What is G6: Angle proofs?

Combine established geometry facts to derive a new result. Equal sides in an isosceles triangle face equal angles, and equal angles face equal sides.

How many marks are G6 questions worth?

In verified past-paper appearances, G6 questions on Edexcel GCSE Maths 1MA1 have been worth 3–5 marks.

Which past exam papers has G6 appeared in?

G6 has appeared in 13 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Apply angle facts, congruence, similarity and quadrilateral properties to derive results about angles and sides, incl. Pythagoras' theorem and isosceles base angles, and obtain simple proofs

Angle proofs

Worked answers, methods and verified real exam appearances for G6 on Edexcel GCSE Maths 1MA1.

Explanation

Combine established geometry facts to derive a new result. Equal sides in an isosceles triangle face equal angles, and equal angles face equal sides.
In a right-angled triangle, Pythagoras' theorem gives $a^2+b^2=c^2$ , where $c$ is always the hypotenuse opposite the right angle.
Angle facts, quadrilateral properties, similarity and congruence can then form a proof chain.
Each line must follow from given information or a named result; a diagram's appearance is not evidence.
For a 'prove' question, show the intermediate equality or congruence and finish with the exact conclusion requested.

Equal sides in an isosceles triangle face equal base angles.

Worked example

An isosceles triangle has equal sides $13$ cm and base $10$ cm. Find its perpendicular height.

1.The perpendicular from the apex bisects the $10$ cm base, giving a right triangle with base $5$ cm.
2.Apply Pythagoras: $h^2+5^2=13^2$ .
3. $h^2=169-25=144$ , so $h=12$ cm, taking the positive root.

Answer: The perpendicular height is $12$ cm.

Common mistakes

Don't use the full $10$ cm base in the right triangle instead of the bisected length $5$ cm.
Don't treat a sloping side as the hypotenuse without checking which side is opposite the right angle.

Exam tip

A simple proof needs a reason beside each claim; a correct final statement without the linked reasons does not earn all proof marks.

Worked practice

Tier 1 · Easy

Triangle

ABC

is isosceles with

AB=AC

. Angle

B

47^\circ

. Work out angle

A

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$86^\circ$	2	Equal sides face equal angles, so angle $C$ is also $47^\circ$ . The angles in a triangle sum to $180^\circ$ , giving $A=180^\circ-47^\circ-47^\circ=86^\circ$ .

Tier 2 · Standard

A right-angled triangle has perpendicular sides

9

cm and

12

cm. Work out the length of its hypotenuse.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$15$ cm	3	By Pythagoras' theorem, $c^2=9^2+12^2=81+144=225$ . Therefore $c=\sqrt{225}=15$ cm.

Tier 3 · Hard

In quadrilateral

ABCD

AB\parallel CD

and

AB=CD

. Diagonal

AC

is drawn. Prove that

BC=AD

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$BC=AD$	4	Because $AB\parallel CD$ , $\angle BAC=\angle DCA$ as alternate angles. Also $AB=CD$ and $AC$ is common. Thus triangles $BAC$ and $DCA$ are congruent by SAS. Corresponding sides $BC$ and $DA$ are therefore equal.

Tier 1 · Easy

A right-angled triangle has hypotenuse

10

cm and one shorter side

6

cm. Work out the other shorter side.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$8$ cm	2	Let the missing side be $x$ . By Pythagoras, $x^2+6^2=10^2$ , so $x^2=100-36=64$ and $x=8$ cm.

Tier 2 · Standard

Two triangles are similar. A side of length

8

cm in the smaller triangle corresponds to a side of length

20

cm in the larger triangle. Another side of the smaller triangle is

14

cm. Work out the corresponding side of the larger triangle.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$35$ cm	2	The scale factor from the smaller triangle to the larger is $20\div8=2.5$ . The corresponding length is $14\times2.5=35$ cm.

Tier 3 · Hard

The diagonals of a rhombus are

10

cm and

24

cm. They bisect each other at right angles. Work out the perimeter of the rhombus.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$52$ cm	4	The half-diagonals are $5$ cm and $12$ cm. They form a right-angled triangle whose hypotenuse is one side of the rhombus, so the side is $\sqrt{5^2+12^2}=\sqrt{169}=13$ cm. The perimeter is $4\times13=52$ cm.

Tier 2 · Standard

A triangle has side lengths

9

cm,

40

cm and

41

cm. Show that the triangle is right-angled, then work out its area.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$9^2+40^2=41^2$ , so the triangle is right-angled; area $=180$ cm $^2$	3	$9^2+40^2=81+1600=1681$ and $41^2=1681$ , so the triangle is right-angled by the converse of Pythagoras' theorem. The perpendicular sides are $9$ cm and $40$ cm, giving area $\frac12\times9\times40=180$ cm $^2$ .

Tier 3 · Hard

Quadrilateral

ABCD

is a rectangle with

AB=12

cm and

BC=5

cm. Point

E

lies on side

DC

and

AE=BE

. Prove that

E

is the midpoint of

DC

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$DE=EC$ , so $E$ is the midpoint of $DC$	4	Compare right-angled triangles $ADE$ and $BCE$ . Their hypotenuses are equal because $AE=BE$ , and $AD=BC$ because opposite sides of a rectangle are equal. The triangles are congruent by RHS, so corresponding sides $DE$ and $EC$ are equal. Since $E$ lies on $DC$ , it is the midpoint of $DC$ .

Tier 3 · Hard

Convex trapezium

ABCD

has

AB\parallel CD

, with the shorter side

CD

directly above

AB

. Given

AB=21

cm,

CD=11

cm and

AD=BC=13

cm, work out the perpendicular height of the trapezium and the length of diagonal

AC

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Height $=12$ cm; $AC=20$ cm	5	The equal non-parallel sides make an isosceles trapezium, so the total overhang $21-11=10$ cm splits equally into $5$ cm at each end. A right triangle at an end has hypotenuse $13$ and base $5$ , giving height $\sqrt{13^2-5^2}=12$ cm. From $A$ to $C$ , the horizontal distance is $5+11=16$ cm, so $AC=\sqrt{16^2+12^2}=20$ cm. Coordinates $A=(0,0)$ , $B=(21,0)$ , $D=(5,12)$ , $C=(16,12)$ verify the unique labelled configuration.

Q10

Tier 3 · Hard

Triangle

ABC

is right-angled at

A

. The perpendicular from

A

meets hypotenuse

BC

D

, where

BD=9

cm and

DC=16

cm. By using similar triangles, work out

AD

AB

and

AC

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$AD=12$ cm; $AB=15$ cm; $AC=20$ cm	5	The altitude creates three similar right-angled triangles. Similarity gives $AD^2=BD\times DC=9\times16=144$ , so $AD=12$ cm. Also $BC=9+16=25$ cm. From the corresponding sides, $AB^2=BD\times BC=9\times25=225$ and $AC^2=DC\times BC=16\times25=400$ . Therefore $AB=15$ cm and $AC=20$ cm. Coordinates $D=(0,0)$ , $B=(-9,0)$ , $C=(16,0)$ and $A=(0,12)$ verify all lengths and the right angle at $A$ , fixing the labelled configuration up to reflection.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	2H	Q18	3	Allowed	Higher	QP MS
2024-06	3F	Q20	5	Allowed	Foundation	QP MS
2022-11	3H	Q20	3	Allowed	Higher	QP MS
2024-11	2F	Q13	4	Allowed	Foundation	QP MS
2022-06	3F	Q20	5	Allowed	Foundation	QP MS
2023-11	1H	Q6	4	Non-calculator	Higher	QP MS
2019-11	2F	Q17	4	Allowed	Foundation	QP MS
2021-11	3H	Q23	4	Allowed	Higher	QP MS
2019-06	2H	Q18	5	Allowed	Higher	QP MS
2024-06	2H	Q22	4	Allowed	Higher	QP MS
2023-06	3F	Q22	5	Allowed	Foundation	QP MS
2022-06	3H	Q15	3	Allowed	Higher	QP MS
2023-06	3H	Q3	5	Allowed	Higher	QP MS

Other points in G Geometry and measures

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