G5, G6 - Triangle congruence and geometrical arguments

Two triangles can match exactly even when one is turned or reflected. You will decide when their information guarantees a match, then use congruence, angle facts and similarity to explain triangle and quadrilateral properties and prove Pythagoras' theorem.

Tests for congruent triangles

Could four identical right triangles explain why $3^2+4^2=5^2$ ? By the end, you will use their areas to prove a result for every right triangle.

You already know

Equal sides of an isosceles triangle face equal angles. Angles in a triangle total $180^\circ$ ; parallel sides give equal alternate angles.

Congruent figures have the same shape and size. One may need moving, turning or reflecting to fit the other. Writing $\triangle ABC\cong\triangle PQR$ matches $A$ to $P$ , $B$ to $Q$ , $C$ to $R$ ; thus $AB=PQ$ , $BC=QR$ , $AC=PR$ .

Test	Information that guarantees congruence	Why it fixes a triangle
SSS	Three corresponding sides equal	Two fixed-radius arcs locate the third vertex, up to reflection.
SAS	Two corresponding sides and their included angle equal	The angle fixes two rays and the lengths fix their endpoints.
ASA	Two corresponding angles and a corresponding side equal	The third angle follows, and the side fixes the size.
RHS	Right angles, equal hypotenuses and one other corresponding side equal	The right angle and two matched lengths fix the triangle.

For ASA, the side can be between the angles or opposite matching angles: the third angle follows from $180^\circ$ . For RHS, identify the side opposite the right angle as the hypotenuse in each triangle. These congruence tests are not on the formula sheet: learn them.

Worked example 1: the included angle

Question. $AB=DE=5\,\mathrm{cm}$ , $AC=DF=7\,\mathrm{cm}$ , and $\angle BAC=\angle EDF=40^\circ$ . Explain why triangles $ABC,DEF$ are congruent and name the side equal to $BC$ .

Plan. List the matching information, checking that the angle lies between the known sides.

\begin{aligned} AB&=DE &&\text{given first side pair}\\ AC&=DF &&\text{given second side pair}\\ \angle BAC&=\angle EDF &&\text{included angle pair}\\ \triangle ABC&\cong\triangle DEF &&\text{SAS} \end{aligned}

Why SAS? The angle is where the two specified sides meet in each triangle.

Answer. The triangles are congruent by SAS, so $BC=EF$ , corresponding sides.

Check. The vertex match $B\leftrightarrow E$ , $C\leftrightarrow F$ gives the same side pair.

Change only the included angle from $40^\circ$ to $60^\circ$ in both triangles. The test should still apply.

Worked example 2: a different included angle

Question. $AB=DE=5\,\mathrm{cm}$ , $AC=DF=7\,\mathrm{cm}$ , and $\angle BAC=\angle EDF=60^\circ$ . Prove $BC=EF$ .

Plan. Establish congruence before claiming the unknown sides equal.

\begin{aligned} AB=DE,\quad AC&=DF &&\text{two given side pairs}\\ \angle BAC&=\angle EDF=60^\circ &&\text{between those sides}\\ \triangle ABC&\cong\triangle DEF &&\text{SAS}\\ BC&=EF &&\text{corresponding sides} \end{aligned}

Why prove congruence first? The unknown sides are a consequence, not part of the given evidence.

Answer. $BC=EF$ , as required.

Check. Changing the shared angle changes the triangles, but both still have the same two lengths and included angle as each other.

ASA offers a different route. If $LM=RS=8\,\mathrm{cm}$ , angles at $L,R$ are $35^\circ$ and at $M,S$ are $65^\circ$ , then the third angles are $80^\circ$ . ASA gives $\triangle LMN\cong\triangle RST$ , hence $LN=RT$ , $MN=ST$ .

AAA establishes similarity, with equal angles and proportional sides, but not necessarily equal size: equilateral triangles of sides 3 and 6 both have three $60^\circ$ angles. SSA is not a general congruence test. Fix $AB=5\,\mathrm{cm}$ , $BC=4\,\mathrm{cm}$ , and $\angle BAC=30^\circ$ : a radius-4 circle about $B$ crosses the ray from $A$ twice, giving two possible triangles. RHS handles the special right-angle case.

Watch out

Wrong: “Two sides and an angle are equal, so SAS.” The angle must be included between those two sides. With a non-included angle, two different triangles may fit. Name the sides and their meeting vertex before choosing SAS.

This time the supplied angle has moved away from the meeting point of the known sides.

Why isosceles base angles are equal

A proof explains why equal base angles follow from equal sides, without relying on a perfectly symmetrical sketch.

Let $AB=AC$ . Bisect the apex angle at $A$ , meeting $BC$ at $D$ . The two smaller triangles have equal outer sides, common $AD$ , and equal included half-apex angles. They are congruent by SAS. Their corresponding base angles are equal; corresponding base halves are equal too. Their equal adjacent angles at $D$ total $180^\circ$ , so each is $90^\circ$ .

[DIAGRAM: asset_name: Congruence derivation in an isosceles triangle; asset_slug: 1ma1_m48_isosceles; caption: Bisecting the apex creates two SAS-congruent triangles and proves the base-angle and midpoint properties.; description: Isosceles triangle ABC has AB and AC marked equal and apex bisector AD. Equal half-apex angles are marked, and D lies on BC. Two lightly filled matching halves identify triangles ABD and ACD; a right-angle square and equal base-half ticks show the derived results.]
Bisecting the apex creates two SAS-congruent triangles and proves the base-angle and midpoint properties.

The diagram records the conclusions as well as the starting information. In a proof, state which facts you were given and which you deduced. The isosceles base-angle result is not on the formula sheet: derive it from SAS congruence when a proof is requested.

Worked example 3: starting with an angle bisector

Question. $AB=AC$ , and $AD$ bisects $\angle BAC$ , with $D$ on $BC$ . Prove $BD=DC$ .

Plan. Compare $ABD$ with $ACD$ using the included half-apex angles.

\begin{aligned} AB&=AC &&\text{given}\\ AD&=AD &&\text{common side}\\ \angle BAD&=\angle DAC &&\text{angle bisector}\\ \triangle ABD&\cong\triangle ACD &&\text{SAS}\\ BD&=DC &&\text{corresponding sides} \end{aligned}

Why not assume $D$ is the midpoint? That is exactly what the proof must establish.

Answer. $BD=DC$ , so $D$ is the midpoint of $BC$ .

Check. The matching order sends $B$ to $C$ , giving the intended base halves.

Change only the condition defining $D$ : it is now given as the midpoint instead of the point where the apex bisector meets the base. The congruence test changes.

Worked example 4: starting with a midpoint

Question. $AB=AC$ , and $D$ is the midpoint of $BC$ . Prove that $AD$ bisects $\angle BAC$ .

Plan. Use three side pairs to prove congruence, then match apex angles.

\begin{aligned} AB&=AC &&\text{given}\\ BD&=DC &&\text{midpoint}\\ AD&=AD &&\text{common side}\\ \triangle ABD&\cong\triangle ACD &&\text{SSS}\\ \angle BAD&=\angle DAC &&\text{corresponding angles} \end{aligned}

Why SSS rather than SAS? We now know the base halves, but have not yet established equal apex angles.

Answer. $AD$ bisects the apex angle.

Check. The conclusion uses an angle pair at $A$ , the required vertex.

The converse base-angle result also holds: two equal triangle angles imply equal opposite sides. Compare the triangle with its reversed vertex order; the equal angles and shared intervening side give ASA. Thus base angles $58^\circ$ imply equal opposite sides and apex angle $64^\circ$ . Given equal sides and apex angle $36^\circ$ , each base angle is $(180^\circ-36^\circ)/2=72^\circ$ .

Watch out

Wrong: “ $AB=AC$ , and $AD$ looks central, so $BD=DC$ .” An unmarked line need not bisect anything. Give a construction or midpoint condition, then use congruence to derive the remaining properties.

Use the midpoint information again, with new labels, to prove another consequence of the same congruence.

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Papers 1, 2 and 3

G5, G6 - Triangle congruence and geometrical arguments

Tests for congruent triangles

Why isosceles base angles are equal

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