Prove Two Triangles Match and Six Facts Come Free

Six equalities at once — three pairs of sides and three pairs of angles. You establish three of them, and the remaining three follow automatically by c.p.c.t., meaning corresponding parts of congruent triangles.

That is why congruence is the engine behind most geometry proofs. This page covers everything in the ICSE Class 7 Mathematics chapter on congruency: congruence by superimposition, vertex correspondence, the four criteria, and proofs using c.p.c.t.

Two figures are congruent if one placed exactly over the other covers it completely, with nothing left over. The symbol is $\cong$.

The practical test is tracing: copy one figure onto tracing paper, then slide, turn or flip it onto the other. If it fits exactly, they are congruent. Those three movements never change a length or an angle, which is why the fit survives them.

- Two line segments are congruent when their lengths are equal. So a 5 cm segment is congruent to any other 5 cm segment.
- Two angles are congruent when their measures are equal, however long the arms are drawn. A $40°$ angle with 2 cm arms is congruent to a $40°$ angle with 10 cm arms.
- Two plane figures are congruent when both shape and size match.

Two identical set-squares from a geometry box stay congruent even when one is turned upside down, and two ₹5 coins are congruent circles.

The point to be clear about is that congruence needs both shape and size. Two squares of side 3 cm and 6 cm have exactly the same shape and are not congruent — they are only similar. Equal area is not enough either: a $2\times 8$ rectangle and a $4\times 4$ square both cover 16 square units without being congruent.

The order of the letters records which vertex matches which, so it carries information and cannot be rearranged.

Writing $△ABC\cong △PQR$ states that

- A matches P, B matches Q, C matches R

and therefore these six parts are equal:

- Sides: $AB=PQ$, $BC=QR$, $CA=RP$
- Angles: $\angle A=\angle P$, $\angle B=\angle Q$, $\angle C=\angle R$

Each pair is read straight off the matching positions. The side $AB$ pairs with $PQ$ because A pairs with P and B with Q.

So $△ABC\cong △PQR$ and $△ABC\cong △QRP$ make different claims, and at most one of them is true for a given pair of triangles.

The practical habit is to write the correspondence before listing any equal parts, and to let the letters generate the list rather than reading them off the diagram. Copying the letters in whatever order the question printed them is the commonest error here — it silently asserts the wrong pairing and every equality that follows is then wrong too.

Each criterion names the three matching parts that are enough to force congruence.

SSS — three pairs of equal sides.

SAS — two pairs of equal sides and the included angle, the one lying between those sides.

ASA (or AAS) — two pairs of equal angles and one pair of equal sides. With two angles known the third is forced, since the angles total $180°$, so the side need not be the included one.

RHS — for right-angled triangles only: equal hypotenuse and one other equal side.

Worked example. In $△ABC$: $AB=5$ cm, $BC=7$ cm, $\angle B=50°$. In $△PQR$: $PQ=5$ cm, $QR=7$ cm, $\angle Q=50°$. The equal angle lies between the two equal sides in both, so they are congruent by SAS, and $CA=RP$ follows by c.p.c.t.

Worked example, RHS. Two right-angled triangles each have hypotenuse 13 cm and one leg 5 cm. They are congruent by RHS, and the third side is forced in both:

$$13^2-5^2=169-25=144=12^2$$

giving 12 cm each time.

Two combinations are not criteria. AAA fixes the shape but not the size — an equilateral triangle of side 2 cm and one of side 20 cm have identical angles and are not congruent. And two sides with a non-included angle can fit two different triangles, which is why SAS insists on the included angle.

Prove two triangles congruent first, then quote c.p.c.t. to claim the part you actually wanted.

Worked proof — the base angles of an isosceles triangle are equal.

Given $△ABC$ with $AB=AC$, and $AD$ the bisector of $\angle A$ meeting $BC$ at D.

In $△ABD$ and $△ACD$:

- $AB=AC$ — given
- $\angle BAD=\angle CAD$ — AD bisects $\angle A$
- $AD=AD$ — common side

So $△ABD\cong △ACD$ by SAS.

Therefore $\angle B=\angle C$ by c.p.c.t.

The same congruence gives two more results free: $BD=DC$ by c.p.c.t., so AD also bisects the base; and $\angle ADB=\angle ADC$, which with their sum of $180°$ makes each $90°$, so AD is perpendicular to BC.

Applied numerically: if the vertex angle is $40°$, the base angles share the rest, so each is $\frac{180-40}{2}=70°$.

The structure of a proof matters as much as the content. List the three matching parts with a reason beside each — given, common, bisector, vertically opposite — then name the criterion, and only then quote c.p.c.t. for the conclusion. Claiming c.p.c.t. before establishing congruence proves nothing at all.