Congruent Triangles – GCSE Maths

1 What "Congruent" Means

Two shapes are congruent if they are exactly the same shape and the same size. One could be cut out and placed perfectly on top of the other.

Congruent — identical in both shape and size. Scale factor $1$.
Similar — the same shape but a different size. Scale factor other than $1$.
Congruent shapes are a special case of similar shapes.

Congruent shapes may be turned over or rotated. A shape and its mirror image are still congruent — you are allowed to flip it before placing it on top.

Notation

$\triangle ABC \cong \triangle PQR$ means "triangle $ABC$ is congruent to triangle $PQR$"

The order of the letters matters. Writing $\triangle ABC \cong \triangle PQR$ states that $A$ corresponds to $P$, $B$ to $Q$ and $C$ to $R$ — so $AB = PQ$, $BC = QR$, $AC = PR$, $\angle A = \angle P$, and so on.

2 The Four Congruence Criteria

You do not need to check all six measurements (three sides and three angles). Any one of these four sets of three is enough.

Code	Stands for	What you must show
SSS	Side, Side, Side	All three pairs of sides are equal
SAS	Side, Angle, Side	Two pairs of sides equal, and the angle between them equal
ASA	Angle, Side, Angle	Two pairs of angles equal, and a pair of corresponding sides equal
RHS	Right angle, Hypotenuse, Side	Both have a right angle, equal hypotenuses, and one other pair of equal sides

3 Why SSA and AAA Do Not Work

SSA is not a congruence criterion. If the equal angle is not between the two equal sides, two genuinely different triangles can be built from the same three measurements — the "ambiguous case".

AAA is not a congruence criterion either. Three equal angles guarantee the shapes are similar, but they could be any size at all. A tiny equilateral triangle and a huge one both have angles $60^\circ$, $60^\circ$, $60^\circ$.

Memory hook: the valid four are SSS, SAS, ASA, RHS. The invalid two are SSA and AAA. Every valid criterion contains at least one side, and in SAS the angle must be the included one.

4 Writing a Congruence Proof

Exam questions ask you to prove congruence. There is a standard four-line layout that scores full marks.

State the first pair of equal parts, with a reason.
State the second pair, with a reason.
State the third pair, with a reason.
Write the conclusion: "therefore $\triangle ABC \cong \triangle PQR$ (SAS)" — naming the criterion.

Useful reasons to have ready: "given", "common side" (a side shared by both triangles), "vertically opposite angles", "alternate angles, $AB \parallel CD$", "base angles of an isosceles triangle", "radii of the same circle", "$M$ is the midpoint".

Worked Example 1 — A full SAS proof

In the diagram, $AB$ and $CD$ are two straight lines crossing at $M$, with $AM = DM$ and $BM = CM$. Prove that $\triangle AMB \cong \triangle DMC$.

①$AM = DM$ (given)

②$\angle AMB = \angle DMC$ (vertically opposite angles)

③$BM = CM$ (given)

④The equal angle lies between the two equal sides.

Therefore $\triangle AMB \cong \triangle DMC$ (SAS).

Worked Example 2 — Using a common side

$ABCD$ is a kite with $AB = AD$ and $CB = CD$. Prove that $\triangle ABC \cong \triangle ADC$, and hence that $\angle ABC = \angle ADC$.

①$AB = AD$ (given, property of the kite)

②$CB = CD$ (given, property of the kite)

③$AC = AC$ (common side — shared by both triangles)

Therefore $\triangle ABC \cong \triangle ADC$ (SSS).

Because the triangles are congruent, all matching parts are equal. $\angle ABC$ in the first triangle matches $\angle ADC$ in the second, so $\angle ABC = \angle ADC$. ∎

This is exactly why a kite has one pair of equal opposite angles.

Worked Example 3 — An ASA proof with parallel lines

$ABCD$ is a parallelogram whose diagonals meet at $O$. Prove that $\triangle AOB \cong \triangle COD$.

①$\angle OAB = \angle OCD$ (alternate angles, $AB \parallel DC$)

②$AB = CD$ (opposite sides of a parallelogram are equal)

③$\angle OBA = \angle ODC$ (alternate angles, $AB \parallel DC$)

Therefore $\triangle AOB \cong \triangle COD$ (ASA).

It follows that $AO = CO$ and $BO = DO$ — which proves that the diagonals of a parallelogram bisect each other. ∎

Worked Example 4 — An RHS proof

Triangle $ABC$ is isosceles with $AB = AC$. $AD$ is the perpendicular from $A$ to $BC$. Prove that $BD = DC$.

①$\angle ADB = \angle ADC = 90^\circ$ (given, $AD \perp BC$) — the R

②$AB = AC$ (given, isosceles) — these are the hypotenuses, the H

③$AD = AD$ (common side) — the S

Therefore $\triangle ABD \cong \triangle ACD$ (RHS).

Matching sides are equal, so $BD = DC$. The perpendicular from the apex of an isosceles triangle bisects the base. ∎

5 Using Congruence to Find Lengths and Angles

The payoff

Once two triangles are congruent, every pair of corresponding sides and angles is equal.

Worked Example 5 — Finding a length

$\triangle PQR \cong \triangle XYZ$. Given $PQ = 8$ cm, $QR = 11$ cm, $\angle P = 47^\circ$ and $XZ = 6$ cm, write down $XY$, $YZ$, $PR$ and $\angle X$.

①The letter order tells you $P \leftrightarrow X$, $Q \leftrightarrow Y$, $R \leftrightarrow Z$.

②$XY$ matches $PQ$, so $XY = 8$ cm.

③$YZ$ matches $QR$, so $YZ = 11$ cm.

④$PR$ matches $XZ$, so $PR = 6$ cm.

⑤$\angle X$ matches $\angle P$, so $\angle X = 47^\circ$.

Congruent shapes have equal areas and equal perimeters. The converse is false: two shapes with the same area need not be congruent — a $2 \times 6$ rectangle and a $3 \times 4$ rectangle both have area $12$ but are quite different.

6 Quick Reference

Congruent

Same shape and same size; reflections still count.

SSS

All three sides equal.

SAS

Two sides and the included angle.

ASA

Two angles and a corresponding side.

RHS

Right angle, equal hypotenuses, one other equal side.

Not valid

SSA (ambiguous) and AAA (only proves similarity).

Proof layout

Three statements with reasons, then the criterion named.

Handy reasons

Common side, vertically opposite, alternate angles, given.

The payoff

Congruence makes all corresponding parts equal.

7 Practice Questions

Question 1

State which congruence criterion (if any) proves each pair of triangles congruent.

(a) Sides $5$, $7$, $9$ and sides $9$, $5$, $7$.
(b) Two sides $6$ and $8$ with the angle between them $50^\circ$, in both.
(c) All three angles $40^\circ$, $60^\circ$, $80^\circ$ in both.

▶ Show solution

(a) SSS — the same three lengths, just listed in a different order.

(b) SAS — the angle is between the two given sides.

(c) Not congruent (AAA). The triangles are similar, but could be any size.

Question 2

Explain why knowing two sides and a non-included angle is not enough to prove congruence.

▶ Show solution

This is the SSA or "ambiguous" case.

If the angle is not between the two known sides, the third vertex can fall in two different places — an arc of the given radius can cut the base line twice.

That produces two triangles with the same three measurements but different shapes, so congruence is not guaranteed.

Question 3

$\triangle ABC \cong \triangle DEF$. Given $AB = 7$ cm, $\angle B = 62^\circ$ and $EF = 9$ cm, write down $DE$, $\angle E$ and $BC$.

▶ Show solution

The letter order gives $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$.

$DE = AB = 7$ cm

$\angle E = \angle B = 62^\circ$

$BC = EF = 9$ cm

Question 4

$M$ is the midpoint of both $AC$ and $BD$. Prove that $\triangle ABM \cong \triangle CDM$.

▶ Show solution

$AM = CM$ ($M$ is the midpoint of $AC$)

$\angle AMB = \angle CMD$ (vertically opposite angles)

$BM = DM$ ($M$ is the midpoint of $BD$)

The equal angle is between the two pairs of equal sides.

Therefore $\triangle ABM \cong \triangle CDM$ (SAS).

Question 5

In triangle $ABC$, $AB = AC$. Prove that $\angle ABC = \angle ACB$ (the base angles of an isosceles triangle are equal).

▶ Show solution

Let $M$ be the midpoint of $BC$, and join $AM$.

$AB = AC$ (given)

$BM = CM$ ($M$ is the midpoint)

$AM = AM$ (common side)

Therefore $\triangle ABM \cong \triangle ACM$ (SSS).

Corresponding angles of congruent triangles are equal, so $\angle ABM = \angle ACM$, i.e. $\angle ABC = \angle ACB$. ∎

Question 6

Two right-angled triangles each have a hypotenuse of $13$ cm and one other side of $5$ cm. Are they congruent? Which criterion applies?

▶ Show solution

Yes — by RHS: both have a right angle, equal hypotenuses ($13$ cm) and one other equal side ($5$ cm).

Confirmation: Pythagoras forces the third side in each to be $\sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm, so in fact all three sides match.

Question 7

$PQRS$ is a parallelogram. Prove that $\triangle PQR \cong \triangle RSP$.

▶ Show solution

$PQ = RS$ (opposite sides of a parallelogram are equal)

$QR = SP$ (opposite sides of a parallelogram are equal)

$PR = RP$ (common side — the diagonal)

Therefore $\triangle PQR \cong \triangle RSP$ (SSS).

This proves that a diagonal cuts a parallelogram into two congruent triangles, and hence that opposite angles are equal.

Question 8

Two triangles have the same area. Must they be congruent? Justify your answer.

▶ Show solution

No.

Counter-example: a triangle with base $12$ cm and height $2$ cm has area $12\text{ cm}^2$, and so does one with base $6$ cm and height $4$ cm. The two shapes are quite different.

Equal area is a consequence of congruence, not a test for it.

Question 9

$AB$ and $CD$ are two chords of a circle with centre $O$, and $AB = CD$. Prove that $\triangle OAB \cong \triangle OCD$, and state what this tells you about the distances of the two chords from the centre.

▶ Show solution

$OA = OC$ (radii of the same circle)

$OB = OD$ (radii of the same circle)

$AB = CD$ (given)

Therefore $\triangle OAB \cong \triangle OCD$ (SSS).

Congruent triangles have equal corresponding heights, so the perpendicular distance from $O$ to $AB$ equals the perpendicular distance from $O$ to $CD$.

In other words, equal chords are equidistant from the centre.

Question 10

$ABCD$ is a square. $P$, $Q$, $R$ and $S$ are points on $AB$, $BC$, $CD$ and $DA$ respectively such that $AP = BQ = CR = DS$.

(a) Prove that $\triangle APS \cong \triangle BQP$. (b) Explain why $PQRS$ has four equal sides. (c) Prove that $PQRS$ is a square.

▶ Show solution

(a) Let $AP = BQ = CR = DS = x$ and let the square have side $s$. Then $SA = DA - DS = s - x$ and $PB = AB - AP = s - x$, so $SA = PB$.

$AP = BQ$ (given)

$\angle PAS = \angle QBP = 90^\circ$ (angles of a square)

$SA = PB$ (shown above)

The right angle lies between the two pairs of equal sides, so $\triangle APS \cong \triangle BQP$ (SAS).

(b) The same argument applies at all four corners, so all four corner triangles are congruent. Their hypotenuses are $SP$, $PQ$, $QR$ and $RS$ — corresponding sides of congruent triangles — so all four are equal.

(c) In $\triangle APS$, $\angle APS + \angle ASP = 90^\circ$. From the congruence, $\angle BPQ = \angle ASP$.

Along the straight line $AB$: $\angle APS + \angle SPQ + \angle BPQ = 180^\circ$.

Substituting, $\angle SPQ = 180 - (\angle APS + \angle ASP) = 180 - 90 = 90^\circ$.

The same holds at each vertex, so $PQRS$ has four equal sides and four right angles — it is a square. ∎

Congruent Triangles (G5) · GCSE Maths Revision · Created with MathJax