# MATHS PROOFS ```Gr 12 Maths
BOOKWORK
(an extract from Gr 12 Maths 2 in 1)
All the proofs you need to know!
Paper 1:
Series
(a maximum of 6 marks)
page 1
Paper 2:
Geometry (6 theorems) and
Trigonometry (3 proofs)
(a maximum of 12 marks)
page 1 to 3
Be sure to study the geometry and trigonometry summaries on
page 4 to 6.
Calculator instructions
page 7
BOOKWORK: PROOFS
PAPER 1
PAPER 2
Circle Geometry Theorems
Arithmetic and Geometric Series
Sn , the sum of n terms
1
Arithmetic Series:
Sn = a + (a + d) + (a + 2d) + . . . + Tn
. . . n terms
&amp; Sn = Tn + (Tn – d) + (Tn – 2d) + . . . + a
. . . n terms
∴ 2 Sn = (a + Tn) + (a + Tn) + (a + Tn) + . . . (a + Tn)
The line segment drawn from the centre of a circle,
perpendicular to a chord, bisects the chord.
Given : ?O with OP ⊥ AB
To prove : AP = PB
Construction : Join OA and OB
s
Proof :
In Δ OPA &amp; OPB
(1) OA = OB . . . radii
(2) Pˆ 1 = Pˆ 2 (= 90&ordm;) . . . given
(3) OP is common
&acirc; ΔOAP ≡ ΔOBP . . . S&oslash;S
&acirc; AP = PB, i.e. OP bisects chord AB
. . . n terms
∴ 2 Sn = n (a + Tn)
∴ Sn = n (a + Tn) 2
Tn = a + (n – 1)d
But
∴ Sn =
n
[ a + a + (n – 1)d ]
2
2
∴ Sn = n [ 2a + (n – 1)d ] Sn = a + ar + ar
% r ) ∴ r Sn =
ar + ar
2
+ ar
+ ar
 –  : ∴ Sn – r Sn = a – ar
3
3
+ . . . . . . + ar
n–2
+ ......
+ ar
+ ar
n –1
n–1
+ ar
... 
n
ˆ at the centre and APB
ˆ at the circumference.
?O, arc AB subtending AOB
P
12
x y
Then Â = x . . . radii equal
&acirc; Oˆ 1 = 2x . . . exterior &oslash; of ΔAOP
... 
Similarly, if Pˆ 2 = y, then Oˆ 2 = 2y
ˆ = 2x + 2y = 2(x + y) = 2 APB
ˆ
&acirc; AOB
the 'middle bit' falls away.
n
B
P
ˆ = 2APB
ˆ
To prove : AOB
Construction : Join PO and produce it to Q.
Pˆ 1 = x
Proof : Let
Geometric Series:
2
1 2
A
The angle which an arc of a circle subtends at the centre is
double the angle it subtends at any point on the circumference.
Given :
2
O
O
1 2
x
y
A
Q
B
n
∴ Sn (1 – r) = a(1 – r )
∴ Sn =
a(1 − r n )
1− r
. . . for r &lt; 1
or
&times;
( − 1)
a(r n − 1)
: Sn =
( − 1)
r −1
3
. . . for r &gt; 1
To prove : Â + Ĉ = 180&ordm; &amp; B̂ + D̂ = 180&ordm;
Construction : Join BO and DO.
− (a − b)
b − a
a − b
=
=
d − c
c − d
− (c − d)
Proof :
Â =
Let
Then
&acirc; Ĉ =
The derivation of the future and present value formulae, Fv and Pv, using the
sum formula of a geometric sequence, must be understood, but is not examinable.
See Topic 5 (Financial Maths) in the Gr 12 Maths 2 in 1 study guide.
&acirc; Â + Ĉ =
&amp; &acirc; B̂ + D̂ =
1
x
2
O
D
1
B
. . . &oslash; at centre = 2 %
&oslash; at circumference
s
360&ordm; – 2x . . . &oslash; about point O
1
(360&ordm; – 2x) = 180&ordm; – x . . . &oslash; at centre = 2 %
2
&oslash; at circumference
180&ordm;
s
180&ordm;
. . . sum of the &oslash; of a
ˆ = 2x
O
1
&acirc; Oˆ 2 =
Annuities:
x
Given : ?O and cyclic quadrilateral ABCD
Choose either depending on the value of r, whichever gives you a positive denominator.
Notice that :
A
The opposite angles of a cyclic quadrilateral are supplementary.
C
4
D
ˆ = APB
ˆ ,
A
1
To prove :
ˆ
= 90&ordm;
DAC
&amp;
ˆ
DPA
= 90&ordm;
B
2
. . . diameter ⊥ tangent
2
. . . &oslash; in semi-?
1
∴ Aˆ 1 + Aˆ 2 = 90&ordm;
and
6
1
P
Construction : Draw diameter AD and join DP.
Proof : We have
x
A
ˆ
ˆ
A
2 = P1
s
. . . &oslash; in same seg.
EC
B
Q
E
F
B̂ = Ê &amp; Ĉ = F̂
s
In Δ DPQ &amp; ABC
. . . construction
(1) DP = AB
(2) DQ = AC . . . construction
(3)
D̂ = Â
C
But
stage 2 :
corresponding &oslash;s
s
corresponding &oslash; equal
DP = AB and
. . . construction
&acirc; AB = AC
DE
DF
= AE
DB
EC
stage 3 :
parallel lines
. . . proportion theorem
DQ = AC
A
. . . S&oslash;S
. . . given
The &acirc; PQ || EF . . .
focal
&acirc; DP = DQ
point
DE
DF
stage 1 :
congruency
. . . given
&acirc; ΔDPQ ≡ ΔABC
E
Given : ΔABC with DE || BC,
D &amp; E on AB &amp; AC respectively.
To prove :
AB
= AC = BC
DE
DF
EF
Proof :
A line parallel to one side of a triangle divides
the other two sides proportionally.
DB
C
To prove :
&acirc; 1̂ = B̂
= Ê
D
B
ΔABC &amp; ΔDEF with Â = D̂
A
i.e. DE || BC &sup2; AD = AE
P 1
Given :
Remember :
Always give
REASONS for
each statement !
ˆ
∴ Aˆ 1 = Pˆ 2 ( = APB)
The Proportion Theorem
If two triangles are equiangular,
then their sides are proportional
and, therefore, they are similar.
Construction : Mark P &amp; Q on DE &amp; DF such that DP = AB &amp; DQ = AC
C
Pˆ 1 + Pˆ 2 = 90&ordm;
But
D
A
ˆ ,
Given : Tangent AC and chord AB subtending APB
with C &amp; P on opposite sides of AB.
5
The Similar Δs Theorem
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.
stage 4 :
proportions
Similarly, by marking S and T on DE and EF such that
Construction : Join DC &amp; BE
1
= 2
Proof :
1
Area of ΔDBE
DB
DB . h
2
1
AE . h′
2
Similarly : Area of ΔADE = AE
1
Area of ΔEDC
EC
EC . h′
2
h′
D
B
SE = AB and ET = BC, it can be proved that :
h
E
AB
BC
=
DE
EF
&acirc; AB = AC = BC DE
C
DF
EF
&acirc; ΔABC and ΔDEF are similar.
s
Similar Δ
s
But :
ΔDBE = ΔEDC
and :
...
Δ are similar if : A : they are equiangular, and
B : their sides are proportional
on the same base DE ;
between || lines, DE &amp; BC
In this proof, we show that A B
s
∴ The Δ are similar . . . Both conditions, A and B, apply
Area of ΔDBE
Area of ΔEDC
AE
&acirc;
=
DB
EC
The converse statement says : B A
s
∴ The Δ are similar
2
TRIGONOMETRY
3
The Cosine rule
Â
Area, Sine &amp; Cosine Rules
1
A
a2 = b2 + c2 – 2bc cos A
b
c D
The Area rule
A
Ĉ acute
Area of ΔABC = 1 ab sin C
Construction :
Proof :
1
Area of ΔABC = ah
2
But
Construction :
B
...
h
= sin C
b
2
C
D
a
2
a = BD + h
∴ a
...
. . . Pythagoras
2
2
= (c – AD) + h
2
2
2
= c – 2c . AD +
2
b
2
2
= b + c – 2c . AD
∴ h = b sin C
2
B
C
a
2
. . . Pythagoras
...
...
= cos A
b
∴ AD = b cos A
1
∴ Area of ΔABC =
ab sin C
2
h
Draw CD ⊥ BA
2
Proof :
b
h
2
in :
acute :
in :
∴ a2 = b2 + c2 – 2bc cos A
2
The Sine rule
= sin C
sin A
= sin B
a
b
Construction :
Proof :
B̂ acute
Always construct the height from a vertex not involved in the formula :
D
i.e. To prove :
b
c
h
B
Draw CD ⊥ AB
h
= sin A
b
∴ h = b sin A
In ΔBDC :
...
a
C
∴
...
sin A
sin B
=
, draw a height from C, not A or B.
a
b
To prove :
a = b + c – 2bc cos A, draw a height from B or C, not A.
30&ordm;
sin A
sin B
=
a
b
Similarly, by drawing a perpendicular from B, one can prove :
∴
To prove :
Remember :
&amp; : ∴ b sin A = a sin B
&divide; ab)
1
ab sin C, draw a height from A or B, not C.
2
Area =
2
2
150&ordm;
2
II
sin (180&ordm; – θ) = sin θ
θ
h
= sin B
a
∴ h = a sin B
Equating
A
180&ordm; – θ
sin 150&ordm; = sin 30&ordm;
cos 150&ordm; = – cos 30&ordm;
cos (180&ordm; – θ) = – cos θ
y
x
x + y = 180&ordm; sin x = sin y
cos x = – cos y
Formulae for Compound Angles
The derivation of the formula : cos (A – B) = cos A cos B + sin A sin B
will not be examined. (However, it is always good to understand how a formula arises ! ).
There is an easy guide to this derivation in Topic 6 in the Gr 12 Maths 2 in 1 Study guide.
You do need to know how to derive all the other formulae from this formula for cos(A – B).
See examples of this, also found in Topic 6 in the Gr 12 Maths 2 in 1 Study guide.
sin A
sin C
=
a
c
sin A
sin B
sin C
=
=
a
c
b
3
3 ways to prove that a quadrilateral is cyclic
GROUPING OF CIRCLE GEOMETRY THEOREMS
I
The grey arrows indicate how
various theorems are used to
prove subsequent ones.
The 'Centre' group
We use 3 converse theorem statements to prove that a quadrilateral is cyclic.
converse of '&oslash;'s in same segment' theorem
Prove that one side
subtends equal angles
at the two other points
(on the same side).
x
2x
diameter
2x
x
x
Equal
II The 'No Centre' group
Equal
chords !
x1
z1
y2
B
Prove that a pair of
opposite angles
is supplementary.
Equal chords subtend
equal angles
and, vice versa,
equal angles are subtended
by equal chords.
Either prove :
Â + Ĉ = 180&ordm;
A
or B̂ + D̂ = 180&ordm;
D
C
converse of 'exterior &oslash; of c.q.' theorem
Prove that an exterior angle
to the interior opposite angle.
Either prove :
5
2
7
1̂ = 2̂
4
8 3
6
1
x
There are 3 ways
to prove that
Either prove :
x1 = x2
or y1 = y2
or z1 = z2
or p1 = p2
p1 x
2
z2
p2
converse of '&oslash;'s of c.q.' theorem
y
y1
or 3̂ = 4̂
or 5̂ = 6̂
or 7̂ = 8̂
2 ways to prove that a line is a tangent to a ?
We use 2 converse theorem statements to prove that a line is a tangent to a ?.
x + y = 180&ordm;
converse of 'tangent ⊥ radius' theorem
IV The 'Tangent' group
O
Prove that the line is perpendicular
to a radius at a point where the
Prove that OP ⊥ AB ;
then AB will be a tangent
A
P
B
converse of 'tan-chord' theorem
Equal
tangents !
Prove that when the line is
drawn through the end point of
a chord, it makes with the chord
an angle equal to an angle in
the alternate segment.
There are 2 ways to prove that
a line is a tangent to a ?.
4
A
y
x
B
Prove that
x = y;
then AB
will be
a tangent
Note :
It is a good idea to
draw a feint circle
(as shown) to see
this converse
theorem clearly.
- pathways of definitions, areas and properties - A Summary
A Rectangle
All you need to know!
f
a
A Trapezium
b
m
A || with one right &oslash;
2
h
Area = ℓ % b
1
d
DIAGONALS are EQUAL
a
s
Sum of the &oslash; of
The Square
DEFINITION :
b
c
e
A Parallelogram
DEFINITION :
Quadrilateral with 1 pair of opposite sides ||
DEFINITION :
Quadrilateral with 2 pairs opposite sides ||
Area = s 2
Area = base % height
Area = Δ 1 + Δ 2
P
= 1 ah + 1 bh
2
2
Sum of the interior angles
= (a + b + c) + (d + e + f)
s
= 2 % 180&ordm;
. . . (2 Δ )
= 360&ordm;
D
B
|| ABCD = ABCQ + ΔQCD
rect. PBCQ = ABCQ + ΔPBA
where ΔQCD ≡ ΔPBA
Properties :
It's all been said 'before' !
Since a square is a rectangle,
a rhombus, a parallelogram,
a kite, . . . ALL the properties
C
m
. . . SS90&ordm;
m
&acirc; || ABCD = rect. PBCQ (in area)
= BC % QC
A Kite
a
2
a
2
b
See how the properties
accumulate as we
move from left to right.
no special properties
and each successive
preceding properties.
Q
A Rhombus
= 1 (a + b) . h
2
'Half the sum of the || sides
% the distance between them.'
The arrows indicate
various ‘ROUTES’
from ‘any’
square (the ‘ultimate
routes, which combine
logic and fact, are
essential to use when
proving specific types
A
Properties :
2 pairs opposite sides equal
2 pairs opposite angles equal
&amp; DIAGONALS BISECT ONE ANOTHER
Area
=
1
product of diagonals (as for a kite)
2
THE DIAGONALS
• bisect one another PERPENDICULARLY
• bisect the angles of the rhombus
• bisect the area of the rhombus
Given diagonals a and b
s
Area = 2Δ = 2 ⎛⎜ 1 b . a ⎞⎟ = ab
2⎠
m
A || with one pair of
or
= base % height (as for a parallelogram)
DEFINITION :
consecutive sides equal
⎝2
DEFINITION :
2
'Half the product of the diagonals'
THE DIAGONALS
• cut perpendicularly
• the LONG DIAGONAL bisects:
the short diagonal, the opposite angles and the area of the kite
5
Note :
x
y y
x
x
x
y y
s
2x + 2y = 180&ordm; . . . &oslash; of Δ, or
&sup2; x + y = 90&ordm;
co-int. &oslash;s suppl.
play a
prominent role
in both Euclidian
and Analytical
Geometry right
through to
2
SUMMARY : TRIGONOMETRY (Gr. 12)
' SPECIAL LS
' THE RATIOS
sin 30E '
&amp; their
' DEFINITIONS :
y
r
O
H
x
r
A
H
y
x
3
1
2
tan 45E ' 1
2
1
and cos 60&deg; '
2
O
A
tan θ '
y
x
&amp; THEIR “FAMILIES” :
' SIGNS :
' GENERAL FORMS
' CRITICAL VALUES :
any ratio
' 90E - θ (an acute angle)
(see denominator ! )
' GRAPHS :
&plusmn;
that SAME
ratio of
'
'
θ
sin(90E + θ)
cos(90E - θ) = sin θ
cos(90E + θ) = - sin θ
AREA '
SINE rule
1
ab sin C
2
1
2
And in RIGHT-ANGLED Δ S :
2
bh as well
2
' cos(A &plusmn; B) = cos A cos B
6
COSINE rule
'
c 2 ' a 2 % b 2 &amp; 2ab cos C
' sin 2x = 2 sin x cos x
' sin(A &plusmn; B) = sin A cos B &plusmn; cos A sin B
&amp; cos θ = 1 - sin θ
sin A
sin B
sin C
'
'
a
b
c
'
use regular trig. ratios &amp; the theorem of Pythagoras !!!
' COMPOUND LS
ˆ sin2 θ = 1 - cos2 θ
= cos θ
For non-right angled Δ S :
But remember : area of ∆=
sin 2 θ % cos 2 θ ' 1
90E + θ (an obtuse angle)
sin(90E - θ) = cos θ
' SOLUTION OF ΔS
sin θ
tan θ '
cos θ
=
' CO-RATIOS (sine and cosine)
' MINIMUM &amp; MAXIMUM VALUES : (sin θ &amp; cos θ)
' MUTUAL RELATIONSHIPS
180&deg; - θ
180&deg; + θ
360&deg; - θ
-θ
K
sin A sin B
' cos 2x = cos2 x - sin2 x
2
= 1 - 2 sin x
2
= 2 cos x - 1
CALCULATOR INSTRUCTIONS
MEAN AND STANDARD DEVIATION
REGRESSION &amp; CORRELATION
There are 3 phases for each calculator procedure :
The equation of the regression line
STEP 1 : how to get there
Brief instructions for the CASIO fx - 82ES
STEP 2 : how to enter the data
STEP 3 : how to find the mean and the standard deviation.
STEP 1 :
Press/Select : MODE ; 2 (STAT) ; 2 (A + Bx )
(and, in the next column, A, B and r)
Grouped Data/Frequency Tables
Ungrouped Data
Casio fx-82ES
Enter the x and y values, each followed by =.
Casio fx-82ES
You'll see :
[MODE] [2 : STAT] [1 : 1 - VAR]
Enter each value, followed by [=]
after the last value : [=] [AC]
1
2
3
X
…
…
…
FREQ
automatically
entered
as 1
,
,
[SHIFT] [STAT] [5 : VAR] [3 : x σ n] [=]
If this is not done, the mean and standard deviation
will be added as entries in the data table.
(These would then need to be deleted.)
Enter each value, followed by [M+]
To find the mean : [RCL] [ x ]
To find the S.D. :
[RCL] [σ x ]
FREQ
…
…
…
Press AC
STEP 3 :
Press
SHIFT
STAT (on '1') ; 7 (Reg) ; then :
For A, press 1
=
or
For B, press 2
=
To find B after A, press AC
and then do the whole of STEP 3.
to move to the top of the right-hand column.
Just like for A and B in the regression function :
Brief instructions for the CASIO fx - 82ES
To find the mean : [SHIFT] [STAT] [5 : VAR] [2 : x ] [=]
To find the S.D. :
You'll see :
Stat
Type in the correct frequencies followed by [=] each
,
time; after the last frequency : [=] [AC]
0.
Sharp EL-531WH
You'll see :
[MODE] [1 : STAT] - to get into stats mode . . .
then [0 : SD]
Enter midpoint of each interval (or each score if
provided), followed by [STO].
Then enter the frequency followed by [M+].
(Enter all the data like this.)
To find the S.D. :
STEP 1 :
Press/Select : MODE ; 2 (STAT) ; 2 (A + Bx )
[SHIFT] [STAT] [5 : VAR] [3 : x σ n] [=]
To find the mean : [RCL] [ x ]
1
2
3
X
…
…
…
The correlation coefficient
It is essential to use [AC] at the end of entering all the data.
[MODE] [1 : STAT] - to get into stats mode . . .
then [0 : SD]
[MODE] [2 : STAT] [1 : 1 - VAR]
[SHIFT] [SETUP] Scroll down (use arrow)
[3 : STAT] [1 : ON]
Use
Note for Casio fx-82ES
Sharp EL-531WH
You'll see :
Enter the midpoint of each interval
(or each score if provided) followed by [=], then . . .
After the last value : [=] - and not [AC] - then . . .
To find the mean : [SHIFT] [STAT] [5 : VAR] [2 : x ] [=]
To find the S.D. :
STEP 2 :
[RCL] [σ x ]
7
Stat
0.
STEP 2 :
Enter the x and y values, each followed by =.
Press AC
STEP 3 :
Press
SHIFT
STAT (on '1') ; 7 (Reg) ; then :
But, now : For r press 3
=
If you have found A and B first, press AC
before going back to the beginning of STEP 3.
```