What is G5: Congruence criteria?

Congruent triangles have exactly the same size and shape, so corresponding sides and angles are equal. Prove congruence using SSS, SAS, ASA or RHS.

How many marks are G5 questions worth?

In verified past-paper appearances, G5 questions on Edexcel GCSE Maths 1MA1 have been worth 3–4 marks.

Which past exam papers has G5 appeared in?

G5 has appeared in 4 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Use the basic congruence criteria for triangles (SSS, SAS, ASA, RHS)

Congruence criteria

Worked answers, methods and verified real exam appearances for G5 on Edexcel GCSE Maths 1MA1.

Explanation

Congruent triangles have exactly the same size and shape, so corresponding sides and angles are equal. Prove congruence using SSS, SAS, ASA or RHS.
SAS needs the angle included between the two known sides. RHS applies only to right-angled triangles and needs equal hypotenuses plus one equal shorter side.
Mark or state the three matching facts, then write triangle names in corresponding order.
Only after proving congruence may you use corresponding parts to establish another equality.
AAA proves similarity rather than congruence, and SSA does not determine a unique triangle, so neither is a valid general congruence test.

Matching two sides and the included angle establishes SAS congruence.

Worked example

Triangles $ABC$ and $DEF$ satisfy $AB=DE$ , $AC=DF$ and $\angle BAC=\angle EDF$ . Prove that $BC=EF$ .

1. $AB=DE$ and $AC=DF$ give two pairs of corresponding equal sides.
2. $\angle BAC=\angle EDF$ is the included angle between those pairs.
3.Therefore $\triangle ABC\cong\triangle DEF$ by SAS, so corresponding sides $BC$ and $EF$ are equal.

Answer: $BC=EF$ by corresponding sides of congruent triangles.

Common mistakes

Don't use SAS with an angle that is not between the two stated equal sides.
Don't write AAA as a congruence proof even though the triangles could have different sizes.

Exam tip

In a proof, state the three matching facts before naming the congruence criterion and the corresponding conclusion.

Worked practice

Tier 1 · Easy

Triangle

ABC

has side lengths

5

cm,

7

cm and

9

cm. Triangle

PQR

has side lengths

5

cm,

7

cm and

9

cm. State the congruence criterion.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	SSS	1	All three side lengths of one triangle match all three side lengths of the other, so the criterion is SSS.

Tier 2 · Standard

Triangles

ABC

and

DEF

are right-angled at

B

and

E

. Also

AC=DF=13

cm and

AB=DE=5

cm. Explain why the triangles are congruent.

(2)

(Total for Question 2 is 2 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	They are congruent by RHS.	2	Both triangles are right-angled, their hypotenuses $AC$ and $DF$ are equal, and one corresponding pair of shorter sides $AB$ and $DE$ is equal. Therefore the triangles are congruent by RHS.

Tier 3 · Hard

In quadrilateral

ABCD

, diagonal

AC

is drawn. Given that

AB=AD

and

\angle BAC=\angle CAD

, prove that

BC=CD

(3)

(Total for Question 3 is 3 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$BC=CD$	3	Compare triangles $BAC$ and $DAC$ . We have $AB=AD$ , $AC$ is common, and the included angles $\angle BAC$ and $\angle CAD$ are equal. The triangles are congruent by SAS, so corresponding sides $BC$ and $DC$ are equal.

Tier 1 · Easy

Triangles

ABC

and

DEF

have

AB=DE

BC=EF

and

\angle ABC=\angle DEF

. A student calls this SSA because the angle was stated last. Write down the correct congruence criterion.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	SAS	1	The equal angle is included between the two equal sides in each triangle. The order in which the facts are stated does not matter, so the correct criterion is SAS.

Tier 2 · Standard

Triangles

JKL

and

MNP

satisfy

\angle J=\angle M

\angle K=\angle N

and

JK=MN

. Explain why the triangles are congruent.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$\triangle JKL\cong\triangle MNP$ by ASA.	2	Two corresponding angles are equal, and the equal side $JK=MN$ lies between those angles. Therefore the triangles are congruent by ASA.

Tier 3 · Hard

Triangle

ABC

is isosceles with

AB=AC

. Point

D

is the midpoint of

BC

, and

AD

is drawn. Prove that

AD

is perpendicular to

BC

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$AD\perp BC$	4	In triangles $ABD$ and $ACD$ , $AB=AC$ , $BD=DC$ because $D$ is the midpoint, and $AD$ is common. The triangles are congruent by SSS, so $\angle ADB=\angle ADC$ . These adjacent angles form a straight line and total $180^\circ$ , so each is $90^\circ$ and $AD\perp BC$ .

Tier 2 · Standard

Triangles

ABC

and

PQR

satisfy

AB=PQ

BC=QR

and

AC=PR

. Write down a congruence statement with the vertices in corresponding order, and write down the angle in triangle

PQR

that corresponds to

\angle ACB

(2)

(Total for Question 7 is 2 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$\triangle ABC\cong\triangle PQR$ ; $\angle PRQ$	2	The side matches give $A\leftrightarrow P$ , $B\leftrightarrow Q$ and $C\leftrightarrow R$ , so $\triangle ABC\cong\triangle PQR$ by SSS. Therefore $\angle ACB$ , whose vertex is $C$ , corresponds to $\angle PRQ$ , whose vertex is $R$ .

Tier 3 · Hard

Line segment

AB

11

cm long. Points

C

and

D

lie on opposite sides of

AB

. Angles

ACB

and

ADB

are right angles, and

AC=AD=7

cm. Prove that

BC=BD

(3)

(Total for Question 8 is 3 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$\triangle ACB\cong\triangle ADB$ by RHS, so $BC=BD$ .	3	Triangles $ACB$ and $ADB$ are right-angled. They have the common hypotenuse $AB$ and the equal shorter sides $AC=AD$ . Therefore the triangles are congruent by RHS, so the corresponding sides $BC$ and $BD$ are equal.

Tier 3 · Hard

The horizontal segment

AB

8

cm long. Point

C

is above

AB

and point

D

is below

AB

. Angles

CAB

DAB

CBA

and

DBA

are each

45^\circ

. Prove that

AB

is the perpendicular bisector of

CD

. Give a reason for each stage.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$\triangle ACB\cong\triangle ADB$ by ASA; $AC=AD$ and $BC=BD$ , so $A$ and $B$ lie on the perpendicular bisector of $CD$ and hence $AB$ is that perpendicular bisector	5	We have $\angle CAB=\angle DAB=45^\circ$ , $\angle CBA=\angle DBA=45^\circ$ , and common included side $AB$ , so $\triangle ACB\cong\triangle ADB$ by ASA. Corresponding sides give $AC=AD$ and $BC=BD$ . Therefore both $A$ and $B$ are equidistant from $C$ and $D$ , so both lie on the perpendicular bisector of $CD$ ; the unique line through them is $AB$ . The configuration exists uniquely: with $A=(0,0)$ and $B=(8,0)$ , the fixed $45^\circ$ rays meet once above at $C=(4,4)$ and once below at $D=(4,-4)$ .

Q10

Tier 3 · Hard

In quadrilateral

ABCD

, the vertices are in order,

B

and

D

lie on opposite sides of

AC

AB=CD

and

BC=AD

. Diagonal

AC

is drawn. Prove that both pairs of opposite sides are parallel, and hence that

ABCD

is a parallelogram.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$\triangle ABC\cong\triangle CDA$ by SSS; $AB\parallel CD$ and $BC\parallel AD$ , so $ABCD$ is a parallelogram	4	Triangles $ABC$ and $CDA$ have $AB=CD$ , $BC=AD$ and common side $AC$ , so they are congruent by SSS. Hence $\angle BAC=\angle DCA$ , making $AB\parallel CD$ by alternate angles. Also $\angle BCA=\angle CAD$ , making $BC\parallel AD$ . A quadrilateral with both pairs of opposite sides parallel is a parallelogram. Coordinates $A=(0,0)$ , $B=(3,1)$ , $C=(5,4)$ and $D=(2,3)$ verify a convex labelled configuration with $B$ and $D$ on opposite sides of $AC$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-11	2H	Q18	3	Allowed	Higher	QP MS
2022-11	3H	Q20	3	Allowed	Higher	QP MS
2019-11	1H	Q22	4	Non-calculator	Higher	QP MS
2019-11	1F	Q29	4	Non-calculator	Foundation	QP MS

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