Year 7 Angles
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
www.drfrostmaths.com
Objectives: Understand notation for angles. Know basic rules of angles
(angles in triangle, on straight line). Recognise alternate, corresponding
and vertically opposite angles. Find angles in isosceles triangles.
Deal with and introduce algebraic angles. Construct diagrams from
written information and form angle proofs.
Last modified: 7th February 2016
For Teacher Use:
Recommended lesson structure:
Lesson 1: Angle notation. Angle basics.
Lesson 2: Z/F/C/v. opposite angles. Angle properties of quadrilaterals.
Lesson 3: Isosceles Triangles
Lesson 4: Algebraic Angles
Lesson 5/6: Constructing diagrams from information. Proof
Lesson 7: Consolidation/Mini-assessment
Angle Notation
𝐴
𝐵
We use capital letters for points.
𝐶
! We can refer to this angle using:
• 𝐴𝐵𝐶 (with a ‘hat’ on the
? middle letter)
• ∠𝐴𝐵𝐶
?
• Or in words: “Angle?𝐴𝐵𝐶”
Quick Starter: How would we refer to each of the following angles?
(Use ∠𝐴𝐵𝐶 notation)
𝑄
1
𝑃
3
𝐶
2
𝑆
𝑅
1
∠𝑆𝑄𝑅 or? ∠𝐶𝑄𝑅
2
∠𝑆𝐶𝑅 ?
3
∠𝑆𝑃𝑄 ?
Angle Basics
30°
128°
𝟓𝟐°?
! “Angles on a straight line
? sum to 180°.”
𝟐𝟒𝟎°
?
(These wordings will be important for angle proofs later)
! “Angles around a point
? sum to 360°.”
85°
𝟒𝟓°
?
50°
! “Angles in a triangle?sum to 180°.”
reflex
acute
obtuse
Name of angle if:
Less than 90:
Acute
?
Between 90 and 180: Obtuse
?
Over 180:
Reflex?angle
Test Your Understanding
i
[JMC 2008 Q4] In this diagram, what
is the value of 𝑥?
A 16
B 36
C 64
D 100 E 144
Solution: C ?
ii
[JMC 2003 Q12] What is the size of
the angle marked 𝑥?
A 42°
B 67°
C 69°
D 71°
E 111°
Solution: A ?
Exercise 1
1
(on provided worksheet)
Find the angles marked with letters.
71°
𝑏
𝑎
45°
𝑏
150°
𝑐
𝑏
105°
𝑑
𝑒
52°
60°
110°
𝒂 = 𝟔𝟒°,
?
? 𝒃 = 𝟕𝟎°,
? 𝒄 = 𝟔𝟎°,
? 𝒅 = 𝟏𝟏𝟐°,
? 𝒆 = 𝟑𝟓°
30°
Exercise 1
(on provided worksheet)
2
[JMC 2000 Q3] What is the value of 𝑥?
Solution: 28
3
[JMC 2015 Q6] What is the value of 𝑥 in
this triangle?
Solution: 50
?
?
4
[JMC 2011 Q6] What is the sum of the
marked angles in the diagram?
Solution: 𝟑𝟔𝟎°
?
5
[JMC 1997 Q17] How big is angle 𝑥?
Solution: 𝟏𝟎𝟒°
?
6
[IMC 2014 Q3] An equilateral triangle is
placed inside a larger equilateral triangle
so that the diagram has three lines of
symmetry. What is the value of 𝑥?
Solution: 𝟏𝟓𝟎°
?
7
[IMC 2011 Q9] In the diagram, 𝑋𝑌 is a
straight line. What is the value of 𝑥?
Solution: 170
?
Exercise 1
8
(on provided worksheet)
[JMC 2010 Q8] In a triangle with angles 𝑥°, 𝑦°, 𝑧° the
mean of 𝑦 and 𝑧 is 𝑥. What is the value of 𝑥?
Solution: 60
?
9
[JMC 1999 Q13] In the diagram, ∠𝑅𝑃𝑀 = 20° and
∠𝑄𝑀𝑃 = 70°. What is ∠𝑃𝑅𝑆?
Solution: 𝟏𝟑𝟎°
?
10
[JMC 2001 Q18] Triangle 𝑃𝑄𝑅 is equilateral. Angle
𝑆𝑃𝑅 = 40°, angle 𝑇𝑄𝑅 = 35°. What is the size of
the marked angle 𝑆𝑋𝑇?
Solution: 𝟏𝟑𝟓°
?
11
[JMC 2007 Q16] What is the sum of the six marked
angles?
Solution: 𝟏𝟒𝟒𝟎°. The sum of the angles at the 7
points is 𝟓 × 𝟑𝟔𝟎 = 𝟏𝟖𝟎𝟎. We’ve excluded the
angles in the 3 triangles, so 𝟏𝟖𝟎𝟎 − 𝟑 × 𝟏𝟖𝟎 =
𝟏𝟒𝟒𝟎.
?
12
[JMC 2015 Q16] The diagram shows a square inside
an equilateral triangle. What is the value of 𝑥 + 𝑦?
Solution: 150
?
Exercise 1
13
(on provided worksheet)
[JMO 2001 A3] What is the value of 𝑥 in the diagram alongside?
Solution: 15
?
14
[Kangaroo Pink 2010 Q9] In the diagram, angle 𝑃𝑄𝑅 is 20°, and the reflex
angle at 𝑃 is 330°.The line segments 𝑄𝑇 and 𝑆𝑈 are perpendicular. What
is the size of angle 𝑅𝑆𝑃?
Solution: 𝟒𝟎°
?
15
[Kangaroo Pink 2006 Q8] The circle shown in the diagram is divided into
four arcs of length 2, 5, 6 and 𝑥 units. The sector with arc length 2 has an
angle of 30° at the centre. Determine the value of 𝑥.
(Hint: An arc is a part of the line that makes up the circle, i.e. part of the
circumference. The arc length grows in proportion to the angle at the
centre)
Solution: 11
?
16
[SMC 2010 Q3] The diagram shows an equilateral triangle touching two
straight lines. What is the sum of the four marked angles?
Solution: 𝟐𝟒𝟎°. Angles at two points some to 𝟑𝟔𝟎°, but we exclude two
angles of 𝟔𝟎°.
?
Angles involving parallel lines
There are three more laws of angles that you need to know, use and quote:
(These arrows indicate the
lines are parallel)
! Alternate angles are equal.
? as ‘Z’ angles)
(Sometimes known
! Corresponding angles are equal.
? as ‘F’ angles)
(Sometimes known
! Vertically opposite angles
? are equal.
Bro Note: The word ‘vertically’ here is the adjective
form of ‘vertex’, which means a point. So “vertically
opposite” means “opposite with respect to a point”.
How to spot them
To identify alternate angles:
Step 1: Identify a line connecting
two parallel lines.
Click >>
Step 2: At each end on your parallel
lines, shoot out along the parallel lines
in opposite directions. Your angles are
the ones wedged between the lines.
Click >>
To identify corresponding angles:
Step 1: Identify an angle between one line
of a parallel pair, and a connecting line.
Click >>
Step 2: This can be shifted along to the
other parallel line of the pair.
Click >>
Examples
(For extra practise outside of class)
Identify the alternate and corresponding angle for each indicated angle (use ∠ notation).
𝑃
𝑊
𝑋
𝐴
𝑄
𝐷
𝐶
𝐵
𝑆
𝐸
𝐺 5
4
𝐹
2
𝐻
Double arrows allows us to match
another pair of parallel lines.
3
𝐽
𝐼
𝐾
1
𝑁
𝑀
𝐿
𝑇
𝑅
𝑈
𝑉
𝑂
#
Angle
Alternate
Corresponding
1
∠𝐽𝑁𝑂
∠𝑃𝐷𝑄 or ∠𝐺𝐿𝑀
2
?
∠DJC
?
∠𝐼𝐶𝐷
?
∠𝐻𝐺𝐿
?
?
∠𝐹𝑀𝑁
?
∠BAC
?
∠𝑇𝑉𝐼
?
∠𝐷𝐽𝐼
?
3
4
5
∠𝑆𝐹𝐻
∠𝐶𝐷𝐽
?
∠𝐿𝑀𝑅
?
∠𝑊𝐴𝑋
?
∠𝑈𝐿𝑀
?
∠𝑁𝐽𝐾
?
Check Your Understanding
ANB is parallel to CMD. LNM is a straight line.
Angle LMD = 68°
(i) Work out the size of the angle marked 𝑦.
𝟏𝟏𝟐°
?
(ii) Give reasons for your answer.
“Alternate angles are equal” OR “Corresponding angles
? line sum to 180”
are equal”, “Angles on straight
i
ii
[JMC 2011 Q11] The diagram shows an
equilateral triangle inside a rectangle. What is
the value of 𝑥 + 𝑦?
A 30
B 45
C 60
D 75
E 90
𝑦
𝑥
Solution:?C
Bro Hint: Is there a line in the
diagram we could add so we
can then use alternate angles?
Cointerior angles
We’ve seen ‘Z’ and ‘F’ angles. There’s also ‘C’ angles!
! Cointerior angles sum to?𝟏𝟖𝟎°
180 − 𝑥
?
𝑥
We can identify them in a similar way
to alternate angles: identify a line
connecting two parallel lines, but this
time go in the same direction at each
end rather than opposite directions.
Application to parallelograms
So we can say:
110°
?
70°
?70°
! Cointerior angles in a parallelogram
add to?𝟏𝟖𝟎°
! Opposite angles in a parallelogram
are equal.
?
Example
Can you come up with an alternative explanation
for why 𝑦 = 112°?
∠𝑴𝑵𝑩 = 𝟏𝟏𝟐° (cointerior angles sum to 180)
𝒚 = 𝟏𝟏𝟐° (vertically opposite
? angles are equal)
One further property of quadrilaterals
What are the sum of the
angles in a quadrilateral?
Draw Hint >>
! Angles in a quadrilateral sum to 𝟑𝟔𝟎°
?
50°
𝑥
35°
45°
?
𝑥 = 𝟏𝟑𝟎°
Exercise 2
1
a
(on provided worksheet)
For each diagram
(i) Find the missing angles and
(ii) List reasons for each answer.
i) 𝟓𝟓°
?
ii) Corresponding
? equal.
angles are
b
? 𝟗𝟓°
i) 𝒙 = 𝟏𝟓𝟎°, 𝒚 =
ii) For x: Corresponding angles
are equal.
For y: Angles on
straight line
?
add to 180. Alternate angles
are equal.
Exercise 2
2
b
a
𝑦
𝑥
115°
122°
80°
𝒙 = 𝟓𝟖°?
𝒚 = 𝟑𝟓°
?
c
30°
d
𝑏
57°
𝑥
𝑎
𝑐
82°
𝒂 = 𝟑𝟎°, 𝒃 =?𝟖𝟐°, 𝒄 = 𝟔𝟖°
41°
𝒙 = 𝟕𝟒°
?
Exercise 2
f
e
85°
110°
𝑥
112°
75°
𝑦
?
𝒙 = 𝟔𝟖°
𝒙 = 𝟗𝟎°
?
Exercise 2
3
For rhombuses are evenly
spaced around a point.
Given the angle shown,
find 𝑥.
Solution: 𝟏𝟐𝟎°
?
30°
𝑥
Exercise 2
4
Find the angles indicated.
100°
41°
𝑐
𝑏
𝑎
𝑑
𝑒
96°
136°
𝒂 = 𝟒𝟏°?
𝒃 = 𝟗𝟔°?
𝒄 = 𝟏𝟑𝟔°
?
𝒅 = 𝟖𝟒°?
𝒆 = 𝟗𝟓° ?
Exercise 2
5
42°
Find the angle 𝛼
Solution: 𝟕𝟏°
𝛼
?
29°
𝐷
6
𝑃
𝐶
The line 𝑃𝐵 bisects the angle 𝐴𝐵𝐶
(this means it cuts it exactly in half).
Determine the angle 𝐵𝑃𝐶.
Solution: 𝟓𝟓° ?
70°
𝐴
𝐵
Exercise 2
7
[JMC 2006 Q7] What is the value of 𝑥?
Solution: 𝟕𝟓°
?
8
[JMC 2007 Q9] In the diagram on the right, 𝑆𝑇 is
parallel to 𝑈𝑉. What is the value of 𝑥?
Solution: 86
?
9
[JMC 2009 Q19] The diagram on the right shows a
rhombus 𝐹𝐺𝐻𝐼 and an isosceles triangle 𝐹𝐺𝐽 in which
𝐺𝐹 = 𝐺𝐽. Angle 𝐹𝐽𝐼 = 111°. What is the size of angle
𝐽𝐹𝐼?
Solution: 𝟐𝟕°
?
10
[IMC 2005 Q7] In the diagram, what is the sum of the
marked angles?
Solution: 𝟑𝟔𝟎°. Sum of angles in 4 triangles is 𝟏𝟖𝟎° ×
𝟒 = 𝟕𝟐𝟎°. The four angles in the quadrilateral sum to
360, thus because of vertically
? opposite angles, the
innermost angles of the triangles sum to 360. Thus
sum of angles is 𝟕𝟐𝟎 − 𝟑𝟔𝟎 = 𝟑𝟔𝟎°.
Exercise 2
11
[JMO 2003 A9] Find the value of 𝑎 + 𝑏 + 𝑐.
Solution: 210
?
12
[SMC 2006 Q3] The diagram shows
overlapping squares. What is the value of 𝑥 +
𝑦?
Solution: 270
?
13
[Cayley 2012 Q2] In the diagram, 𝑃𝑄 and 𝑇𝑆
are parallel. Prove that 𝑎 + 𝑏 + 𝑐 = 360.
Suppose we add a line horizontal with 𝑹 as
shown, where 𝒃 is split into 𝒃𝟏 and 𝒃𝟐 .
𝒂 and 𝒃𝟏 are cointerior thus 𝒂 + 𝒃𝟏 = 𝟏𝟖𝟎°.
𝒃𝟐 and 𝒄 are cointerior thus 𝒃𝟐 + 𝒄 = 𝟏𝟖𝟎°
?
Thus
𝒂 + 𝒃 + 𝒄 = 𝒂 + 𝒃𝟏 + 𝒃𝟐 + 𝒄
= 𝟏𝟖𝟎° + 𝟏𝟖𝟎° = 𝟑𝟔𝟎°
𝒚
𝒃𝟏
𝒃𝟐
Exercise 2
14
[Kangaroo Grey 2005 Q20] Five straight lines intersect at a common point
and five triangles are constructed as shown. What is the total of the 10
angles marked in the diagram?
A 300° B 450° C 360° D 600° E 720°
Answer: 𝟕𝟐𝟎°. Sum of angles in 5 triangles is 𝟏𝟖𝟎° × 𝟓 = 𝟗𝟎𝟎°.
Since angles at the central point add to 𝟑𝟔𝟎°, but each angle in a triangle at the
centre can be paired with a vertically opposite angle not in a triangle, the central
angles in the triangles sum to 𝟏𝟖𝟎°. 𝟗𝟎𝟎° − 𝟏𝟖𝟎° = 𝟕𝟐𝟎°
?
Isosceles Triangles
What do these marks mean?
The lines are of the
? same length.
𝟒𝟎°
?
𝟔𝟓°
?
50°
70°
?
𝟕𝟎°
! “Base angles of an
isosceles triangle are equal.”
Warning!
Sometimes diagrams are drawn in such a way that it’s not visually obvious what the
two angles the same are. You can use the ‘finger slide method’ to identify these.
Diagram not
drawn accurately.
Click for
Broanimation >
Put your fingers on the two
marks.
Slide your fingers in the same
direction but away from each
other.
These two corners are where
the angles are the same.
Test Your Understanding
i
[JMC 2013 Q3] What is the value of 𝑥?
A 25
B 35
C 40
D 65
E 155
Solution: C
?
ii
[IMC 2012 Q11] In the diagram, 𝑃𝑄𝑅𝑆 is a parallelogram;
∠𝑄𝑅𝑆 = 50°; ∠𝑆𝑃𝑇 = 62° and 𝑃𝑄 = 𝑃𝑇. What is the size
of ∠𝑇𝑄𝑅?
A 84° B 90° C 96° D 112° E 124°
Solution:
?C
Using angles to give you information about sides
You’ve so far used sides which are equal to find angles.
But we can do the opposite too! If two angles are equal, then two sides are equal.
[Kangaroo Pink 2004 Q6] In
the diagram 𝑄𝑅 = 𝑃𝑆.
What is the size of ∠𝑃𝑆𝑅?
75°
65°
1. Use information provided.
2. Work out some initial angles.
3. Use angles to give us
information about side lengths.
4. Use new knowledge of side
lengths to work out more
angles…
Go >
Go >
Go >
Go >
Angle Wall Challenge
How far can you get down the angle challenge
wall? (do in order, and draw the diagram first)
Hint: You may want to add extra lines.
𝟔𝟎°
60°
𝟑𝟎°
𝟏
𝟕𝟓°
𝟒𝟓°
𝟑𝟎°

𝟔𝟎°
𝟕𝟓°
?
∠𝐷𝐴𝐶 = 30°
?
∠𝐴𝐷𝐵 = 45° (Δ𝐴𝐷𝐶 is?isosceles)
∠𝐴𝐷𝐶 = 75° (Δ𝐴𝐷𝐶 is?isosceles)
∠𝐵𝐷𝐶 = 30° (75° − 45°)
?
∠𝐷𝐵𝐶 = 15° (angles within
? Δ𝐷𝐶𝐵)
 ∠𝐵𝐴𝐶 = 60°
𝟏𝟓°

This was a Junior Maths Olympiad problem!
(to prove that ∠𝐵𝐷𝐶 = 2 × ∠𝐷𝐵𝐶)
Exercise 3
1
(on provided worksheet)
Find the value of each variable.
Solution: 𝒙 = 𝟒𝟔°, 𝒚 = 𝟕𝟒°, 𝒛 = 𝟐𝟎°
?
2
[JMC 2014 Q10] An equilateral triangle is
surrounded by three squares, as shown. What is
the value of 𝑥?
Solution: 30
?
3
[JMC 2005 Q13] The diagram shows two equal
squares. What is the value of 𝑥?
Solution: 140
?
4
[IMC 2010 Q6] In triangle 𝑃𝑄𝑅, 𝑆 is a point on
𝑄𝑅 such that
𝑄𝑆 = 𝑆𝑃 = 𝑃𝑅 and ∠𝑄𝑃𝑆 = 20°. What is the
size of ∠𝑃𝑅𝑆?
Solution: 𝟒𝟎°
?
5
[IMC 1999 Q8] In the diagram 𝑃𝑄 = 𝑃𝑅 = 𝑅𝑆.
What is the size of angle 𝑥?
Solution: 𝟏𝟎𝟖°
?
Exercise 3
(on provided worksheet)
6 [JMC 2008 Q19] In the diagram on the right, 𝑃𝑇 = 𝑄𝑇 = 𝑇𝑆,
𝑄𝑆 = 𝑆𝑅, ∠𝑃𝑄𝑇 = 20°. What is the value of 𝑥?
Solution: 35
?
7 [Kangaroo Grey 2010 Q9] The diagram shows a quadrilateral
𝐴𝐵𝐶𝐷, in which 𝐴𝐷 = 𝐵𝐶, ∠𝐶𝐴𝐷 = 50°, ∠𝐴𝐶𝐷 = 65° and
∠𝐴𝐶𝐵 = 70°. What is the size of ∠𝐴𝐵𝐶?
Solution: 𝟓𝟓°
?
8 [Cayley 2004 Q1] The “star” octagon shown in the diagram is
beautifully symmetrical and the centre of the star is at the
centre of the circle. If angle 𝑁𝐴𝐸 = 110° (as indicated), how big
is the angle 𝐷𝑁𝐴? (i.e. 𝑥)
Solution: 𝟐𝟎°
?
9 [IMC 2003 Q8] Lines 𝐴𝐵 and 𝐶𝐷 are parallel and 𝐵𝐶 = 𝐵𝐷.
Given that 𝑥 is an acute angle not equal to 60°, how many
other angles in this diagram are equal to 𝑥?
Solution: 4
?
10 [IMC 1997 Q11] In the quadrilateral 𝐴𝐵𝐶𝐷, ∠𝐴𝐵𝐶 = 90°,
∠𝐵𝐴𝐷 = 70°, and 𝐴𝐵 = 𝐵𝐷 = 𝐵𝐶. What is the size of ∠𝐵𝐷𝐶?
Solution: 𝟔𝟓°
?
𝑁
𝑥
𝑊
𝐷
𝐴
𝐶
𝐵
𝑆
110°
𝐸
Exercise 3
11
(on provided worksheet)
[Kangaroo Pink 2013 Q9] The diagram shows an
equilateral triangle 𝑅𝑆𝑇 and also the triangle 𝑇𝑈𝑉
obtained by rotating triangle 𝑅𝑆𝑇 about the point 𝑇.
Angle 𝑅𝑇𝑉 = 70°. What is angle 𝑅𝑆𝑉?
Solution: 𝟑𝟓°
?
12
[IMC 2000 Q8] In the triangle 𝐴𝐵𝐶, 𝐴𝐷 = 𝐵𝐷 = 𝐶𝐷.
What is the size of angle 𝐵𝐴𝐶?
Solution: 𝟗𝟎° ?
13 In the diagram 𝐴𝐵 is parallel to 𝐶𝐷 and 𝐴𝐸 = 𝐴𝐶.
Determine ∠𝐶𝐴𝐸.
𝐴
𝐵
30°
Solution: 𝟏𝟐° ?
𝐸
54°
𝐶
𝐷
Exercise 3
14
(on provided worksheet)
[TMC Regional 2008 Q9] If 𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷 =
𝐷𝐸 = 𝐸𝐹 and angle 𝐴𝐸𝐹 = 75°, what is the
size of angle 𝐸𝐴𝐹?
Solution: 𝟐𝟏°
?
STARTER : Algebraic Angles
What are the angles in each case, in terms of the variables given?
(Hint: Just think what you’d do usually – it’s no different here!)
i
180?− 𝑥
𝑥
ii
iii
90 ?− 𝑦
𝑦
𝑧
180 −?2𝑧
!
Overview
There are two types of problems you’ll have which involve algebraic angles:
1. Angles given
…and you have to find an
expression for a given angle.
Example:
2. No angles given
…and you have to introduce variables
yourself, either so that you can prove
two angles have some relationship, or
so you can form an equation and
hence find an angle.
Example:
Given that 𝐴𝐵 = 𝐴𝐶 and 𝑧 < 90, which of
the following expressions must equal 𝑧?
A 𝑥−𝑦
B 𝑥+𝑦
C 𝑥 + 𝑦 − 180
D 180 + 𝑥 − 𝑦
E 180 − 𝑥 + 𝑦
[JMO 2010 A10] In the diagram, 𝐽𝐾 and 𝑀𝐿 are
parallel. 𝐽𝐾 = 𝐾𝑂 = 𝑂𝐽 = 𝑂𝑀 and 𝐿𝑀 = 𝐿𝑂 =
𝐿𝐾. Find the size of angle 𝐽𝑀𝑂.
Finding remaining angle in triangle/on line
Angle 1
Angle 2
?
If Angle 1 is 90 + 𝑥, what is
angle 2?
180 − ?90 + 𝑥
(Expand
brackets)
= 180? −
90 −
𝑥
= 90 − 𝑥? (Simplify)
Bro Tip: An easier way to do the
subtraction is to think what we
have to do to 90 + 𝑥 to get to
180. Adding 90 gets us from 90
to 180, and subtracting 𝑥
cancels out the 𝑥.
Quickfire Questions:
Angle 1
Angle 2
𝑥
180?− 𝑥
𝑥 + 20
160?− 𝑥
𝑥 − 10
190?− 𝑥
2𝑥 + 60
120?− 2𝑥
𝑥−𝑦
180 −?𝑥 + 𝑦
180 − 2𝑥
2𝑥
?
160 − 𝑥
20?+ 𝑥
Finding remaining angle in triangle/on line
Angle 2
Angle 1
?
Angle 1
Angle 2
Remaining
𝑥
𝑥
180?− 2𝑥
𝑥 + 20
𝑥 + 50
110?− 2𝑥
2𝑥
90 − 𝑥
90?− 𝑥
90 − 3𝑥
90 − 2𝑥
5𝑥
?
𝑥−𝑦
160 + 𝑥
20 − ?
2𝑥 + 𝑦
30 + 2𝑥 − 𝑦
50 + 𝑥 + 𝑦
100?− 3𝑥
Full example
[JMC 2002 Q21] Given that 𝐴𝐵 = 𝐴𝐶 and 𝑧 < 90, which of the
following expressions must equal 𝑧?
A 𝑥−𝑦
B 𝑥+𝑦
C 𝑥 + 𝑦 − 180
D 180 + 𝑥 − 𝑦
E 180 − 𝑥 + 𝑦
Try to gradually work
out angles in the
diagram (in terms of 𝑥
and 𝑦)
𝐷
Two possible ways:
∠𝐴𝐵𝐶 = 𝑦 (Δ𝐴𝐵𝐶 is isosceles)
∠𝐴𝐷𝐵 = 180 − 𝑥 (angles on straight line sum to 180)
Therefore using angles in Δ𝐴𝐵𝐷:
𝑧 = 180 − 𝑦 − 180 − 𝑥
= 180 − 𝑦 − 180 + 𝑥
=𝑥−𝑦
?
∠𝐴𝐵𝐶 = 𝑦 (Δ𝐴𝐵𝐶 is isosceles)
∠𝐵𝐴𝐶 = 180 − 2𝑦
(angles in triangle 𝐴𝐵𝐶 sum to 180)
∠𝐷𝐴𝐶 = 180 − 𝑥 − 𝑦
(angles in triangle 𝐴𝐷𝐶 sum to 180)
Therefore 𝑧 = ∠𝐵𝐴𝐶 − ∠𝐷𝐴𝐶
= 180 − 2𝑦 − 180 − 𝑥 − 𝑦
=𝑥−𝑦
?
Check Your Understanding
How far can you get down the angle challenge
wall? (do in order, and draw the diagram first)
𝐶
𝐷
 ∠𝐴𝐶𝐵 = 180 − 2𝑥°
?
 ∠𝐵𝐶𝐷 = 2𝑥 − 90°
?
 ∠𝐴𝐵𝐷 = 180 − 𝑥
𝑥
𝐴
𝐵
?
Modelling restrictions on angles
We have already seen we can represent the two base angles of an
isosceles triangle both as 𝑥 to ensure they have the same value.
But how can we model other descriptions?
𝐶
2𝑥
?
𝑥?
𝐴
“Angle 𝐶𝐵𝐴 is twice the
value of angle 𝐶𝐴𝐵.”
𝐵
𝐷
“The line 𝐴𝐶 bisects the
angle 𝐷𝐴𝐵.”
𝐶
?𝑥
𝐴
𝑥?
Bro Note: To ‘bisect’ means
to cut in half.
𝐵
Modelling restrictions on angles
𝐴
In a triangle 𝐴𝐵𝐶, ∠𝐶𝐴𝐵 = 2 ×
∠𝐵𝐶𝐴 and ∠𝐴𝐵𝐶 = 7 × ∠𝐵𝐶𝐴
𝐵
a) Suggest suitable expressions for
each of the angles.
𝒙, 𝟐𝒙,?𝟕𝒙
b) Hence determine ∠𝐵𝐶𝐴.
𝒙 + 𝟐𝒙 + 𝟕𝒙 = 𝟏𝟎𝒙
𝟏𝟎𝒙 = 𝟏𝟖𝟎
?
So 𝒙 = 𝟏𝟖°
𝐶
Exercise 4
1
(on provided worksheet)
Given the value of ∠𝐷𝐵𝐴 indicated, work out the value of ∠𝐷𝐵𝐶.
∠𝑫𝑩𝑨
∠𝑫𝑩𝑪
𝑤
180 − 𝑤
𝑥 + 10
170?− 𝑥
2𝑥 − 30
160?− 2𝑥
90 + 2𝑥
90 ?
− 2𝑥
130 + 𝑥 − 2𝑦
𝐷
?
𝐴
𝐵
𝐶
50 − ?
𝑥 + 2𝑦
Given the values of ∠𝐴𝐵𝐶 and ∠𝐵𝐴𝐶 indicated, work out the value of ∠𝐴𝐶𝐵.
2
𝐴
𝐵
𝐶
∠𝑨𝑩𝑪
∠𝑩𝑨𝑪
∠𝑨𝑪𝑩
𝑥
𝑥
180?− 2𝑥
𝑥
10
170?− 𝑥
𝑥 + 10
150
20?− 𝑥
2𝑥 + 30
20 − 5𝑥
130?+ 3𝑥
4𝑥 − 20
7𝑥 − 30
230 ?
− 11𝑥
Exercise 4
3
(on provided worksheet)
Determine all the angles in each diagram in
terms of 𝑥 and/or 𝑦.
a
c
𝑦
180?− 2𝑦
𝑥
?𝑦
𝑦
b
𝟏𝟖𝟎 −?𝒙 − 𝒚
d
90?− 𝑥
180 − 2𝑥
𝑥
𝟐𝒙 ?
𝒙 +?𝒚
Exercise 4
(on provided worksheet)
h
e
𝟏𝟔𝟎 −?𝒙
20 + 𝑥
150 − 𝑦
𝟑𝟎 − 𝒙?+ 𝒚
𝑥
f
i
𝑥
?
𝒙 − 𝟏𝟖𝟎
?
𝟏𝟖𝟎 − 𝟐𝒙
𝟐𝟕𝟎
?−𝒙
?𝒙
𝑥
g
𝑦 − 10
𝑥 + 20
𝟏𝟕𝟎 −?𝒙 − 𝒚
?
𝟏𝟖𝟎 − 𝟐𝒙
𝟏𝟖𝟎 − 𝒙
?
Exercise 4
(on provided worksheet)
4
Write suitable expressions in terms of 𝑥 for each missing angle. Hence
determine the smallest angle in the diagram in each case.
a
Angle 𝐴𝐵𝐷 is 3 times bigger than angle 𝐷𝐵𝐶.
𝐷
?
3𝑥
b
𝑥
4𝒙 = 𝟏𝟖𝟎
?
∴ 𝒙 = 𝟒𝟓
?
𝐴
𝐵
𝐶
Angle 𝐵𝐶𝐴 is twice as large as angle 𝐵𝐴𝐶.
𝐴
𝑥?
𝟗𝟎° + 𝟑𝒙 = 𝟏𝟖𝟎°
𝒙 = 𝟑𝟎°?
?
2𝑥
𝐵
c
The angles 𝐴𝐶𝐵, 𝐴𝐵𝐶 and 𝐵𝐴𝐶
are in the ratio 3: 4: 5.
𝐴
𝟏𝟐𝒙 = 𝟏𝟖𝟎
∴ 𝒙 = 𝟏𝟓°?
Smallest angle = 𝟒𝟓°
5𝑥
?
4𝑥?
𝐵
𝐶
?3𝑥
𝐶
Exercise 4
5
(on provided worksheet)
[Based on JMO 2005 B4] In this
figure 𝐴𝐷𝐶 is a straight line and
𝐴𝐵 = 𝐵𝐶 = 𝐶𝐷. Also, 𝐷𝐴 = 𝐷𝐵. We
wish to find the size of ∠𝐵𝐴𝐶. Let
∠𝐷𝐴𝐵 = 𝑥.
?
𝑥
?
2𝑥 or 180 − 3𝑥
?
180 − 2𝑥
𝑥
?
180 − 3𝑥
?
𝑥
a) By using the information provided,
determine all the remaining angles in
the diagram in terms of 𝑥.
b) By considering the three angles in
Δ𝐷𝐶𝐵, hence determine ∠𝐵𝐴𝐶.
Solution: 𝟑𝟔°
6
[Based on JMC 2011 Q23] The points 𝑆, 𝑇, 𝑈 lie
on the sides of the triangle 𝑃𝑄𝑅, as shown, so
that 𝑄𝑆 = 𝑄𝑈 and 𝑅𝑆 = 𝑅𝑇. ∠𝑇𝑆𝑈 = 40°.
a) Let ∠𝑇𝑆𝑅 = 𝑥° and ∠𝑈𝑆𝑄 = 𝑦°. What is
the value of 𝑥 + 𝑦?
Solution: 𝟏𝟒𝟎°
b) Using the information provided, find
expressions for ∠𝑇𝑅𝑆 and ∠𝑈𝑄𝑆.
(See diagram)
c) Hence find an expression for ∠𝑅𝑃𝑄.
(See diagram)
d) Using your answer to part (a), hence find
the value of ∠𝑅𝑃𝑄.
𝟐𝒙 + 𝟐𝒚 − 𝟏𝟖𝟎 = 𝟐𝟖𝟎 − 𝟏𝟖𝟎 = 𝟏𝟎𝟎°
?
?
?
2𝑥 + 2𝑦
− 180
?
180 − 2𝑥
𝑥
𝑦
?
180 − 2𝑦
Exercise 4
7
(on provided worksheet)
[Based on TMC Final 2014 Q4] The triangle 𝐴𝐵𝐶 is
isosceles with 𝐴𝐶 = 𝐵𝐶 as shown.
The point 𝐷 lies on the line 𝐵𝐶 such that the triangle
𝐴𝐵𝐷 is isosceles with 𝐴𝐵 = 𝐵𝐷 as shown. ∠𝐵𝐴𝐶 =
2 × ∠𝐵𝐶𝐴 and we wish to work out the value of
∠𝐴𝐷𝐶.
Suppose we let ∠𝐵𝐶𝐴 = 𝑥. Hence determine:
a) ∠𝐵𝐴𝐶
𝟐𝒙 (as specified)
b) ∠𝐶𝐵𝐴
𝟐𝒙
c) Hence by considering the triangle 𝐶𝐴𝐷,
determine 𝑥.
𝟓𝒙 = 𝟏𝟖𝟎° therefore 𝒙 = 𝟑𝟔°
d) Use the remaining information to determine
∠𝐴𝐷𝐶.
𝟏𝟐𝟔°
?
?
?
8
[JMO 2004 B1] In the rectangle
ABCD, M and N are the
midpoints of AB and CD
respectively; AB has length 2
and AD has length 1.
Given that ∠𝐴𝐵𝐷 = 𝑥°,
calculate ∠𝐷𝑍𝑁 in terms of 𝑥.
?
45°
(because Δ𝑀𝐵𝐶 is isosceles
and right-angled)
?
𝑥
Solution: 𝟏𝟑𝟓
? −𝒙
Exercise 4
9
(on provided worksheet)
[Based on JMO 2006 B3] In this diagram, Y lies on the line AC,
triangles ABC and AXY are right angled and in triangle ABX, AX = BX.
The line segment AX bisects angle BAC and angle AXY is seven
times the size of angle XBC.
Let ∠𝑋𝐵𝐶 = 𝑥. In terms of 𝑥, find expressions for:
a) ∠𝐴𝑋𝑌 (see diagram)
?
b) ∠𝐵𝑋𝑌 (hint: the lines 𝑋𝑌 and 𝐵𝐶 are parallel) (see diagram)
?
c) ∠𝑋𝐴𝑌 (see diagram)
?
d) ∠𝐵𝐴𝑋 (see diagram: AX?bisects ∠𝑩𝑨𝑪)
e) ∠𝐵𝑋𝐴 (using the fact that Δ𝐵𝑋𝐴 is isosceles) (see diagram)
?
f) Optional: By considering angles around a suitable point,
hence find ∠𝐴𝐵𝐶.
𝟏𝟒𝒙 + 𝟕𝒙 + 𝟏𝟖𝟎 − 𝒙 = 𝟑𝟔𝟎
𝟐𝟎𝒙 + 𝟏𝟖𝟎 = 𝟑𝟔𝟎
𝟐𝟎𝒙 =
? 𝟏𝟖𝟎
𝒙 = 𝟗°
14𝑥
∠𝑨𝑩𝑪 = 𝟗 + 𝟐𝟕 = 𝟑𝟕°
7𝑥
180 − 𝑥
𝑥
90 − 7𝑥
Exercise 4
(on provided worksheet)
10 [JMC 2003 Q23] In the diagram alongside, 𝐴𝐵 = 𝐴𝐶 and
𝐴𝐷 = 𝐶𝐷. How many of the following statements are true for
the angles marked?
𝑤=𝑥
𝑥 + 𝑦 + 𝑧 = 180
𝑧 = 2𝑥
A all of them
B two
C one
D none of them
E it depends on 𝑥
Solution: A
?
[JMC 2005 Q23] In the diagram, triangle 𝑋𝑌𝑍 is isosceles, with
11 𝑋𝑌 = 𝑋𝑍. What is the value of 𝑟 in terms of 𝑝 and 𝑞?
1
A
𝑝−𝑞
2
D 𝑝+𝑞
Solution: B
1
B
𝑝+𝑞
C 𝑝−𝑞
2
E Impossible to determine
?
12 [JMC 2015 Q25] The four straight lines in the diagram are such
that 𝑉𝑈 = 𝑉𝑊. The sizes of ∠𝑈𝑋𝑍, ∠𝑉𝑌𝑍 and ∠𝑉𝑍𝑋 are
𝑥°, 𝑦° and 𝑧°.
Which of the following equations gives 𝑥 in terms of 𝑦 and 𝑧?
A 𝑥 =𝑦−𝑧
B 𝑥 = 180 − 𝑦 − 𝑧
𝑧
𝑦−𝑧
C 𝑥=𝑦−
D 𝑥 = 𝑦 + 𝑧 − 90 E 𝑥 =
2
2
Solution: E
?
Exercise 4
13
(on provided worksheet)
[JMO 2008 B3] In the diagram ABCD and APQR are congruent rectangles. The
side PQ passes through the point D and ∠𝑃𝐷𝐴 = 𝑥°. Find an expression for
∠𝐷𝑅𝑄 in terms of x.
𝟏
𝟐
Solution: 𝒙
1
𝑥
2
𝑥°
1
90 − 𝑥
2
1
90 − 𝑥
2
?
Proof
Now that we have a number of angle skills, including introducing algebraic angles, we
now have all the skills to form a ‘proof’.
A ‘proof’ is a sequence of justified statements that results in the desired conclusion.
! Examples (which we’ll do later)
𝐴
Using the diagram, prove that
∠𝐴𝐶𝐷 = ∠𝐴𝐵𝐶 + ∠𝐵𝐴𝐶
𝐵
𝐶
𝐷
[JMO 2015 B2] The diagram shows triangle 𝐴𝐵𝐶, in
which ∠𝐴𝐵𝐶 = 72° and ∠𝐶𝐴𝐵 = 84°. The point 𝐸
lies on 𝐴𝐵 so that 𝐸𝐶 bisects ∠𝐵𝐶𝐴. The point 𝐹 lies
on 𝐶𝐴 extended. The point 𝐷 lies on 𝐶𝐵 extended so
that 𝐷𝐴 bisects ∠𝐵𝐴𝐹.
Prove that 𝐴𝐷 = 𝐶𝐸.
STARTER: Constructing diagrams
Sometimes you are not given a diagram, but have to construct one given information.
Can you form a suitable diagram given each of the descriptions?
Make sure that you use marks/arrows to indicate when sides are the same length or parallel, or where angles are equal.
𝐴
“𝐴𝐵𝐶 is an
equilateral triangle.”
𝐵
“𝐴𝐵𝐶 is a triangle such
that 𝐴𝐵 = 𝐵𝐶. A point 𝑃
lies outside the triangle
such that 𝑃𝐴 = 𝑃𝐵.
𝐶
𝑃
𝐵
?
𝐴
𝐴
“𝐴𝐵𝐶 is a triangle where
∠𝐴𝐵𝐶 = 90°. A point 𝐷
lies on 𝐴𝐶 such that
∠𝐴𝐷𝐵 = 90°.”
𝐶
𝐷
?
𝐶
𝐵
𝐴
“The point 𝐹 lies inside
the regular pentagon
𝐴𝐵𝐶𝐷𝐸 so that 𝐴𝐵𝐹𝐸 is a
rhombus.”
𝐸
?
𝐹
𝐷
𝐶
𝐵
Bro Note: For
parallelograms/
rhombuses, we
need not indicate
parallel sides
because it is
implied by the
lengths.
Bro Tip: We name
‘vertices’ (i.e. corners)
using capital letters,
and go either
clockwise or
anticlockwise around
the shape. e.g. If
“𝐴𝐵𝐶𝐷 is a square”,
then the first two
diagrams are OK, but
the last is not.
𝐴
𝐵 𝐶
𝐵


𝐷
𝐷
𝐶
𝐴
𝐴
𝐵

𝐶
𝐷
A harder one for discussion…
Two of the angles of triangle 𝐴𝐵𝐶 are given by ∠𝐶𝐴𝐵 = 2𝛼 and
∠𝐴𝐵𝐶 = 𝛼, where 𝛼 < 45°. The bisector of angle 𝐶𝐴𝐵 meets 𝐵𝐶 at
𝐷. The point 𝐸 lies on the bisector, but outside the triangle, so that
∠𝐵𝐸𝐴 = 90°. When produced, 𝐴𝐶 and 𝐵𝐸 meet at 𝑃.
Construct the diagram.
Bro Note: To ‘bisect’ an
angle is to cut it in half.
So a ‘bisector’ of an
angle is a line which
bisects the angle.
?
Simple Proofs
The simplest proofs just require you to find an angle, but you need to give a
reason for each step.
𝐴𝐵 is parallel to 𝐶𝐷 and 𝐿𝑁𝑀 is a
straight line. Prove that 𝑦 = 112°.
∠𝐿𝑁𝐵 = 68° (corresponding angles are equal)
?
∠𝐿𝑁𝐴 = 112° (angles on a straight line sum to 180°)
Bro Tip: A helpful way to write out each step of the
proof (when referring to angles) is in the form:
∠𝒂𝒏𝒈𝒍𝒆 = 𝒗𝒂𝒍𝒖𝒆
(reason)
Test Your Understanding
i
𝑇𝑃 = 𝑄𝑃 and 𝑃𝑄𝑅 is a straight line.
Prove that ∠𝑇𝑃𝑄 = 40°
∠𝑇𝑄𝑃 = 70° (angles on straight line sum to 180°)
∠𝑄𝑇𝑃 = 70°
? triangle are equal)
(base angles of isosceles
∠𝑇𝑃𝑄 = 40° (angles in triangle sum to 180°)
ii (if you finish)
𝐴
𝐵𝐶𝐷 is a straight line. Let ∠𝐴𝐵𝐶 = 𝑥 and
∠𝐵𝐴𝐶 = 𝑦. Prove that ∠𝐴𝐶𝐷 = 𝑥 + 𝑦.
𝑦
𝑥
𝐵
𝐶
𝐷
∠𝐴𝐶𝐷 = 180 − 𝑥 − 𝑦° (angles in triangle sum to 180°)
∠𝐴𝐶𝐷 = 𝑥 + 𝑦 (angles on a ?
straight line sum to 180°)
Exercise 5a
(on provided worksheet)
1
a
Prove ∠𝑅𝑄𝐶 = 55°. Your proof should only
require one line and be in the form
“∠𝑎𝑛𝑔𝑙𝑒 = 𝑣𝑎𝑙𝑢𝑒 𝑟𝑒𝑎𝑠𝑜𝑛 ”
∠𝑹𝑸𝑪 = 𝟓𝟓° (corresponding angles are equal)
?
𝐴
b
85°
𝐵
𝐸
?
Prove that ∠𝐸𝐹𝐻 = 85°. (Your proof should
consist of two lines)
𝐷
𝐶
(Lots of possible proofs, but here’s one…)
∠𝑪𝑭𝑮 = 𝟖𝟓° (corresponding angles are equal)
∠𝑬𝑭𝑮 = 𝟖𝟓° (vertically opposite angles are equal)
?
𝐺
𝐹
𝐻
c
Prove that ∠𝐵𝐶𝐷 = 2𝑥. (Your proof should consist of
three lines)
𝐵
∠𝑨𝑩𝑪 = 𝒙 (base angles of isosceles triangle 𝑨𝑩𝑪 are equal)
∠𝑩𝑪𝑨 = 𝟏𝟖𝟎 − 𝟐𝒙 (angles in triangle sum to 𝟏𝟖𝟎°)
∠𝑩𝑪𝑫 = 𝟐𝒙 (angles on straight line sum to 𝟏𝟖𝟎°)
?
𝑥
𝐴
?
𝐶
𝐷
Exercise 5a
(on provided worksheet)
2 Draw diagrams which satisfy the following criteria, ensuring you note where lines are of equal
length or parallel. You do NOT need to find any angles.
𝐶
a
𝐴𝐵𝐶 is a right-angled triangle such that ∠𝐶𝐴𝐵 = 90°.
𝐷 is a point on 𝐵𝐶 such that 𝐴𝐷 = 𝐵𝐷.
𝐷
?
b
𝐵
𝐵
𝐴
𝐴𝐵𝐶 is an isosceles triangle where 𝐴𝐵 = 𝐴𝐶. 𝐷 is a
point that lies inside the triangle such that 𝐵𝐶𝐷 is
equilateral.
?𝐷
𝐴
𝐶
𝐷
c
25°
?
𝐴
d
𝐵
𝐶
Points 𝐴, 𝐵 and 𝐶 lie in that order on a straight line.
A point 𝐷, not on the line, is placed such that
𝐴𝐷 = 𝐴𝐵, 𝐵𝐷 = 𝐵𝐶, and ∠𝐵𝐷𝐶 = 25°.
𝑄
𝑃
Points 𝑇 and 𝑈 lie outside a parallelogram 𝑃𝑄𝑅𝑆, and
are such that triangles 𝑅𝑄𝑇 and 𝑆𝑅𝑈 are equilateral
and lie wholly outside the parallelogram.
𝑆
?
𝑈
𝑇
𝑅
Exercise 5a
3
a
(on provided worksheet)
𝐵
𝐴
𝐶
Prove that ∠𝐵𝐶𝐴 = 20°
40°
∠𝑩𝑨𝑫 = 𝟒𝟎° (base angles of isosceles 𝚫 are equal)
∠𝑨𝑩𝑫 = 𝟏𝟎𝟎° (angles in triangle sum to 180)
∠𝑪𝑩𝑫 = 𝟖𝟎° (angles on straight line sum to 180)
∠𝑩𝑫𝑪 = 𝟖𝟎° (base angles of isosceles 𝚫 are equal)
∠𝑩𝑪𝑫 = 𝟐𝟎° (angles in triangle sum to 180)
?
𝐷
b
𝐵
𝐴
70°
𝐷
𝐶
𝐹
In the diagram, 𝐴𝐵 and 𝐶𝐸 are parallel, and
𝐷𝐹 = 𝐷𝐸. Prove that ∠𝐷𝐸𝐹 = 35°
𝐸 ∠𝑩𝑫𝑬 = 𝟕𝟎° (alternate angles are equal)
∠𝑭𝑫𝑬 = 𝟏𝟏𝟎° (angles on a straight line sum to 180)
? of isosceles 𝚫𝑫𝑬𝑭 are
∠𝑫𝑬𝑭 = 𝟑𝟓° (as base angles
equal and angles in triangle sum to 180)
Other Types of Proof
𝐵
𝐵𝐶𝐷 is a straight line and 𝐴𝐵 = 𝐴𝐶.
Prove that 𝐴𝐶 bisects the angle 𝐵𝐴𝐷.
70°
𝐶
𝐴
Before we start this proof, what
specifically are we trying to show?
That ∠𝑩𝑨𝑪?= ∠𝑪𝑨𝑫
30°
𝐷
Bro Note: You need a conclusion so it’s
clear that your proof is complete..
Refer to the provided ‘cheat sheet’.
∠𝐵𝐶𝐴 = 70° (base angles of isosceles
triangle are equal)
∠𝐴𝐶𝐷 = 110° (angles on straight line sum to
180°)
∠𝐵𝐴𝐶 = 40° (angles in triangle sum to 180°)
∠𝐶𝐴𝐷 = 40° (angles in triangle sum to 180°)
∠𝐵𝐴𝐶 = ∠𝐶𝐴𝐷 therefore 𝐴𝐶 bisects ∠𝐵𝐴𝐷.
Proof ?
𝐵
Similarly, what would we need to do in order to:
𝐷
…prove that 𝐴𝐵𝐶 is a straight line?
…prove that 𝐴𝐵𝐶
is isosceles?
?
Show that
∠𝐵𝐴𝐶 = ∠𝐵𝐶𝐴
Show that ∠𝐷𝐵𝐴 + ∠𝐷𝐵𝐶 = 180°
𝐴
𝐵
𝐶
𝐴
?
𝐶
Test Your Understanding
i
𝐶𝐷𝐵 is a straight line and 𝐴𝐶 = 𝐴𝐷 = 𝐷𝐵.
∠𝐷𝐴𝐵 = 72° as shown.
Prove that triangle 𝐴𝐵𝐶 is isosceles.
𝐶
𝐷
∠𝑫𝑩𝑨 = 𝟑𝟔° (base angles of isosceles triangle 𝑨𝑫𝑩 are equal)
𝐵
36°
∠𝑪𝑫𝑨 = 𝟕𝟐° (exterior angle of triangle is sum of two other
interior angles)
Proof ?
∠𝑫𝑪𝑨 = 𝟕𝟐° (base angles of isosceles triangle 𝑨𝑪𝑫 are equal)
𝐴
∠𝑪𝑨𝑫 = 𝟑𝟔° (angles in triangle sum to 𝟏𝟖𝟎°)
∠𝑪𝑨𝑩 = 𝟑𝟔° + 𝟑𝟔° = 𝟕𝟐°
ii (if you finish)
∠𝑩𝑪𝑨 = ∠𝑪𝑨𝑩 = 𝟕𝟐° ∴ triangle 𝑨𝑩𝑪 is isosceles.
𝐷
In the diagram, 𝐴𝐵 = 𝐵𝐶 = 𝐵𝐷 and ∠𝐴𝐷𝐶 =
90°. Prove that 𝐴𝐵𝐶 is a straight line.
𝐶
𝐵
𝐴
Bro Hint: You should choose a
suitable angle to be 𝑥 and start by
writing “Let ∠𝑚𝑦𝑎𝑛𝑔𝑙𝑒 = 𝑥”
Let ∠𝑫𝑨𝑩 = 𝒙.
∠𝑨𝑩𝑫 = 𝒙 (base angles of isosceles triangle are equal)
∠𝑨𝑩𝑫 = 𝟏𝟖𝟎 − 𝟐𝒙 (angles in triangle sum to 𝟏𝟖𝟎°)
∠𝑩𝑫𝑪 = 𝟗𝟎 − 𝒙 (as ∠𝑨𝑫𝑪 = 𝟗𝟎°)
∠𝑩𝑪𝑫 = 𝟗𝟎 − 𝒙 (base angles of isosceles triangle are equal)
∠𝑫𝑩𝑪 = 𝟐𝒙 (angles in triangle 𝑩𝑪𝑫 sum to 𝟏𝟖𝟎°)
∠𝑫𝑩𝑨 + ∠𝑫𝑩𝑪 = 𝟏𝟖𝟎 − 𝟐𝒙 + 𝟐𝒙 = 𝟏𝟖𝟎° therefore 𝑨𝑩𝑪 is a straight line.
Proof ?
Exercise 5b
(on provided worksheet)
𝐷
𝐵
1
a
𝐴
𝐴𝐷𝐶 and 𝐴𝐵𝐶 are right-angled triangles where
∠𝐴𝐶𝐷 = 10° and ∠𝐷𝐶𝐵 = 70°. Prove that triangle
𝐴𝐵𝐶 is isosceles.
10°
∠𝑪𝑨𝑫 = 𝟖𝟎° (angles in triangle 𝑨𝑪𝑫 add to 𝟏𝟖𝟎°).
∠𝑨𝑪𝑩 = ∠𝑨𝑪𝑫 + ∠𝑫𝑪𝑩 = 𝟖𝟎°
∠𝑫𝑨𝑪 = ∠𝑨𝑪𝑩 therefore 𝚫𝑨𝑩𝑪 is isosceles.
70°
?
𝐶
𝐸
b
𝐵
𝐴𝐶 = 𝐵𝐶, 𝐴𝐶𝐷 is a straight line, ∠𝐵𝐶𝐴 = 40°
and ∠𝐵𝐶𝐸 = 70° as per the diagram. Prove
that 𝐴𝐵 and 𝐶𝐸 are parallel.
40° 70°
𝐴
𝐶
𝐷
∠𝑩𝑨𝑪 = 𝟕𝟎° (base angles in isosceles
triangle 𝑨𝑩𝑪 are equal)
∠𝑬𝑪𝑫 = 𝟕𝟎° (angles on straight line at 𝑪
sum to 𝟏𝟖𝟎°)
∠𝑩𝑨𝑪 = ∠𝑬𝑪𝑫 which are corresponding
angles, therefore 𝑨𝑩 and 𝑪𝑬 are parallel.
?
Exercise 5b
(on provided worksheet)
2
As before, 𝐴𝐶 = 𝐵𝐶. Suppose also that 𝐶𝐸
bisects the angle 𝐵𝐶𝐷, but unlike before, no
angles are known.
By introducing a suitable variable (you may wish
to start your proof “Let ∠𝑎𝑛𝑔𝑙𝑒 = 𝑥”) prove that
𝐴𝐵 and 𝐶𝐸 are parallel.
𝐸
𝐵
𝐴
𝐶
𝐷
Let ∠𝑬𝑪𝑫 = 𝒙.
∠𝑩𝑪𝑬 = 𝒙 (as 𝑪𝑬 bisects ∠𝑩𝑪𝑫).
∠𝑩𝑪𝑨 = 𝟏𝟖𝟎 − 𝟐𝒙 (angles on straight line
at 𝑪 sum to 𝟏𝟖𝟎°)
?
𝟏𝟖𝟎− 𝟏𝟖𝟎−𝟐𝒙
∠𝑩𝑨𝑪 =
= 𝒙 (base angles of
𝟐
isosceles triangle 𝑨𝑩𝑪 are equal)
∠𝑩𝑨𝑪 = ∠𝑬𝑪𝑫 = 𝒙, and 𝑨𝑪𝑫 is a straight
line, therefore 𝑨𝑩 and 𝑪𝑬 are parallel.
Exercise 5b
3
(on provided worksheet)
[JMO 2001 B3] In the diagram, B is the midpoint of AC and the lines AP,
BQ and CR are parallel. The bisector of ∠𝑃𝐴𝐵 meets BQ at Z.
Draw a diagram to show this, and join Z to C.
(i) Given that ∠𝑃𝐴𝑍 = 𝑥°, find ∠𝑍𝐵𝐶 in terms of x.
(ii) Show that CZ bisects ∠𝐵𝐶𝑅.
?
Exercise 5b
4
(on provided worksheet)
[JMO 2015 B2] The diagram shows triangle 𝐴𝐵𝐶, in which ∠𝐴𝐵𝐶 = 72° and
∠𝐶𝐴𝐵 = 84°. The point 𝐸 lies on 𝐴𝐵 so that 𝐸𝐶 bisects ∠𝐵𝐶𝐴. The point 𝐹 lies
on 𝐶𝐴 extended. The point 𝐷 lies on 𝐶𝐵 extended so that 𝐷𝐴 bisects ∠𝐵𝐴𝐹.
Prove that 𝐴𝐷 = 𝐶𝐸.
?
Exercise 5b
5
(on provided worksheet)
[JMO 2011 B4] In a triangle ABC, M lies on AC and N lies on AB so that ∠𝐵𝑁𝐶 =
4𝑥°, ∠𝐵𝐶𝑁 = 6𝑥° and ∠𝐵𝑀𝐶 = ∠𝐶𝐵𝑀 = 5𝑥°. Prove that triangle ABC is
isosceles.
?
Exercise 5b
6
(on provided worksheet)
[JMO 2015 B4] The point 𝐹 lies inside the regular pentagon 𝐴𝐵𝐶𝐷𝐸 so that
𝐴𝐵𝐹𝐸 is a rhombus. Prove that 𝐸𝐹𝐶 is a straight line.
(Hint: the interior angle of a pentagon is 108°)
?
Exercise 5b
7
(on provided worksheet)
[Hamilton 2006 Q2] In triangle ABC, ∠𝐴𝐵𝐶 is a right angle. Points P and Q lie
on AC; BP is perpendicular to AC; BQ bisects ∠𝐴𝐵𝑃.
Prove that CB = CQ.
?
Exercise 5b
(on provided worksheet)
8
𝐴𝐵𝐶 is a triangle such that ∠𝐴𝐵𝐶 = 90° and 𝐷 is a
point on the line 𝐴𝐶 such that 𝐵𝐷 = 𝐶𝐷. By letting
∠𝐷𝐶𝐵 = 𝑥 or otherwise, prove that 𝐶𝐷 = 𝐷𝐴.
(hint: prove that Δ𝐴𝐵𝐷 is isosceles first).
𝐴
𝐷?
Diagram
𝐶
𝐵
Let ∠𝑫𝑪𝑩 = 𝒙.
∠𝑫𝑩𝑪 = 𝒙 (base angles of isosceles 𝚫 are equal)
∠𝑪𝑫𝑩 = 𝟏𝟖𝟎 − 𝟐𝒙 (angles in triangle sum to 180)
∠𝑨𝑫𝑩 = 𝟐𝒙 (angles on straight line sum to 180)
∠𝑨𝑩𝑫 = 𝟗𝟎 − 𝒙 (as ∠𝑫𝑩𝑪
? + ∠𝑫𝑩𝑨 = 𝟗𝟎°)
∠𝑪𝑨𝑩 = 𝟗𝟎 − 𝒙 (as angles in 𝚫𝑨𝑩𝑪 add to 180)
Proof= ∠𝑫𝑨𝑩
∴ 𝑨𝑩𝑫 is isosceles as ∠𝑨𝑩𝑫
∴ 𝑨𝑫 = 𝑩𝑫 = 𝑫𝑪
□