3 ways to
prove that a quad. is
a cyclic quad '.
y
x + y = 180º
IV
The
'Tangent'
group
Equal
tangents !
There are ' 2 ways to prove that
a line is a tangent to a ?'.
Copyright © The Answer Series
Equal
radii !
4
CIRCLE GEOMETRY THEOREMS
PAPER 2: GEOMETRY
3
Circle Geometry Theorems
1
Given :
ˆ at the centre and APB
ˆ at the circumference.
?O, arc AB subtending AOB
P
Construction : Join PO and produce it to Q.
Given : ?O with OP ⊥ AB
Proof :
To prove : AP = PB
Let
Then
Construction : Join OA and OB
O
In Δ OPA & OPB
. . . radii
(2) Pˆ 1 = Pˆ 2 (= 90º)
A
. . . given
1 2
P
Pˆ 1 =
x
 =
x
12
x y
. . . øs opposite equal radii
â Oˆ 1 = 2x
s
(1) OA = OB
ˆ = 2APB
ˆ
AOB
To prove :
The line segment drawn from the centre of a circle,
perpendicular to a chord, bisects the chord.
Proof :
The angle which an arc of a circle subtends at the centre is
double the angle it subtends at any point on the circumference.
O
1 2
x
y
A
. . . exterior ø of ΔAOP
Q
B
Similarly, if Pˆ 2 = y, then Oˆ 2 = 2y
B
ˆ
= 2 x + 2y
â AOB
= 2( x + y)
(3) OP is common
ˆ = 2 APB
. . . RHS
â ΔOAP ≡ ΔOBP
â AP = PB, i.e. OP bisects chord AB 2
The line drawn from the centre of a circle that
bisects a chord is perpendicular to the chord.
...
4
This proof has
been added in the
2021 Exam Guidelines.
Given : ?O and cyclic quadrilateral ABCD
Given : ?O with AP = PB
To prove :
To prove : OP ⊥ AB
(1) OA = OB
(2) AP = PB
O
. . . radii
A
. . . given
1 2
P
Proof :
â ΔOAP ≡ ΔOBP
B
O
ˆ = 2x
O
1
B
...
ø at centre = 2 %
ø at circumference
2
â Â + Ĉ = x + 180º – x = 180º
s
. . . ø on a straight line
& â B̂ + D̂ = 180º â Pˆ 1 = Pˆ 2 = 90º
i.e. OP  AB i
C
. . . øs about point O
â Ĉ = 1 (360º – 2x) = 180º – x
. . . SSS
D
1
= x
â Oˆ 2 = 360º – 2x
Pˆ 1 = Pˆ 2
Pˆ 1 + Pˆ 2 = 180º
Let Â
Then
(3) OP is common
Copyright © The Answer Series
x
B̂ + D̂ = 180º
2
In Δs OPA & OPB
But,
 + Ĉ = 180º &
A
Construction : Join BO and DO.
Construction : Join OA and OB
Proof :
The opposite angles of a cyclic quadrilateral are supplementary.
...
...
ø at centre = 2 %
ø at circumference
sum of the øs of a
quadrilateral = 360º
5
The angle between a tangent to a circle and a
chord drawn from the point of contact is equal to the
angle subtended by the chord in the alternate segment.
These proofs are
logical & easy to follow.
Method 2
Method 1
We use 2 'previous' facts involving right øs
Draw radii and use
'ø at centre' theorem.
Given:


ø in semi-? = 90º
. . . so, draw a diameter!
. . . so, join RK!
?O with tangent at N and
chord NM subtending K̂
at the circumference.
Given: ?O with tangent at N and
M
K
RTP:
tangent ⊥ diameter
ˆ
MNQ
= K̂
O
chord NM subtending K̂
at the circumference.
Q
x
ˆ
= 90º
ONQ
. . . radius ⊥ tangent
ˆ
∴ MON
= 2x
∴ K̂ = x
ˆ
RNQ
= 90º
Proof:
ˆ
& RKN
= 90º
ˆ
∴ ONM
= 90º – x
ˆ
∴ OMN
= 90º – x
. . . øs opposite equal radii
. . . ø at centre = 2 % ø at circumference
Let
. . . tangent ⊥ diameter
. . . ø in semi-?
ˆ
MNQ
= x
ˆ
â RNM
= 90º – x
ˆ
â RKM
= 90º – x
ˆ
â MKN
= x
ˆ
ˆ â MNQ
= MKN
ii
Q
x
Then . . .
. . . sum of øs in Δ
ˆ
â MNQ
= K̂ Copyright © The Answer Series
O
N
Construction: diameter NR; join RK
P
ˆ
= x
Proof: Let MNQ
K
M
ˆ
ˆ
MNQ
= MKN
RTP:
N
Construction: radii OM and ON
R
. . . øs in same segment
P
PROVING THEOREMS
x y
x y
y
x
2x
Copyright © The Answer Series
6
2y
x
y
2x 2y
2x
2y
y
x
? THEOREM PROOFS: A Visual presentation
The Situation
Construction
Examinable
The LOGIC . . .
radii
1
a
perpendicular
line
centre
chord
Theorem
Statement
The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord.
radii
2
midpoint
of chord
centre
chord
Theorem
Statement
The line drawn from the centre of a circle that bisects a chord is perpendicular to the chord.
Note:
The bolded words in the statements are the approved 'reasons' to use.
Copyright © The Answer Series
The Situation
The LOGIC . . .
Construction
3
radius produced
inscribed ∠
xy
x y
centre
central ∠
x
arc
x 2x 2y
equal base ∠ s
Theorem
Statement
4
y
y
exterior ∠ s of Δs
The angle which an arc of a circle subtends at the centre is
double the angle it subtends at any point on the circumference.
radii
y
x
opposite ∠ s in a
cyclic quadrilateral
Theorem
Statement
Copyright © The Answer Series
y
x
2y
2x
∠ at centre = 2 % ∠ at circumference
& ∠ s about a point = 360º.
The opposite angles of a cyclic quadrilateral are supplementary.
? THEOREM PROOFS: A Visual presentation, continued
The Situation
5
inscribed ∠
Examinable
The LOGIC . . .
Construction
Method 1:
radii
90° – x
chord
x
x
x
s
base ∠ of isosceles Δ
radius ⊥ tangent
tangent
90° – x
x
2x
2x
x
x
sum of ∠ s of Δ
∠ at centre = 2 % ∠ at circumference
Method 2:
diameter & join . . .
This theorem is known
as the
90° –
x
Tan Chord theorem
x
x
diameter ⊥ tangent
& ∠ in semi-?
Theorem
Statement
Copyright © The Answer Series
x
90° –
x
x
∠ s in the
same segment
The angle between a tangent to a circle and a chord drawn from the point of
contact is equal to the angle subtended by the chord in the alternate segment.
ix
x
FURTHER ? THEOREM PROOFS: A Visual presentation
The Situation
6
Theorem
Statements
The LOGIC . . .
?
The ∠ in a semi-?
is a right ∠.
180º
diameter
Central angle is
a straight angle
7
Construction: radii
∠ s subtended
by an arc
Inscribed angle is
a right angle
The LOGIC . . .
x
x
(at the
circumference
of the circle)
arc
8
2x
2x
∠ subtended by an arc (or chord)
(at the centre of the circle)
∠ s subtended by an arc (or chord)
(at the circumference)
x
cyclic
quadrilateral
Copyright © The Answer Series
∠ s subtended by
the same arc
are equal.
x
The exterior ∠ of a
cyclic quadrilateral
exterior
angle
y
y
Opposite ∠ s :
x + y = 180º
Adjacent ∠ s :
x
x + y = 180º
= the interior opposite ∠.
FURTHER ? THEOREM PROOFS: A Visual presentation, continued . . .
The Situation
The LOGIC . . .
Construction
Method 1:
radii
Theorem
Statement
9
radii ⊥ tangents
congruent Δs
Method 2:
chord and inscribed ∠
x
x
x
x
x
x
tan-chord theorem 4
Proofs 6
to 9
sides opposite
equal base ∠ s in Δ
are not examinable, but,
the LOGIC is crucial when studying geometry.
Copyright © The Answer Series
xi
x
Tangents to a circle
from a common point
are equal.
Again, the
bolded words
are the ‘approved
reasons' to use.
EXAMPLE 1
PROBLEM-SOLVING: An Active Approach
This is an excellent example testing so many
?-geometry theorems (excluding tangents).
ACT !
A
Be Active . . .
Mark all the information on the drawing :
equal or parallel sides equal øs right øs
K
Calculate, giving reasons,
the sizes of :
Use all the Clues . . .
ˆ
(c) M
4
ˆ +N
ˆ
(d) N
1
2
parallel lines alternate, corresponding & co-interior øs
ˆ
(e) M
1
(f) Prove that KG = GM
There are
3 theorems
in ? geometry
which involve
right angles.
F
76º
K
1
O
2
1
3 2
H
G
4
ˆ
ˆ = 1O
(c) M
4
1
12
N
2
ˆ +N
ˆ = 180º - 76º
(d) N
1
2
= 104º sum of the interior øs of a quadrilateral = 360º ;
types & properties of all quadrilaterals
. . . opp. øs of c.q. are suppl.
ˆ = 90º
(e) KML
. . . ø in semi-?
ˆ = 38º in 4.3
ˆ &M
ˆ
â M 1 = 180º - (90º + 38º) . . . str. NMP
4
= 52º Recall the Theory systematically . . .
Previous geometry
Revise previous geometry under the headings
angles lines triangles quadrilaterals
(f)
ˆ = corresponding KML
ˆ = 90º
G
3
. . . ON || LM
i.e. line from centre, OG ⊥ chord KM
Circle geometry
â KG = GM Recall and apply the facts by group.
11
L
1 2
3 2 1
= 38º . . . ø at centre = 2 % ø at circumference
isosceles Δs ; congruency, similarity, theorem of Pythagoras
Copyright © The Answer Series
M
N
ˆ = corr. Lˆ . . . ON || LM
(b) O
1
1
= 76º Δs sum of the interior øs = 180º ; exterior ø of Δ ; equilateral Δs ;
s
1 2
(a) Lˆ 1 ( = Fˆ )
= 76º . . . øS in same segment
diameter ø in semi-? = 90º ; diameter ⊥ tangent
3 2 1
Answers
equal chords equal øs
90º angle the Theorem of Pythagoras, or . . .
P
1 2
4
intersecting lines adj. suppl. & vert. opp. øs ; øs about a point.
L
H
1
3 2
G
ˆ
(b) O
1
equal radii or tangents (in triangles) equal base øs
O
2
1
(a) Lˆ 1
The information provides clues to facts :
T
76º
ON || LM and F̂ = 76º.
radii diameter tangents
C
F
O is the centre of the circle
and diameter KL is produced
to meet NM produced at P.
M
P
EXAMPLE 2
3 ways to prove that a quadrilateral is cyclic
(a) Complete the following by writing the appropriate missing word.
If a chord of a circle subtends a right angle on the circumference,
then this chord is a . . . . . .
We use the 3 converse statements for cyclic quadrilaterals.
M
B
(b) A, B, C and D are points on the circle.
1
converse theorem
2
Prove that one side subtends
equal angles at the two other points
(on the same side).
2
BC produced and AD produced
meet at N.
C
A
1
AB produced and DC produced
meet at M.
1
D
x1
z1
y2
z2
p2
B
converse theorem
2
If M̂ = N̂, prove that AC is a diameter of the circle.
Prove that a pair of
opposite angles
is supplementary.
N
or B̂ + D̂ = 180º
D
(b) Let M̂ = N̂ = x
. . . given
ˆ = C
ˆ = y . . . vert. opp. øs
& C
1
2
1
x+y
Then :
Bˆ 1 = x + y
. . . ext. ø of ΔBMC
ˆ = x+y
& D
1
. . . ext. ø of ΔDCN
ˆ = 180º
But, Bˆ 1 + D
1
â 2(x + y) = 180º
â x + y = 90º
ˆ
ˆ = 90º
i.e. B = D
x
B
1
D
1̂ = 2̂
8 3
or 3̂ = 4̂
or 5̂ = 6̂
or 7̂ = 8̂
2 ways to prove that a line is a tangent to a ?
2
x
N
We use the 2 converse theorem statements to prove that a line is a tangent to a ? if . . .
converse theorem
1
. . . it is perpendicular to a radius
at a point where the radius meets
the circumference.
â AC is a diameter . . . subtends a right ø
ACT!
O
Prove that OP ⊥ AB;
then AB will be a tangent
A
P
B
converse theorem
A: Be Active
. . . when drawn through
the end point of a chord,
it makes with the chord
an angle equal to an angle
A
in the alternate segment.
C: Use all your Clues
T: Apply the Theory systematically, recalling
each group, and fact, one at a time.
Copyright © The Answer Series
Either prove :
4
6
1
y
2
C
y1
x+y
. . . opp. ø of c.q.
7
C
A
s
2
5
2
Prove that an exterior angle
of a quadrilateral is equal
to the interior opposite angle.
M
Either prove :
 + Ĉ = 180º
converse theorem
(a) . . . diameter.
Either prove :
x1 = x2 or y1 = y2
or z1 = z2 or p1 = p2
A
Answers
1
p1 x
2
y1
12
y
x
B
Prove that
x = y;
then AB
will be
a tangent
Note :
It is a good idea to
draw a faint circle
(as shown) to see
this converse
theorem clearly.
Gr 12 Maths National November 2018: Paper 2
2018
QUESTION 10
QUESTION 9
EUCLIDEAN GEOMETRY
Give reasons for your statements in QUESTIONS 8, 9 and 10
QUESTION 8
In the diagram, ABCD is a cyclic quadrilateral such that
AC  CB and DC = CB. AD is produced to M such that
AM  MC.
M
Prove the theorem
which states that
Jˆ + Lˆ = 180º.
8.1 PON is a diameter of the circle centred at O.
TM is a tangent to the circle at M, a point on the circle.
R is another point on the circle such that OR || PM.
ˆ = 66º.
NR and MN are drawn. Let M
J
9.1 In the diagram, JKLM is
a cyclic quadrilateral and
the circle has centre O.
O
(5)
1
A
K
L
T
M
2
1 2
R
1 66º
9.2 In the diagram, a smaller circle ABTS and a bigger circle
BDRT are given. BT is a common chord.
Straight lines STD and ATR are drawn.
Chords AS and DR are produced to meet in C, a point
outside the two circles.
BS and BD are drawn.
ˆ = y.
 = x and R
2
1
1
Let B̂ = x.
N
2
O
P
D
1
1
x
2
M
2
B
1
C
Calculate, with reasons, the size of EACH of the following
angles :
ˆ
8.1.2 M
(2)(2)
8.1.1 P̂
8.1.3
ˆ
N
1
8.1.5
ˆ
N
2
8.1.4
ˆ
O
2
S
(1)(2)
1
2
3
(3)
B
D
(5)
10.1.2 ∆ACB ||| ∆CMD
(3)
CM 2
AM
=
AB
DC 2
(6)
10.2.2 AM = sin2x
(2)
10.2.1
4
2
1
R
C
1
2
AB
[16]
D
TOTAL : 150
9.2.1 Name, giving a reason, another angle equal to :
C
F
10.1.1 MC is a tangent to the circle at C.
10.2 Hence, or otherwise, prove that:
2 3
y
M
20
1
T
8.2 In the diagram, ∆AGH is drawn. F and C are points on
AG and AH respectively such that AF = 20 units,
FG = 15 units and CH = 21 units. D is a point on FC
such that ABCD is a rectangle with AB also parallel to GH.
The diagonals of ABCD intersect at M, a point on AH.
A
1 2
x
2
10.1 Prove that :
B
A
(a) x
21
15
H
8.2.2 Calculate, with reasons, the length of DM.
(2)(2)
9.2.2 Prove that SCDB is a cyclic quadrilateral.
G
8.2.1 Explain why FC || GH.
(b) y
(1)
(3)
ˆ = 100º.
9.2.3 It is further given that Dˆ 2 = 30º and AST
Prove that SD is not a diameter of circle BDS.
(5)
[16]
Q26
(4)
[16]
Copyright © The Answer Series: Photocopying of this material is illegal
EXAMPLE 4
EXAMPLE 5
P
Prove that PA
is a tangent to ?M.
Don't be put off by this drawing !
Direct your focus to one situation at a time 
8
T
12
M
3
A
A
1 2
4
B
Q
Answers
1
1
1 2
X
2
1
B
2
2
In right-ød MAQ :
Q
Y
AM = 5 units
. . . TM = AM = radii
â TM = 5 units
3
P
. . . 3:4:5  ; Pythag.
â PM = 13 units
P
8
12
M
3
â PAM is a 5 : 12 : 13  !
T
T
A
4
Q
i.e. PM 2 = PA2 + AM 2
Make statements, with reasons,
ˆ = 90º
â PAM
1. In ?XPBA : about P̂1 and B̂1
. . . converse of Pythag.
â PA is a tangent to ? M . . . converse tan chord theorem
ˆ
2. In ?ABYQ : about B̂1 and AQY
ˆ
3. In quadrilateral APTQ : about P̂1 and AQT
4. What can you conclude about quadrilateral APTQ?
Mark the øs on the drawing as you proceed.
See Gr 11 Maths 3 in 1
Study Guide
Answers
1. In c.q. XPBA : P̂1 = B̂1
. . . arc XA subtends øs in same segment
ˆ
2. In c.q. ABYQ : B̂1 = AQY
. . . exterior ø of cyclic quad.
ˆ
3. In quad. APTQ : P̂1 = AQT
4.  APTQ is a cyclic quad.
Copyright  The Answer Series
. . . both = Bˆ1 above
Module 9b: Circle Geometry
Notes
. . . converse of exterior ø of cyclic quad.
15
Exercises
Full Solutions
B
Answers
EXAMPLE 7
AB is a tangent to the circle at B and BD is a chord.
B̂1 = x
(a) Let
AD cuts the circle in E.
C is a point on BD so that ABCE is a cyclic quadrilateral.
AC, BE and CE are joined.
then D̂ = x
. . . tan chord theorem
and Ĉ2 = x
. . . øs in same segment
1 2
y
Let B̂2 = y
&
A
A
then
 2 = y
. . . øs in same segment
1 2
Ĉ3 = D̂ + Â 2
B
B
2
1
2
3
E
ˆ = x+y
& ABC
3 21
C
 AB = AC
. . . B̂1 = x and
. . . sides opp = øs
. . . both = x in (a)
. . . converse tan chord thm
 AC is a tangent to ?ECD at C.
Prove that :
(a) AB = AC
(b) AC is a tangent to circle ECD at point C.
Copyright  The Answer Series
B̂2 = y
D
(b) Ĉ2 = D̂
16
1
x
2
y
x
3 21
C
. . . ext. ø of Δ
= x+y
1
1
2
3
E
x
D
GRADES 8 - 12
ALL MAJOR SUBJECTS IN
ENGLISH & AFRIKAANS
S
A
WWW.THEANSWER.CO. ZA
T
TAS FET EUCLIDEAN GEOMETRY COURSE
BOOKLET 4
Proportionality, Similarity
& The Theorem of Pythagoras
by The TAS Maths Team
Proportionality,
Similarity &
The Theorem of
Pythagoras
Geometry FET
Course Booklets Set
WWW.THEANSWER.CO. ZA
PROPORTIONALITY, SIMILARITY &
THE THEOREM OF PYTHAGORAS
Proportion
PROPORTIONALITY
When two ratios are equal, e.g. a = c ,
Ratio
b
Example 1 : Determine the ratio BC : AB
A
B
3 cm
Answer
BC
= 3 = 1 AB
2
6
AB
1
& sin A =
2
²
 = 30º
B
b
1
Q
P
b
d
d
C
A
The lengths of a, b, c and d, could be as follows :
a = 4 units ; b = 2 units ; c = 6 units ; d = 3 units
3
60º
c
B
30º
2
a
i.e. If PQ || BC, then a = c .
A
In Trigonometry, we have the sin, cos and tan ratios.
e.g. BC = sin A
If a line, PQ, is drawn parallel to one side of a triangle,
it divides the other two sides proportionally.
Both BC and AB
are lengths,
measured in
the same unit.
Ratios can also be written as fractions :
C
A
The Proportion theorem states :
The ratio BC : AB = 3 : 6 = 1 : 2 6 cm
d
we say that : a, b, c and d are in proportion, or,
that a and b are in the same proportion as c and d.
We use Ratio to compare two quantities
of the same kind (in the same unit).
Then :
a
= 4 = 2
2
b
c
= 6 = 2
3
d
and
b
6
Q
P
2
â a = c
C
4
B
d
C
Example 2 : Divide a 30 cm line in the ratio 2 : 3.
Note : A proportion can be written in many ways :
Answer
30 cm
A
B
â 5k = 30 cm
â k = 6 cm
3 parts
2 parts
A
B
Alternatively,
let AP = 2k and PB = 3k;
then AB = 5k
P
If
4
= 6,
2
3
then
2
= 3 (invertendo)
4
6
AP 2
=
AP = 2 PB,
3
PB 3
Copyright © The Answer Series
but,
4
= 2 (alternando).
6
3
These forms are all equivalent and imply that : 2 % 6 = 4 % 3.
â AP = 12 cm and PB = 18 cm Note :
or
To summarise:
a
c
b
d
²
=
=
a
c
b
d
AP 2
2
=
AP = AB
AB 5
5
1
or
a
b
=
c
d
or
3
ad = bc
Worked Example 3
THE PROPORTION THEOREM
The Theorem Statement:
i.e. PQ || BC ²
2
In the figure, RT | | AB, RS | | AC and DR = .
A
A line parallel to one side of a triangle
divides the other two sides proportionally.
P
R
5
RA
D
3.1 Write down the values of the following ratios :
Q
B
AP AQ
=
PB QC
A
(b) DS =
(a) DT =
TB
C
T
B
DC
S
3.2 Prove that TS || BC
The Converse Theorem Statement:
Answers
If a line divides two sides of a triangle
proportionally, the line is parallel to the third side.
i.e.
P
AP
AQ
=
² PQ || BC
PB
QC
B
TB
C
(b)
Worked Example 2
Find x :
Find x :
a
x
3
RA
DS
DR
2
=
=
DC
DA
7
3a
Answer
Answer
Because of the || lines, and the
proportion theorem above, we know
that the ratio x : 2 will equal the ratio 6 : 3
Because of the || lines, and the
proportion theorem above, we know that
the ratio x : 5 will equal the ratio 3a : a (= 3 : 1)
â By inspection : x = 4 units
â By inspection : x = 15 units
% 2)
x = 6
2
3
â x = 2%2
â x = 4 units or, by calculation :
% 5)
In ΔDBA:
RT || AB ; prop. theorem
. . . In ΔDCA:
RS || AC ; prop. theorem
RA
5
5
. . . converse of proportion theorem
Remember :
x = 3a
A
a
â x = 5%3
AB : BC = 2 : 5, while AB : AC = 2 : 7
2
i.e. AB =
B
â x = 15 units AC
C
Note : When applying these statements, focus on one triangle
at a time, and apply either one fact or the other.
Copyright © The Answer Series
...
. . . both = DR = 2
DS
DT
=
TB
SC
â TS || BC or, by calculation :
5
3.2 In ΔDBC :
5
6
2
2
3.1 (a) DT = DR =
Q
Worked Example 1
x
C
A
2
â AB =
7
2
AC
7
Question
Solution
Question
Worked Example 5
P
In PQR the lengths of PS, SQ, PT and TR are
3, 9, 2 and 6 units respectively.
S
5.1 Give a reason why ST || QR.
B
5.2 If AB || QP and RA : AQ = 1 : 3,
calculate the length of TB.
Solution
T
Q
R
A
Answers
5.1 In PQR :
PS
3
1
=
=
SQ
9
3
&
PT
2
1
=
=
TR
6
3
P
3
PS
PT
=

SQ
TR
2
S
T
9
â ST || QR . . . converse of proportion thm
Q
5.2 In RPQ :
RA
RB
=
= 1
4
RP
RQ
â RB =
1
RP
4
= 2 units
â TB = 4 units Copyright © The Answer Series
4
B
3 parts
A1 part R
. . . proportion theorem ; AB || QP
RA:AQ = 1:3
. . . RP = PT + TR = 8 units
6
Proving the Proportion theorem
THE PROOF OF THE PROPORTION THEOREM
Be sure to revise the following two concepts involving areas of triangles.
These concepts are used in the proof of the proportion theorem which follows.
Given : ΔABC with DE || BC, D & E on AB & AC respectively.
A
IMPORTANT CONCEPTS REQUIRED
To prove :
Δs on the same base
AD
= AE
DB
EC
Construction : Join DC & BE
and between the same || lines
have equal areas.
Proof :
A
ΔABC = ΔDBC in area
B
These Δs have the same base, BC,
and the same height (since they lie
between the same || lines).
D
C
Area of ΔADE
=
Area of ΔDBE
Similarly:
1
2 AD.h
1
2 DB.h
D
B
= AD
Area of ΔADE
= AE
Area of ΔEDC
EC
â Area of ΔADE = Area of ΔADE
the ratio of their bases.
Area of ΔDBE
A
h
B
x
C
Copyright © The Answer Series
y
D
C
1
′
2 AE.h
1
′
2 EC .h
and: ΔADE is common
=
E
DB
the ratio of their areas equals
Area of ΔABC
=
Area of ΔACD
h
But: ΔDBE = ΔEDC . . . on the same base DE ; between || lines, DE & BC
When Δs have the same height,
1
x .h
2
1
y.h
2
h′
Area of ΔEDC
â AD = AE DB
x
y
These  s have a common vertex, A,
and therefore the same height.
5
EC
PROPORTION THEOREM PROOF: A Visual presentation
The Situation
a
Heights of Δs
The Construction
base a
c
base c
h'
h
d
b
height (H)
Create 2 Δs
Parallel lines in a Δ
These 2 Δs have
common height h'
These 2 Δs have
common height h
These 2 Δs have
c
a
h
c
a
h'
Bases
Base
d
b
equal areas
(same base &
same height H)
BASE
b
d
H
The
same
=
1
2 ah
1
2 bh
=
a
b
=

Copyright © The Answer Series
a
c
=
b
d
1
2 ch
1
2 dh
'
c
=
d
'
But :
=
The Theorem Statement:
A line drawn parallel to one side of a triangle
divides the other two sides proportionally.
6
2.4
EXERCISE 1
PROPORTION THEOREM AND APPLICATIONS
A
x cm
1. In the figure A and B are points on PQ and PR
such that AB || QR.
E
D
Answers on page 17
Determine the length of AD,
i.e. find x.
3 cm
B
P
In ΔABC, DE || BC and AB = 6 cm.
AD = x cm, AE = 5 cm and EC = 3 cm.
5 cm
C
AR and BQ are drawn.
Answer the following questions, which refer to
a theorem. You need to redraw the sketch.
1.1 Complete :
area ΔAPB
= ....
area ΔAQB
....
1.2 Complete:
area ΔAPB
= ....
area ΔABR
....
A
Q
R
(a) AG
GH
2.1 Complete the following theorem by writing down the missing word(s) only.
"A line parallel to one side of a triangle divides the two other sides . . ."
2.2 In the accompanying figure, MS || QR.
x
2.3 The following measurements
are known in the given sketch :
R
E
AD = 3x - 1
W
AE = 3
E
B
C
Determine the integral value(s) of x for which DE || BC.
7
R
S
X
D
H
Z
If S is a point on diagonal XZ such that
SR || XW, prove that ST || XY.
A
BD = 7x - 6
G
DE
R is a point on WZ and T is a point
on YZ such that RT is parallel to
diagonal WY.
3
Q
D
(b) AD
3.2 Quadrilateral WXYZ is given.
S
2
C
If a line is parallel to one side of a triangle, then the line divides the other
two sides in proportion, and conversely if . . . . . . . , then . . . . . . .
x+2
M
Determine, without giving reasons,
the value of x.
F
3.1 Complete the following theorem statement :
P
Furthermore, PM = x cm,
MQ = 2 cm, PS = (x + 2) cm
and SR = 3 cm.
1
2.5.2 Determine the length of DE.
1.5 Give the wording of the theorem which is under consideration here.
1
3
2.5.1 Write down the values of the following ratios :
1.4 What can you deduce from 1.1, 1.2 and 1.3?
Copyright © The Answer Series
B
Furthermore, AB = 1 cm, BC = 3 cm and CD = 1 cm.
1.3 What can you say about the area of ΔAQB and the area of ΔABR, and why?
CE = x + 2
A
2.5 In the accompanying figure, BF || CG || DH
and DG || EH.
B
T
Y
4.
In the diagram below HJKL is a parallelogram, with the diagonals
intersecting at M.
H
N
SIMILARITY
L
When polygons are similar, they have the same shape, but NOT necessarily
F
the same size. One figure is an enlargement or reduction of the other.
M
J
K
S
The definition of similarity
ˆ = 90º.
JHK
The conditions for polygons to be similar are :
JK is produced to S. N is a point on HL. NS intersects JL at F.
A : the polygons must be equiangular,
HJ = 6 units ; HK = 8 units ; KS = 5 units ; FL = 13 units
4.1
B : their corresponding sides must be in proportion.
Determine, with reasons, the following ratios in simplified form :
(a) JK : KS
We will show that, for triangles :
(b) JM : MF
4.2
If A holds, then B holds â similar
Hence, prove that HK || NS
i.e.
A
5.
AND
In the accompanying figure, DF || BC
 = P̂ , B̂ = Q̂ and Ĉ = R̂ ,
F
and AF = FC .
FE EB
A
P
then AB = BC = AC .
PR
PQ
QR
C
D
Prove that ADEF is a trapezium.
If, in ΔABC & ΔPQR,
â ΔABC ||| ΔPQR
E
B
x
C
Q
x
R
and, conversely . . .
B
If B holds, then A holds â similar
ACT!
If, in ΔABC & ΔPQR,
C: Use all your Clues
P
 = P̂ , B̂ = Q̂ and Ĉ = R̂.
T: Apply the Theory systematically
Copyright © The Answer Series
A
AB
BC
=
= AC , then
PQ
QR
PR
A: Be Active
â ΔABC ||| ΔPQR
8
B
C
Q
R
Worked Example 1
Worked Example 2
D
A
In the figure, BA || ED, AB = 8, AC = 4 and DE = 10 units.
1.1 Name a pair of similar triangles (in the correct order).
Explain why they are similar.
1.2 Calculate the length of EC.
8
Say, with reasons, whether the following pairs of triangles are similar or not.
4
C
B
L
2.1
10
E
17
23
F
Answers
1.1 In Δs ACB and ECD
...
Aˆ = Eˆ
. . . alternate øs ; BA || ED
ˆ
Bˆ = D
. . . alternate øs ; BA || ED
ˆ = ECD
ˆ
Also, ACB
s
. . . vert. opposite ø
â ΔACB ||| ΔECD 2.2
18
Note : As soon as you have shown Δ's to be similar by showing they are
equiangular, you can claim that their sides are in proportion
and write this down without having to look back at the diagram.
CD
2.3
P
9
12
C
R
ED
N
75
A
. . . AAA
EC
H
25
M
It is essential when naming the Δs,
to order the letters according
to the equal øs, especially for
writing out the proportional
sides which follow.
, then : AC = AB = CB
||| Δ ECD
i.e. if Δ ACB
69
51
E
9
A
E
y
x
6
x
B
3
B
y
C
Q
Answers
2.1 ΔLMN ||| ΔEFH or : the same sides,
but inverted.
∴
NB : The triangles must be named in the correct order !
because :
LM
MN
LN
=
=
,
EF
FH
EH
2.2 ΔPQR ||| ΔABC It is then useful to mark off what you are looking for with a ?
(or its letter if it has a letter) and a  for the sides you have lengths for.
51
69
75
equals 3;
equals 3 and
equals 3
17
23
25
i.e. the sides are in proportion.
because :
while
PQ
12
PR
18
=
= 2 and
=
= 2,
AB
6
AC
9
QR
9
=
= 3, which is ≠ 2
BC
3
â The sides are not in proportion.
1.2 ∴
EC ?
ED 
CD
=
=
AC 
AB 
CB
â
ED
EC
=
AB
AC
...
10
EC
â
=
8
4
% 4)
. . . ACB ||| ECD
2.3 ΔBAC ||| ΔEDC ; because, in these triangles :
B̂ = Ê
When creating your equation, place what
you are looking for, EC in this case, on
the top LHS. It eases the calculation.
Note : These 2 Δs are not congruent.
 = D̂ . . . both = y
[â 3rd ø equal too ! ]
â ΔBAC ||| ΔEDC â EC = 5 units Copyright © The Answer Series
. . . both = x
9
. . . AAA
Even though there are 2 øs
and a side equal, the sides do
not correspond.
D
Worked Example 3
Similar Δs vs.
Proportion Theorem Application
A
Find the values of x and y in the
figure alongside.
10 cm
D
B
x
5 cm
y
E
3 cm
12 cm
A
Worked Example 4
C
12
Find x and y in the sketch alongside
% 10)
y
D
Answer
In ΔABC :
15
E
8
x
10
=
5
B
. . . DE || BC; proportion theorem
8
4
% 12)
In ΔABC : DE || BC
y
12
=
30
5
The
unknown
. . . DE = AE ; proportional sides
8
BC
AC
∴ y = 7 1 cm 2
% 15)
x
15
= 8
12
â x = 10 units
The proportion theorem
does NOT refer to the
lengths of the parallel lines,
only to AB and AC
and their segments.
Similar triangles theorem (finding y)
In the same figure above ,
ΔABC can be seen as
an enlargement of ΔADE
and the sides of these
triangles are proportional.
In Δs ADE and ABC :
Note : The Theorem of Pythagoras can be proved using
(1) Â is common
similar ∆s (see page 13 & 14). Consider this proof an ideal
ˆ = B̂ . . . corresponding øs; DE || BC
(2) ADE
'worked example' of the application of similar triangles.
ˆ = Ĉ . . . corresponding øs; DE || BC ]
[& AED
â DE = AD or AE
BC
Note:
â
Distinguish between the applications of
the similar ∆ s and proportion theorems!
% 30)
(See next column.)
AB
y
12
=
20
30
â
y
15 ⎞
⎛ or
= 12
⎜
⎟
30
12 + 8 ⎝ 15 + 10 ⎠
. . . equiangular Δs
â ΔADE ||| ΔABC
Copyright © The Answer Series
C
The Proportion theorem (finding x)
∴ x = 6 1 cm ∆ADE ||| ∆ABC ²
x
AC
...
Note:
AD
AE
DE
= DB or EC
BC
because BC is a side of ΔABC,
while DB and EC are not.
â y = 18 units
It is only by using the similarity of the triangles,
that we can relate the lengths of the parallel sides to
the lengths of the other 2 sides of the triangles.
10
SIMILAR ΔS : STATEMENTS & PROOF
STATEMENT: Equiangular Δs proportional sides SIMILAR Δs
PROOF:
A
Given: ABC & DEF with  = D̂, B̂ = Ê & Ĉ = F̂
Required to prove:
B
AB
AC
BC
=
=
DE
DF
EF
If two triangles are
equiangular, then their sides
are proportional and, therefore,
they are similar.
D
C
P 1
Q
E
F
Construction: Mark P & Q on DE & DF such that DP = AB & DQ = AC
Proof: In s DPQ & ABC
(1) DP = AB
. . . construction
(2) DQ = AC
. . . construction
(3) D̂ = Â
. . . given
â DPQ ≡ ABC
â Pˆ = B̂
. . . SøS
1
Stage 2 :
corresponding øs
. . . given
= Ê
Stage 3 :
parallel lines
The â PQ || EF . . . corresponding øs equal
focal
DP
DQ
=
. . . (prop thm; PQ || EF)
â
point
DE
DF
But DP = AB and DQ = AC
â
Stage 1 :
congruency
. . . construction
AB
AC
=
DE
DF
Stage 4 :
proportions
Similarly, by marking P and R on DE and EF such that
PE = AB and ER = BC, it can be proved that: AB = BC
DE
â AB = AC = BC DE
DF
EF
â ABC and DEF are similar
Copyright © The Answer Series
11
EF
THE CONVERSE: Proportional sides equiangular Δs SIMILAR Δs
Triangles with sides in proportion, are equiangular and,
therefore, they are similar.
PROOF (Optional):
Given: s ABC & DEF with
D
A
BC
AB
AC
=
=
EF
DE
DF
1
B
Required to prove: ABC & DEF are equiangular
ˆ (xˆ ) = B̂ & GFE
ˆ (yˆ ) = Ĉ
Construction: GEF
1
1
3
y2
x
2
C E
y
1
x1
2
F
G & D on opposite sides of EF
G
Proof: In s ABC & GEF
(1) xˆ 1 = B̂
. . . construction
(2) yˆ 1 = Ĉ
. . . construction
Stage 1 :
equiangular ∆s
(∆1 & ∆2)
â ABC & GEF are equiangular
â AB = BC = AC
EF
GE
GF
The
focal
point
proportional sides of equiangular s
(applying the 'original' theorem)
...
Comparing this with the 'given' :
AB AB
AC AC
=
and
=
GE DE
GF DF
â GE = DE and GF = DF
â
BC 
. . .  all =


EF 
Stage 2 :
proportions
compared
â In s DEF & GEF
(1) GE = DE
. . . proved above
(2) GF = DF
. . . proved above
Stage 3 :
congruency
(∆2 & ∆3)
(3) EF is common
â DEF ≡ GEF . . . SSS
â xˆ 2 = xˆ 1
&
= B̂
â also D̂ = Â
yˆ 2 = yˆ 1
. . . construction
= Ĉ
. . . 3rd ø of the two s
i.e. s ABC & DEF are equiangular
â ABC and DEF are similar
Copyright © The Answer Series
12
. . . (∆1 & ∆3)
. . . as for 1st proof!
Stage 4 :
return to the start !
(∆3 ! ∆2 ! ∆1)
NOT listed as
examinable in the 2021
Exam Guidelines.
SIMILAR ΔS THEOREM PROOF: Visualised
The Situation
The LOGIC . . .
Construction
Imagine copying the
small Δ on the big Δ
A
D
1
2
B
C
3
E
The 2 shaded Δs are congruent (SS)
F
 1̂ = 2̂
2 Equiangular Δs : ΔABC & ΔDEF
You actually mark the lengths of
But
the small Δ onto the sides of the big Δ.
Then join their endpoints.
x
p
y
But:
q
AB =
x
 The horizontal lines are parallel
Similarly, it can be proved that:
D
x
 The sides of this Δ are proportional.
y
i.e. x =
p
q
B
. . . Proportion theorem
x
y
AB = BC
DE
EF
y
 AB = BC = AC
DE
EF
DF
C
 ΔABC ||| ΔDEF
E
F
 AB = AC
DE
DF
Copyright © The Answer Series: Photocopying of this material is illegal
. . . given
 1̂ = corresponding 3̂
& AC = y
A
2̂ = 3̂
13
The Statement:
Equiangular Δs are similar
EXAMPLE 3 (National November 2017 P2, Q10)
(b)
ˆ
= 90º . . . tangent ⊥ radius
OWS
ˆ
ˆ
â OVN = OWS
(= 90º)
(i)
In the diagram, W is a point on the
circle with centre O.
â MN || TS
. . . corresp. øs equal
V is a point on OW.
(ii)
Chord MN is drawn such that
MV = VN.
The tangent at W meets
OM produced at T and
ON produced at S.
O
T
M
1
2 1
(a) Give a reason why OV ⊥ MN.
2
O
V
W
M
T
1
1 2
1
2
MN || TS
(ii)
TMNS is a cyclic quadrilateral
Shade the
quadrilateral
TMNS
V
2 N
(b) Prove that:
(i)
The most 'basic'
way to prove
lines || is:
alt. or corresp.
øs equal or
co-int. øs suppl.
W
S
1
2 N
S
(iii) OS . MN = 2ON . WS (a Grade 12 question)
There are 3 ways to prove that a quadrilateral is
a cyclic quadrilateral – choose 1:
Answers
(a) Line (OV) from centre to midpoint of chord (MN) y
x
In this case, the midpoint of the chord is given, and
we can conclude that OV ⊥ MN because of that.
Prove that:
x + y = 180º
Prove that:
Ext. ø = int. opp. ø
s
ˆ = N
ˆ
M
1
1 . . . ø opposite equal radii
Note: Analyse the information and the diagram.
= Ŝ
So far, we have used and applied a 'centre' theorem, in (a).
Another clue is the 'tangent' at W.
Think about tangent facts . . . .
. . . corresp. øs ; MN || TS
â TMNS is a cyclic quadrilateral
Copyright © The Answer Series
13
Prove that:
A side subtends equal øs
at 2 other vertices
We chose and proved
that the exterior ø of
quadrilateral TMNS
= the interior opposite ø
converse ext. ø
. . . of cyclic quad.
(iii)
This question looks like ratio and proportion.
Mark the sides on the diagram.
The sides appear to involve OWS, which has VN || WS,
(even though MN = 2VN)
. . . Maybe apply the proportion theorem in this ?
But, the sides in the question involve the horizontal sides
WS and VN.
So, proportion theorem is excluded.
We will use similar s !
In s OVN and OWS
ˆ is common
 O
2
ˆ = OWS
ˆ
 OVN
. . . corresp. øs ; MN || TS
â OVN ||| OWS
. . . øøø
Let's 'arrange' the sides to suit the question.
â
OS
WS  OW 
=
=

VN  OV 
ON
. . . equiangular  s
â OS . VN = ON . WS
But VN = 1 MN
2
â OS .
. . . V midpoint MN
1
MN = ON . WS
2
% 2)â OS . MN = 2ON . WS Copyright  The Answer Series
14
THE THEOREM OF PYTHAGORAS
THE STATEMENT
Proving the Theorem of Pythagoras, using similar triangles
In ΔABC: Ĉ = 90º ² c 2 = a 2 + b 2
A
ˆ = 90º
Given : ΔABC with ABC
A
D
Required to prove : AC 2 = AB 2 + BC 2
Construction : Draw BD  AC
c
1
b
B
2
C
PROOF:
B
C
a
 = x ; then Bˆ 1 = 90º – x
Let
. . . ø sum of ABD
â Bˆ 2 = x
In a right-angled Δ, the square on the hypotenuse
equals the sum of the squares on the other two sides.
â Ĉ = 90º – x
:
In ΔABC:
=
a2
+
b2
In Δs ABD and ACB :
A
(1) Â is common
THE CONVERSE STATEMENT
c2
. . . ø sum of BCD
(2) Bˆ 1 = Ĉ
. . . = 90º – x
ˆ = 90º]
ˆ = ABC
[& ADB
² Ĉ = 90º
â ABD ||| ACB
A
â AB = AD
AC
AB

D
C
B
. . . AAA
 = BD 


 BC 
AB is the common side
of the 2 triangles :
 AB2 = . . .
2
c
â AB = AC . AD
b
 : Similarly, by proving BCD ||| ACB :
B
a
C
â BC2 = AC . DC
:
If the square on one side of a triangle equals the
sum of the squares on the other two sides,
then the angle between these two sides is a right angle.
â AB2 + BC2 = AC . AD + AC . DC
= AC (AD + DC)
= AC . AC
= AC2
Copyright © The Answer Series: Photocopying of this material is illegal
14
BCD ||| ACB
² BC = CD
AC
BC
A
D
2
â BC = AC . CD B
Here, BC is the common side :
 BC2 = . . .

C
The Theorem of Pythagoras: The proof in stages
The Theorem Statement
The Situation
Construction
A
In a right-angled Δ,
the square on the hypotenuse
equals the sum of the squares
on the other two sides.
The Method
D
b
c
a
B
C
B
A
A
D
D
b
c
Similarly:
b
c
C
B
B
a
C
In ΔAB D & ΔAC B : Common  & a right  each
In ΔCB D & ΔCA B : Common Ĉ & a right  each
 ΔABD ||| ΔACB
 ΔCBD ||| ΔCAB
 AB = AD
AC
 CB = CD
AB
CA
 AB2 = AC. AD
i.e. c2 = AC. AD
The Conclusion

CB
 CB2 = CA.CD
...
&

i.e. a2 = AC.CD
:
A
D
 c2 + a2 = AC. AD + AC. CD
c
= AC(AD + CD)
= b.b
b
. . . AC = b
B
= b2
Copyright © The Answer Series
NOT listed as examinable in the
2021 Exam Guidelines.
15
a
C
...

Similar s : Advice for problem-solving
When asked to prove a proportion, e.g. AB = BC ,
The symbol for similar Δs is: |||
DE
|||
& the symbol for congruent Δs is:
mark the 4 sides (AB, DE, BC & EF) on the figure.
It will then be clear which triangles need to be proved similar.
If we write ΔABC ||| ΔDEF, with the letters in this order, it means
If asked to prove PQ2 = PR.PS, we need to mark PQ twice,
because it would be a common side of the two triangles
– see the proof of the Theorem of Pythagoras illustrating this.
that  = D̂ , B̂ = Ê and Ĉ = F̂ and the sides will also correspond . . .
AB
= BC = AC
DE
EF
DF
...
ABC
EF
DEF
Always remember to accumulate facts as you work through
a sum, i.e. mark the facts on the figure as you prove them.
This is very important because it means we don't have to read off
the sides from the drawing, provided we have the letters in
the correct order !
Distinguish between the applications of the
proportion and the similar Δs theorems:
It is usually a good idea to write down all three ratios,
Given: AQ : QC = 5 : 2 and PQ || BC
maybe bracketing the one not immediately required
– it may well be needed later in the question.
THE PROPORTION THEOREM
AP : PB = 5 : 2
We can prove that triangles are similar
either by:
Also:
proving them equiangular (2 øs are sufficient) . . . The Theorem
AP : AB = AQ : AC = 5 : 7
&
P
A
5 parts
Q
2 parts
B
C
PB : AB = QC : AC = 2 : 7
whereas
or, by:
. . . because PQ and BC are the sides of similar Δs APQ & ABC
proving that their sides are proportional. . . . The Converse Theorem
PQ : BC = 5 : 7
SIMILAR TRIANGLES THEOREM
When two pairs of angles of two triangles have been
proved equal, mark the third angles as being equal too.
Be sure to know and apply the
theorem statements accurately !
It could be required later in the question.
16
Copyright © The Answer Series: Photocopying of this material is illegal
6.1
EXERCISE 2: SIMILAR TRIANGLES
P
A
1.2
36
32
B
C
Q
B
R
24
E
C
28
ABC |||  . . . .
2.
D
D
A
B
21
A
E
x
In the figure, P is a point on SQ such
ˆ
ˆ = TQS.
that STP
TS = 51 mm, PS = 32,6 mm and
TP = 29 mm.
C
ˆ = ABC
ˆ , AD = 5 and DC = 4.
In the figure, CDB
5.
5
A
4
C
S
4
Q
9.1
a
= .....
b
9.2
c
= .....
d
E
7
a
c
b
3
d
For more examples, see the
Topic Guide at the start of Section 2
in The Answer Series
Gr 12 Maths 2 in 1
2
y
5
x
Complete with reasons:
B
P
Find, with reasons, the lengths of
x and y.
Copyright © The Answer Series
D
T
3
R
17
D
6
4,5
8.2 Ê = . . . . .
S
4
E
B
8.1 In the figure alongside, prove that GDF ||| GED.
9.
4.1 Complete: CBA ||| . . . . .
C
7.3 Hence calculate the length of BE.
51 mm
P 32,6 mm
3.2 Calculate QT (answer correct to one decimal place).
14
6
7.4 Determine the value of BD .
AD
29 mm
Q
D
7.2 Complete : BE = DE
...
...
T
(Show all working details.)
A
7.1 BDE ||| . . .
2.3 Hence calculate the area of rectangle ABCD to the nearest cm .
4.2 Calculate the length of BC.
The accompanying figure shows ABC with DE || AC.
AC = 14, DE = 6 and EC = 4 units.
2
4.
Prove: AB2 = AC.AD
Prove: BA . BD = BC . BE
D
2.2 If BC = 18 cm and BE = 12 cm,
B
calculate the length of
2.2.1 AE
2.2.2 AB correct to two decimals.
3.1 Prove that STP ||| SQT.
C
C
7.
2.1 Complete the following statement:
∆ABE ||| ∆ . . . ||| ∆ . . .
3.
E
F
27
CAB |||  . . . .
Make a neat copy of this sketch and fill in
all the other angles in terms of x.
Reasons are not required.
B
D
Answers on page 18
1.1
A
6.2
A
3
5
F
4
G
ANSWERS TO EXERCISES
EXERCISE 1: PROPORTION THEOREM
AND APPLICATIONS
2.3 DE || BC ²
1.3
â 3x + 5x - 2 = 21x - 18
B
1.5
Q
R
7x - 6
&
AP
PB
=
AQ BR
". . . proportionally"
% 5)
2
3
â 3x = 2x + 4
â x = 4 x
M
2
Q
Copyright © The Answer Series
3x
C
4
D
â DE = 5 1
G
4
3.1
Theorems
3.2
In ZWY :
ZT
= ZR
TY
RW
= ZS
A
x cm
= 33
4
x+2
SX
x
H
. . . RT || WY; proportion theorem
B
C
3
ZT
= ZS
TY
SX
â ST || XY R
18
proportion thm.
S
â In ZXY we have
3 cm
Z
W R
. . . RS || WX;
5 cm
E
D
S
= 1
which, in ZWX,
8
P
F
3
 x 4
3
â x = 30
. . . proportion theorem
x
E
. . . proportion thm.; DE || BC
8
4x
2
Note: 'integral' value(s) required
Bookwork
6
x
5
1
RHS = 3 = 3 = 1
6
2
x+2
 Solution : x = 4
x = 5
22
B
DE
= GH
AD
AG
â DE =
Check x = 4 : LHS = 3x - 1 = 11 = 1
They’re equal because they lie on the same base (AB)
x = x+2
. . . prop thm; DG || EH
C
B
â 8x = 30
2.2
AD
= 4 DE
2.5.2
E
3
2.4
2.1
(b)
A
D
â x = 4 or 4
and between the same || lines (AB & QR). 1.4
. . . prop thm; CG || DH
A
â (3x - 4)(x - 4) = 0
PB
BR
AG
= 4 GH
â 3x2 - 16x + 16 = 0
A
1.2
2.5.1 (a)
2
P
AP
AQ
. . . proportion thm.
â (3x - 1)(x + 2) = 3(7x - 6)
Questions on page 7
1.1
3x - 1
= 3
7x - 6
x+2
X
T
Y
. . . converse of proportion theorem
H
4.1
N
L
5.
A
EXERCISE 2: SIMILAR TRIANGLES
F
Questions on page 16
F
M
J
K
C
D
S
E
(a) In JHK :
JK = 10 units
...
3 : 4 : 5 = 6 : 8 : 10
HM = 4 units
. . . diagonals of a parallelogram
JM2 = 62 + 42
ˆ = 90º; Pythagoras
. . . JHM
AF
= FE
FC
EB
²
But, in ABC : AF = AD
FC
CAB ||| FED
. . . a= c ² a= b
. . . sides in ratio 8 : 7 : 9
b
d
c
A
2.
d
D
x
12
. . . prop. thm.; DF || BC
DB
E
90º - x
x
â FE = AD
EB
= 52
DB
90 º- x
90º - x
B
x
18
C
. . . converse of prop. thm.
 JM = 52 =
4  13 = 2 13
â In ABF, DE || AF
 ML = 2 13
. . . diagonals of ||m bisect one another
â ADEF is a trapezium . . . FL =
1.2
B
AF
= FC
FE
EB
(b) In JHM :
13
 ABC |||  RPQ
ˆ = 90º; Pythagoras
JHK
JK : KS = 10 : 5 = 2 : 1
 MF =
1.1
. . . 1 pair opp. sides ||
13
2.1
NB: The order of the letters
must correspond
with the equal angles
ABE ||| ECB ||| DEC
2.2.1 ABE ||| ECB
 JM : MF = 2 13 : 13 = 2 : 1
Now, try some
4.2
In JFS : JK : KS = JM : MF
 MK || FS
. . . in (a) & (b)
AE = BE
²
CHALLENGING QUESTIONS . . .
BE
BC
. . . sides in proportion
2
2
% BE) â AE = BE = 12 = 8 cm
BC
. . . converse of proportion theorem
18
 HK || NS
2.2.2
See Section 3
â AB =
Page 255 in
ACT!
AB2 = 122 - 82 = 80
. . . Thm. of Pythagoras
80
j 8,94 cm
The Answer Series
Gr 12 Maths 2 in 1
A: Be Active
2.3
Area of rectangle ABCD = BC % AB
C: Use all your Clues
= 18 % 8,94
T: Apply the Theory systematically
j 161 cm2
Copyright © The Answer Series
19
...
length %
breadth
3.1
In s STP and SQT
6.1
T
(1) Ŝ is common
ˆ = Q̂
(2) STP
. . . given
29 mm
. . . equiangular Δs
â STP ||| SQT
â
â
QT
TP
= ST
SP
j 45,4 mm
BC
CD
â BC
2
= CD.AC
= 4.9
= 36
3
6.2
2
4
D
â AB
S
â x = 6 units
4
Q
2
7.2
y
5
x
T
3
R
7.3
% 5)
y
= 6
5
4
â y = 7 1 units
2
Copyright © The Answer Series
C
AB
BDE ||| BAC . . . equiør ∆s . . .
BC
AC
9.1
a
= 7
b
3
. . . || lines ; proportion theorem !
9.2
c
= 7
d
10
. . . proportional sides of similar s !
BD
= BE = 3
AD
EC
4

CHALLENGING QUESTIONS . . .
A
D
14
6
B
NB: BD
AD
. . . øs in similar ∆s in 8.1
Now, try some
â BE = 6 . . . sides of sim. ∆s
BE + 4
14
â BE = 3 units
7.4
ˆ
Ê = GDF
common ø &
corresponding øs
. . . prop. sides
â 8BE = 24
s
8.2
= AC.AD
â BE = DE
. . . proportional sides
. . . equiangular Δs
â 14BE = 6BE + 24
common ø &
QRP ||| QTS . . . equiø ∆ . . .
corresponding øs
â
. . . given
B
7.1
G
D
Note : Side AB is common to the two  s
A
ST || PR ;
prop. thm.
PQ
â PR =
QS
ST
2
â GDF ||| GED
B
(1) Â is common
5
r
A
In  ABD and ACB
C
4
GF
= 4 = 2
GD
6
3
â AB = AD
proportional
...
sides
F
DF
= 3 = 6 = 2 ; and
ED
9
3
4,5
We need  s ABD & ACB
AC
5
GD
= 6 = 2;
GE
9
3
BD
ˆ = Ĉ
(2) ABD
6
3
E
. . . equiangular Δs
s
P
...
C
E
. . . given
â ABD ||| ACB
In PQR:
x = 4
B
â BA.BD = BC.BE
= AC
â BC = 6 units
5.
BE
must be correct !
BC
4,5
â BA = BC
. . . The order of the letters
CBA ||| CDB
â
ˆ
(2) Ĉ = BDE
â BCA ||| BDE
Note :
It is a good idea to
place the required
length (QT) in the
top left position of
the proportion.
QT
= 51
29
32,6
32,6
4.2
P 32,6 mm
S
D
D
(1) B̂ is common
. . . sides in proportion
% 29) â QT = 51 % 29
4.1
51 mm
8.1
A
In s BCA and BDE
Q
3.2
We need  s BCA & BDE
E
4
. . . DE || AC ; prop. theorem
C
See Section 3
Page 256 to 262 in
The Answer Series
Gr 12 Maths 2 in 1
DE
AC
20
See The Answer Series
Gr 12 Maths 2 in 1
Study Guide
For more practice, see the TOPIC GUIDE (on Page 148)
at the start of SECTION 2: The Exam Paper 2s
and
The Challenging Questions (Pp 255 – 262)
in the
Copyright © The Answer Series
NEW
20
SECTION 3
PLEASE NOTE
These Geometry materials (Booklets 1 to 4) were created and
produced by The Answer Series Educational Publishers (Pty) (Ltd)
to support the teaching and learning of Geometry in high schools
in South Africa.
They are freely available to anyone who wishes to use them.
This material may not be sold (via any channel) or used for
profit-making of any kind.
GRADES 8 - 12
ALL MAJOR SUBJECTS IN
ENGLISH & AFRIKAANS
S
A
WWW.THEANSWER.CO. ZA
T
2021
Q
9. In the diagram, PQRS is a cyclic
93º
quadrilateral. PS is produced to W.
TR and TS are tangents to the circle at
R and S respectively.
P
T̂ = 78º and Q̂ = 93º.
9.1 Give a reason why ST = TR.
(1)
9.2 Calculate, giving reasons, the size of:
9.2.1 Ŝ 2
9.2.2 Ŝ 3
R
1
(2)(2) [5]
2
1
S
2
3
78º
T
W
Copyright © The Answer Series
2021
Q
9. In the diagram, PQRS is a cyclic
93º
quadrilateral. PS is produced to W.
TR and TS are tangents to the circle at
R and S respectively.
P
T̂ = 78º and Q̂ = 93º.
9.1 Give a reason why ST = TR.
(1)
9.2 Calculate, giving reasons, the size of:
9.2.2 Ŝ 3
9.2.1 Ŝ 2
R
1
(2)(2) [5]
2
MEMOS
9.1
Tangents from a common point.
9.2.1 Ŝ 2 = R̂ 2
=
. . . øs opposite equal sides
1
(180º
2
– 78º)
. . . ø sum of Δ
1
S
2
3
78º
= 51º 9.2.2 Ŝ 3 + Ŝ 2 = Q̂
. . . ext. ø of cyc. quad.
∴ Ŝ 3 + 51º = 93º
ˆ = 42º ∴ S
3
Copyright © The Answer Series
T
W
2021
10. In the diagram, BE and CD are diameters
E
C
of a circle having M as centre. Chord AE
is drawn to cut CD at F. AE  CD.
x
Let Ĉ = x.
10.1 Give a reason why AF = FE.
(1)
F
10.2 Determine, giving reasons, the size
of M̂1 in terms of x.
(3)
10.3 Prove, giving reasons, that AD is
a tangent to the circle passing
through A, C and F.
(4)
10.4 Given that CF = 6 units and
AB = 24 units, calculate, giving
reasons, the length of AE.
1
2
1
A
M
3
2
3
(5)
[13]
B
Copyright © The Answer Series
D
2021
E
C
10. In the diagram, BE and CD are diameters
of a circle having M as centre. Chord AE
F
2
1
Let Ĉ = x.
2
A
10.1 Give a reason why AF = FE.
1
3
& EB = DC = 2(6 + 12)
= 36 units
...
AE2 = EB2 – AB2
= 362 – 242
= 720
M
∴ AE =
OR :
10.3 Prove, giving reasons, that AD is a
tangent to the circle passing through
A, C and F.
(4)
10.3
AC is a diameter of ☼ACF
...
...
10.2
M̂1 = 2Â 1
line from centre ⊥ to chord
...
& Â 1 = 90º – x
ø at centre = 2 % ø at circ.
. . . ø sum of Δ
∴ M̂1 = 2(90º – x)
= 180º – 2x Copyright © The Answer Series
. . . ø in semi-☼
. . . øøø
ø in semi-☼; diameter CD
2
...
2
ME2
FM2
–
∴ EF =
= 182 – 122
= 180
i.e. line AD ⊥ diameter AC
∴ AD is a tangent to ☼ACF EA
∴ FM = 12 units
∴ ME = 6 + 12 = 18 units
conv tan ⊥ rad
...
. . . radii equal
Theorem of Pythagoras
∴ EF = 6 5
∴ AE = 12 5 ≈ 26,83 units
ˆ = x
A
2
. . . ø in semi-☼
OR :
In ΔEAB :
F is the midpoint of EA
conv tan chord thm
&
M is the midpoint of EB
∴ FM = 1 AB = 12 units
2
∴ MC = 18
∴ MD = 18
Theorem of Pythagoras
720 = 12 5  26,83 units ˆ = 90º
EAB
AB
conv ø in semi-☼
∴ AD is a tangent to ☼ACF . . .
MF ⊥ AE, i.e.
diameter
∴ FM = EF = 1
ˆ = Ĉ
∴ A
2
10.1
...
∴ ΔEFM | | | ΔEAB
OR :
MEMOS
midpt theorem
(1) Ê is common
ˆ = EAB
ˆ
(2) EFM
D
B
ˆ = 90º
& DAC
(5)
[13]
line from centre of ☼
In ΔEFM & ΔEAB
(3)
10.4 Given that CF = 6 units and
AB = 24 units, calculate, giving
reasons, the length of AE.
...
3
(1)
...
∴ FM = 1 AB = 12 units
2
∴ In right ød ΔEAB :
10.2 Determine, giving reasons, the size
of M̂1 in terms of x.
F is the midpoint of EA
& M is the midpoint of EB
x
is drawn to cut CD at F. AE ⊥ CD.
10.4 In ΔEAB :
∴ FA = CF
AF
FD
CF . FD
6 . 30
180
6 5
∴ AE = 12 5
...
. . . radii equal
ΔCFA ||| ΔAFD
∴ AF2 =
=
=
∴ AF =
...
. . . øøø
line from centre of ☼
midpt theorem
2021
MEMOS
11.1 In the diagram, chords DE, EF and DF are
drawn in the circle with centre O.
11.1
The angle between a tangent to a circle and a chord drawn from the point of
contact is equal to the angle subtended by the chord in the alternate segment.
Method 1
KFC is a tangent to the circle at F.
Draw radii and use 'ø at centre' theorem.
Given : ?O with tangent at F and chord FE subtending D̂ at the circumference.
ˆ = Ê
DFK
RTP :
E
D
Construction : radii OF and OD
D
ˆ = x
DFK
ˆ
OFK = 90º . . . radius ⊥ tangent
ˆ = 90º – x
∴ OFD
ˆ = 90º – x . . . øs opposite equal radii
∴ ODF
Proof :
Let
O
ˆ = 2x
∴ DOF
F
x
Method 2
These proofs are logical
& easy to follow.
We use 2 'previous' facts involving right øs
(5)
 tangent ⊥ diameter . . . so, draw a diameter !
. . . so, join RK !
 ø in semi-? = 90º
Given : ?O with tangent at F and chord FE subtending D̂ at the circumference.
M
ˆ = Ê
DFK
RT P :
E
Construction : diameter FM; join ME
Proof :
ˆ = 90º
MFK
ˆ = 90º
& MEF
. . . tangent ⊥ diameter
D
. . . ø in semi-?
Then . . .
Let
Copyright © The Answer Series
C
F
K
. . . ø at centre = 2 % ø at circumference
ˆ = Ê â DFK
C
Prove the theorem which states that
ˆ = Ê .
DFK
O
. . . sum of øs in Δ
∴ Ê = x
K
E
ˆ = x
DFK
ˆ
â MFD = 90º – x
ˆ = 90º – x
â MED
ˆ = x
â DEF
ˆ = DEF
ˆ â DFK
O
. . . øs in same segment
x
K
F
C
E
D
O
K
Copyright © The Answer Series
F
C
2021
11.2 In the diagram, PK is a tangent to the
circle at K. Chord LS is produced to P.
N and M are points on KP and SP
respectively such that MN || SK.
Chord KS and LN intersect at T.
K
11.2.1 Prove, giving reasons, that:
ˆ
(a) K̂ 4 = NML
(4)
(b) KLMN is a cyclic
quadrilateral.
(1)
1
11.2.2 Prove, giving reasons, that
ΔLKN ||| ΔKSM.
N
(5)
2
2
T
2
1
3KN = 4SM, determine the
2
2
(4)
1
11.2.4 If it is further given that
NL = 16 units, LS = 13 units
and KN = 8 units, determine,
with reasons, the length of LT. (4)
[23]
P
Copyright © The Answer Series
1
3
3
11.2.3 If LK = 12 units and
length of KS.
4
3
M
1
S
L
2021
MEMOS
11.2.1 (a) K̂ 4 = Ŝ 1
11.2 In the diagram, PK is a tangent to the
ˆ = NML
circle at K. Chord LS is produced to P.
Chord KS and LN intersect at T.
converse
ext ø of c.q.
...
ext ø of cyclic quad. KLMN
= Ŝ 2
...
corresp øs ; MN || SK
1
(4)
ˆ = Ŝ
(1) LKN
2
ˆ = M
ˆ
(2) N
3
3
∴
(1)
11.2.3
11.2.2 Prove, giving reasons, that
(5)
11.2.3 If LK = 12 units and
3KN = 4SM, determine the
(4)
...
øs in same segment
ΔLKN | | | ΔKSM KS
SM
=
LK
KN
∴
...
. . . øøø
|||Δ
s
and KN = 8 units, determine,
with reasons, the length of LT. (4)
[23]
3
1
P
LT
LS
=
...
LN
LM
LT
13
∴
=
16
13 + 6
13
% 16
∴ LT =
19
208
=
19
prop thm ; MN || ST
≈ 10,95 units Copyright © The Answer Series
2
2
KS
3
∴
=
12
4
36
∴ KS =
4
= 9 units In ΔLNM :
2
2
3
SM
=
4
KN
∴ SM = 6 units
NL = 16 units, LS = 13 units
2
1
11.2.4 4SM = 3KN = 3 % 8 = 24
11.2.4 If it is further given that
3
1
3KN = 4SM
∴
1
T
3
N
L
4
K
∴ In Δs LKN & KSM
(b) KLMN is a cyclic
length of KS.
...
ˆ = M̂
11.2.2 LKN
1
11.2.1 Prove, giving reasons, that:
ΔLKN ||| ΔKSM.
corresp øs ; MN || SK
∴ KLMN is a cyclic quad. respectively such that MN || SK.
quadrilateral.
...
ˆ
(b) K̂ 4 = NML
N and M are points on KP and SP
ˆ
(a) K̂ 4 = NML
tan chord thm
...
M
S
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