GCSE Congruence Practice Questions

Write down the correspondence the congruence statement gives

Writing triangle $ABC$ is congruent to triangle $PQR$ names the vertices in matching order, so the pairs of equal parts can be read straight off.

Match the part you want to the part you know

The part asked for corresponds to a part that is known, so the two are equal.

State the answer

So $QR = BC = 5$ cm.

Write down the correspondence the congruence statement gives

Writing triangle $ABC$ is congruent to triangle $PQR$ names the vertices in matching order, so the pairs of equal parts can be read straight off.

Use the angle sum of a triangle to find the missing angle

The three angles of a triangle add up to $180^\circ$, which gives the angle that was not stated.

State the answer

So $\angle PRQ = \angle ACB = 65^\circ$.

Recall the five conditions that prove congruence

A set of information proves congruence only if it matches one of SSS, SAS, ASA, AAS or RHS.

Read off the pattern of the set that works

Written in order round the triangle, one of the sets reads A, A, S (the side is not between the two angles) — that is the pattern AAS.

Check the first wrong set

The set $\angle EDF = \angle QPR$, $\angle DEF = \angle PQR$ and $\angle DFE = \angle PRQ$ is AAA — angles alone fix the shape but not the size, so it cannot prove the two triangles are congruent.

Check the second wrong set

The set $DE = PQ$, $EF = QR$ and $\angle DFE = \angle PRQ$ is SSA — two sides and an angle that is not between them leaves two possible triangles, so it cannot prove the two triangles are congruent.

Say why the failing sets fail

SSA leaves two possible triangles and AAA leaves a whole family of similar triangles, so neither can prove congruence.

State the answer

So the set that is enough is the AAS set: $\angle EDF = \angle QPR$, $\angle DEF = \angle PQR$ and $EF = QR$

Recall the five conditions that prove congruence

A set of information proves congruence only if it matches one of SSS, SAS, ASA, AAS or RHS.

Say what all five have in common

Each of the five fixes all three side lengths — by the cosine rule, the sine rule or Pythagoras — and three sides fix a triangle.

Read off the pattern of the set that works

Written in order round the triangle, one of the sets reads R, H, S (right angle, hypotenuse, one other side) — that is the pattern RHS.

Check the first wrong set

The set $AB = RS$, $BC = ST$ and $\angle ACB = \angle RTS$ is SSA — two sides and an angle that is not between them leaves two possible triangles, so it cannot prove the two triangles are congruent.

Check the second wrong set

The set $CA = TR$, $AB = RS$ and $\angle ACB = \angle RTS$ is SSA — two sides and an angle that is not between them leaves two possible triangles, so it cannot prove the two triangles are congruent.

Check the third wrong set

The set $\angle BAC = \angle SRT$, $\angle ACB = \angle RTS$ and $\angle ABC = \angle RST$ is AAA — angles alone fix the shape but not the size, so it cannot prove the two triangles are congruent.

Check the last wrong set

The set $AB = RS$ and $\angle BAC = \angle SRT$ is not enough — there are simply not enough parts to fix a triangle, so it cannot prove the two triangles are congruent.

Say why the failing sets fail

SSA leaves two possible triangles and AAA leaves a whole family of similar triangles, so neither can prove congruence.

Note the exception

The one time two sides and a non-included angle do work is when that angle is a right angle. That case has its own name: RHS.

State the answer

So the set that is enough is the RHS set: $\angle ABC = \angle RST = 90^\circ$, $CA = TR$ and $BC = ST$

Recall the five conditions that prove congruence

A set of information proves congruence only if it matches one of SSS, SAS, ASA, AAS or RHS.

Say what all five have in common

Each of the five fixes all three side lengths — by the cosine rule, the sine rule or Pythagoras — and three sides fix a triangle.

Read off the pattern of the set that works

Written in order round the triangle, one of the sets reads S, S, S (three sides) — that is the pattern SSS.

Mark the correct set on a diagram

Marking the equal parts on both triangles is the quickest way to see which condition the information matches.

Check the first wrong set

The set $AB = UV$, $CA = WU$ and $\angle ABC = \angle UVW$ is SSA — two sides and an angle that is not between them leaves two possible triangles, so it cannot prove the two triangles are congruent.

Check the second wrong set

The set $\angle BAC = \angle VUW$, $\angle ABC = \angle UVW$ and $\angle ACB = \angle UWV$ is AAA — angles alone fix the shape but not the size, so it cannot prove the two triangles are congruent.

Check the third wrong set

The set $BC = VW$ and $\angle BAC = \angle VUW$ is not enough — there are simply not enough parts to fix a triangle, so it cannot prove the two triangles are congruent.

Check the last wrong set

The set $AB = UV$ and $BC = VW$ is not enough — there are simply not enough parts to fix a triangle, so it cannot prove the two triangles are congruent.

Say why the failing sets fail

SSA leaves two possible triangles and AAA leaves a whole family of similar triangles, so neither can prove congruence.

Note the exception

The one time two sides and a non-included angle do work is when that angle is a right angle. That case has its own name: RHS.

Note that the order of the letters matters

SAS and SSA use the same three parts; only the position of the angle differs, and that is the difference between a proof and no proof.

Write the congruence statement for the correct set

With that information triangle $ABC$ and triangle $UVW$ are congruent, so every remaining pair of parts is equal as well.

Say what congruence then gives you

Once two triangles are proved congruent, all three pairs of sides and all three pairs of angles are equal — that is what makes congruence worth proving.

Summarise the method

List the equal parts, see which of the five conditions they match, and only then write the conclusion.

State the answer

So the set that is enough is the SSS set: $AB = UV$, $BC = VW$ and $CA = WU$