Congruence of Triangles: ICSE Class 7 Guide

A student in my Thursday batch once brought me two triangle cutouts from her craft class and asked if they were the same. I turned one over and they matched perfectly, edge for edge, corner for corner. 'Yes,' I said, 'those are congruent.' She then asked the question every Class 7 student eventually asks: 'If they are the same, why do we need letters and tests and proofs? Can't we just cut them and check?'

The honest answer is that cutting works for paper but not for buildings, for steel beams, for computer graphics or for exam questions where you only get a sketch. In real life you almost never get to hold both triangles in your hands. You get a figure, some measurements, and a question. The congruence criteria, SSS, SAS, ASA and RHS, let you confidently say 'these two triangles are identical' without touching them.

This chapter is short in the Selina textbook, only about twenty pages, but it punches above its weight. Everything you learn here comes back in Class 8 for similar triangles, in Class 9 for coordinate geometry, and in Class 10 for proofs involving circles. Get congruence right now and the next three years of geometry become much easier.

Two figures are congruent if they have exactly the same shape and exactly the same size. You can slide one onto the other, maybe rotate or flip it, and they match perfectly.

The symbol for congruence is $\cong$. So $\triangle ABC \cong \triangle PQR$ reads 'triangle ABC is congruent to triangle PQR'.

When we write this, the order of the letters matters. It tells us which vertices correspond to which.

* $A \leftrightarrow P$
* $B \leftrightarrow Q$
* $C \leftrightarrow R$

Because of this correspondence, we automatically know six equalities: three pairs of equal sides and three pairs of equal angles.

* $AB = PQ$
* $BC = QR$
* $CA = RP$
* $\angle A = \angle P$
* $\angle B = \angle Q$
* $\angle C = \angle R$

The magic is that you do not need all six to claim congruence. You only need three, and the chapter is about which three work and which three do not.

Once you prove two triangles are congruent, you get a bonus: every remaining pair of sides and angles automatically becomes equal. This is called CPCT, which stands for Corresponding Parts of Congruent Triangles.

In practice, you use three measurements to prove congruence, then use CPCT to claim the other three for free. This is the standard way ICSE wants you to solve 'prove that' questions.

For example, if you prove $\triangle ABC \cong \triangle PQR$ by SAS using $AB = PQ$, $\angle B = \angle Q$ and $BC = QR$, then by CPCT you immediately get $CA = RP$, $\angle A = \angle P$ and $\angle C = \angle R$. No extra work needed.

ICSE school papers typically ask questions that follow this exact pattern: given some data, prove the triangles are congruent, then use CPCT to conclude that some specific pair of sides or angles is equal. Learn to recognise this two step structure.

Problem: In $\triangle ABC$ and $\triangle DEF$, $AB = DE$, $BC = EF$ and $CA = FD$. Prove that $\triangle ABC \cong \triangle DEF$.

Solution: This is the direct application of the SSS criterion.

Step 1: State the given.
In $\triangle ABC$ and $\triangle DEF$:
$AB = DE$ (given)
$BC = EF$ (given)
$CA = FD$ (given)

Step 2: Apply the criterion.
By the SSS congruence criterion, $\triangle ABC \cong \triangle DEF$.

Step 3: Write the conclusion.
Hence, the two triangles are congruent.

Notice the format. Selina and ICSE school papers expect exactly this structure: given, criterion, conclusion. Writing in this three step format helps you get full marks even if you are unsure.

Problem: In $\triangle PQR$ and $\triangle XYZ$, $PQ = XY$, $\angle Q = \angle Y$ and $QR = YZ$. Prove they are congruent.

Solution:

Step 1: Given.
$PQ = XY$ (given)
$\angle Q = \angle Y$ (given)
$QR = YZ$ (given)

Step 2: Check that the angle is included between the two sides. Here, $\angle Q$ is between sides $PQ$ and $QR$, so yes, it is included. SAS applies.

Step 3: Apply.
By the SAS congruence criterion, $\triangle PQR \cong \triangle XYZ$.

If the angle had been $\angle P$ instead of $\angle Q$, it would not have been included between $PQ$ and $QR$, and SAS would not apply. You would have to try another criterion, or prove that additional equalities exist.

Problem: In $\triangle ABC$, $AB = AC$. The bisector of $\angle A$ meets $BC$ at $D$. Prove that $BD = DC$.

Solution: This is a classic Selina question. We want to show that the angle bisector from $A$ also bisects the opposite side. We will use ASA.

Step 1: Given.
$AB = AC$ (given, isosceles triangle)
$AD$ bisects $\angle A$, so $\angle BAD = \angle CAD$ (given)
$AD$ is common to both smaller triangles.

Step 2: Consider $\triangle ABD$ and $\triangle ACD$.

$AB = AC$ (given)
$\angle BAD = \angle CAD$ (given)
$AD = AD$ (common side)

Step 3: This is SAS with the included angle $\angle BAD$ between sides $AB$ and $AD$. So

Step 4: By CPCT, $BD = DC$.

Hence proved.

This type of proof, where you identify two triangles inside a larger figure, prove congruence and then use CPCT, is the single most common kind of question in ICSE Class 7 exams. Master it with five or six examples.

Problem: In two right triangles, the hypotenuse and one leg of the first are equal to the hypotenuse and one leg of the second. Prove they are congruent.

Solution: Let $\triangle ABC$ and $\triangle PQR$ be right triangles with $\angle B = \angle Q = 90°$. Given $AC = PR$ (hypotenuse) and $BC = QR$ (one leg).

Step 1:
$\angle B = \angle Q = 90°$ (given, right angles)
$AC = PR$ (hypotenuse, given)
$BC = QR$ (one leg, given)

Step 2: By the RHS criterion, $\triangle ABC \cong \triangle PQR$.

Step 3: By CPCT, all other corresponding parts are equal.

RHS is the quickest criterion to apply because right triangles carry a built in $90°$ equality. Just make sure you are explicitly given the right angle and the hypotenuse.

Problem: In the figure, $AB = AD$ and $BC = DC$. Prove that $\triangle ABC \cong \triangle ADC$ and hence that $\angle BAC = \angle DAC$.

Solution:

Step 1: Given.
$AB = AD$ (given)
$BC = DC$ (given)
$AC = AC$ (common side)

Step 2: Consider $\triangle ABC$ and $\triangle ADC$. All three sides of one are equal to all three sides of the other, so by SSS,

Step 3: By CPCT, $\angle BAC = \angle DAC$.

Hence proved.

This problem also shows that $AC$ lies on the angle bisector of $\angle BAD$, which is a pretty geometric fact. One short proof, two insights.

The six errors that cost students the most marks in congruence problems.

1. Wrong vertex order in the congruence statement. Writing $\triangle ABC \cong \triangle PQR$ when actually $A$ corresponds to $R$, not $P$. The order in the statement tells the examiner who maps to whom.

2. Using SSA or AAA. These are not valid criteria. Two triangles sharing two sides and a non included angle may not be congruent. Three equal angles make triangles similar but not necessarily congruent.

3. Missing the 'included' check in SAS and ASA. The angle in SAS must be between the two sides. The side in ASA must be between the two angles. If not, the criterion does not apply.

4. Forgetting to mention the common side or angle. In a figure where two triangles share a side, you must explicitly write 'common side' as a reason.

5. Using CPCT before proving congruence. You cannot claim corresponding parts are equal until after you have proven the triangles are congruent. Order matters.

6. Skipping the conclusion line. Always end with 'hence proved' or 'therefore the triangles are congruent'. ICSE examiners want a clear final statement.

* Two triangles are congruent if they have the same shape and size. Symbol $\cong$.
* Four criteria: SSS, SAS, ASA and RHS. No SSA, no AAA.
* In SAS and ASA, the angle or side in the middle must be included between the other two.
* Use CPCT to claim the remaining corresponding parts are equal after proving congruence.
* Always write proofs in a clear given, criterion, conclusion format.
* RHS works only for right triangles and needs the hypotenuse plus one leg.
* The order of vertices in the congruence statement tells the examiner the correspondence. Write it carefully.