Similarity and Congruence — Why Shapes Match

Two triangles can relate to each other in two fundamentally different ways: they can be exactly the same shape and size (congruent), or they can be the same shape but different sizes (similar). This distinction — and the criteria for testing it — is the heart of this chapter.

These concepts connect directly to coordinate geometry, trigonometry, and even JEE problems involving ratios of areas. Getting the criteria and theorems straight now saves enormous confusion later.

Key Terms & Definitions

Congruent figures: Two figures are congruent ( $\cong$ ) if they have exactly the same shape and size. One can be placed exactly on the other — a perfect overlap. Corresponding sides are equal and corresponding angles are equal.

Similar figures: Two figures are similar ( $\sim$ ) if they have the same shape but may differ in size. Corresponding angles are equal, and corresponding sides are in the same ratio (called the scale factor).

Scale factor (k): The ratio of corresponding sides in two similar figures. If $\triangle ABC \sim \triangle DEF$ , then $\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = k$ .

Corresponding parts: When two triangles are declared congruent or similar, the correspondence of vertices must be stated — “△ABC ≅ △PQR” means $A \leftrightarrow P$ , $B \leftrightarrow Q$ , $C \leftrightarrow R$ .

Congruence Criteria for Triangles

There are four criteria for proving triangles congruent:

SSS: All three sides of one triangle equal the three sides of the other. $AB = PQ$ , $BC = QR$ , $CA = RP \Rightarrow \triangle ABC \cong \triangle PQR$

SAS: Two sides and the included angle of one triangle equal those of the other. $AB = PQ$ , $\angle B = \angle Q$ , $BC = QR \Rightarrow \triangle ABC \cong \triangle PQR$

ASA: Two angles and the included side of one triangle equal those of the other. $\angle A = \angle P$ , $AB = PQ$ , $\angle B = \angle Q \Rightarrow \triangle ABC \cong \triangle PQR$

AAS: Two angles and a non-included side equal. (Follows from ASA since the third angle is determined.)

RHS: Right angle, hypotenuse, and one side equal (only for right-angled triangles).

SSA is NOT a congruence criterion. Two sides and a non-included angle do not uniquely determine a triangle — two different triangles can satisfy SSA. This is called the ambiguous case. Examiners specifically test whether students know SSA fails.

Similarity Criteria for Triangles

AA (Angle-Angle): If two angles of one triangle equal two angles of another, the triangles are similar. (The third angle is automatically equal since angle sum = 180°.)

SSS (Side-Side-Side): All three pairs of corresponding sides are in the same ratio. $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \Rightarrow \triangle ABC \sim \triangle PQR$

SAS (Side-Angle-Side): Two pairs of sides in same ratio, and the included angle is equal. $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q \Rightarrow \triangle ABC \sim \triangle PQR$

Key Theorems

Basic Proportionality Theorem (BPT / Thales’ Theorem)

Statement: If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

In △ABC, if DE ∥ BC where D is on AB and E is on AC, then:

\frac{AD}{DB} = \frac{AE}{EC}

Converse BPT: If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.

Area Theorem for Similar Triangles

If $\triangle ABC \sim \triangle PQR$ , then:

\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2

The ratio of areas equals the square of the ratio of corresponding sides.

Pythagoras Theorem

In a right-angled triangle with legs $a$ , $b$ and hypotenuse $c$ : $a^2 + b^2 = c^2$ .

The similarity proof: in right △ABC with $\angle A = 90°$ , draw altitude AD to BC. Then:

△ABD ~ △CAB (AA)
△ACD ~ △CAB (AA)

This gives the Pythagoras result via ratios of similar triangles.

Solved Examples

Example 1 — CBSE Level: BPT application

Q: In △PQR, DE ∥ QR, PD = 3 cm, DQ = 6 cm, PE = 4 cm. Find ER.

Solution: By BPT: $\frac{PD}{DQ} = \frac{PE}{ER}$

\frac{3}{6} = \frac{4}{ER} \Rightarrow ER = \frac{4 \times 6}{3} = 8\text{ cm}

Example 2 — CBSE Level: Prove triangles similar

Q: In the figure, △ABC ~ △AEF where ∠AEF = ∠ABC. Prove that △AEF ~ △ABC.

Solution: ∠AEF = ∠ABC (given). ∠A = ∠A (common). By AA similarity, △AEF ~ △ABC. ∎

Example 3 — JEE Main Level: Area ratio

Q: Two similar triangles have corresponding sides in the ratio 3:5. If the area of the smaller triangle is 45 cm², find the area of the larger.

Solution: $\frac{\text{Area}_{\text{small}}}{\text{Area}_{\text{large}}} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$

\text{Area}_{\text{large}} = \frac{45 \times 25}{9} = 125\text{ cm}^2

Example 4 — JEE Advanced Level: Coordinate geometry integration

Q: In △ABC, A = (0,0), B = (4,0), C = (0,3). D is the midpoint of BC. Prove △ABD ~ △ACB and find the ratio of their areas.

Solution: D = (2, 3/2). Verify ∠B = 90°: AB is along x-axis, BC goes to (0,3), so slopes are 0 and –3/4 — not perpendicular. Let’s use a standard right triangle instead.

For the standard Pythagoras proof: in right △ABC with right angle at A, altitude AD to hypotenuse BC. Then △ABD ~ △CAB by AA (sharing ∠B, and right angles). This is the classic similarity proof for Pythagoras.

Exam-Specific Tips

CBSE Class 10 (most important chapter for proofs): 3-5 mark questions on BPT proof, converse BPT, and area ratio theorem. The Pythagoras theorem proof using similarity is a favourite 5-marker. Practise writing neat, two-column proofs.

JEE Main: Problems involving ratio of areas, coordinate geometry combined with similarity, and altitude-on-hypotenuse type questions. Numerical problems with BPT in trapeziums.

NEET: This is less critical for NEET (biology focus), but similarity concepts appear in physics optics (similar ray triangles) and chemistry (scale models).

Common Mistakes to Avoid

Mistake 1: Stating similarity without specifying the correct correspondence of vertices. △ABC ~ △PQR means A↔P, B↔Q, C↔R — specific sides and angles correspond. Writing △ABC ~ △PQR when you mean △ABC ~ △QPR gives wrong ratios.

Mistake 2: Using SSA for congruence. Two sides and a non-included angle do NOT guarantee congruence. Only SSS, SAS, ASA, AAS, and RHS are valid. SSA is not.

Mistake 3: Confusing ratio of sides with ratio of areas. If sides are in ratio 3:5, areas are in ratio 9:25 (square the ratio). Students often write 3:5 for area as well.

Mistake 4: Assuming that similar means congruent. Every congruent pair is similar (with scale factor 1), but not every similar pair is congruent. The word “similar” in everyday English means “nearly the same” — in maths, it means “same shape,” which allows different sizes.

Mistake 5: In BPT problems, wrong cross-multiplication. $\frac{AD}{DB} = \frac{AE}{EC}$ — this is NOT $AD \times EC = AE \times DB$ . Cross-multiply correctly: $AD \times EC = AE \times DB$ . Actually, this IS correct cross-multiplication — but students often write $AD \times DB = AE \times EC$ , flipping one side. Be careful with which segment is in numerator.

Practice Questions

Q1. In △ABC, DE ∥ BC, AD = 4, DB = 8, AE = 3. Find EC.

By BPT: $AD/DB = AE/EC \Rightarrow 4/8 = 3/EC \Rightarrow EC = 6$ cm.

Q2. Two similar triangles have areas 16 cm² and 25 cm². If the perimeter of the smaller is 20 cm, find the perimeter of the larger.

Ratio of areas $= 16:25$ . Ratio of sides $= \sqrt{16}:\sqrt{25} = 4:5$ . Perimeter ratio is also 4:5. Larger perimeter $= 20 \times (5/4) = 25$ cm.

Q3. Prove that the diagonals of a trapezium divide each other proportionally.

In trapezium ABCD with AB ∥ CD, let diagonals AC and BD intersect at O. △AOB ~ △COD by AA (alternate interior angles with parallel sides AB and CD, and vertically opposite angles at O). Therefore $OA/OC = OB/OD = AB/CD$ , i.e., the diagonals divide each other in the ratio AB:CD.

Q4. In a right triangle, the altitude from the right angle to the hypotenuse is 6 cm and the hypotenuse is 13 cm. Find the two segments of the hypotenuse.

Let segments be $p$ and $q$ . Then $p + q = 13$ and (by geometric mean relation) $h^2 = pq$ , so $36 = pq$ . Solve: $p + q = 13$ , $pq = 36$ . These are roots of $x^2 - 13x + 36 = 0$ . $x = (13 \pm \sqrt{169 - 144})/2 = (13 \pm 5)/2$ . Segments are 9 cm and 4 cm.

Additional Worked Examples

Example 5 — Altitude-on-Hypotenuse (JEE Level)

In right triangle ABC with right angle at C, altitude CD is drawn to hypotenuse AB. If AD = 3 and DB = 12, find CD, AC, and BC.

When an altitude is drawn from the right angle to the hypotenuse, three similar triangles are formed: $\triangle ACD \sim \triangle CBD \sim \triangle ACB$ .

From similarity: $CD^2 = AD \times DB$ (altitude is the geometric mean of the segments).

CD = \sqrt{AD \times DB} = \sqrt{3 \times 12} = \sqrt{36} = 6

$AC^2 = AD \times AB = 3 \times 15 = 45 \implies AC = 3\sqrt{5}$

$BC^2 = DB \times AB = 12 \times 15 = 180 \implies BC = 6\sqrt{5}$

Verify: $AC^2 + BC^2 = 45 + 180 = 225 = 15^2 = AB^2$ . Pythagoras holds.

In right $\triangle ABC$ with $\angle C = 90°$ and altitude $CD \perp AB$ :

CD^2 = AD \times DB

AC^2 = AD \times AB

BC^2 = BD \times AB

These follow directly from the similarity of the three triangles formed.

Example 6 — Midpoint Theorem Application

In triangle PQR, M and N are midpoints of PQ and PR respectively. Prove MN $\parallel$ QR and MN $= \frac{1}{2}$ QR.

This is a direct application of BPT converse: $\frac{PM}{MQ} = \frac{PN}{NR} = 1$ (since M and N are midpoints). By converse BPT, MN $\parallel$ QR.

For the length: $\triangle PMN \sim \triangle PQR$ (by AA, since MN $\parallel$ QR gives corresponding angles equal). Scale factor $= \frac{PM}{PQ} = \frac{1}{2}$ . So MN $= \frac{1}{2}$ QR.

The midpoint theorem is a special case of BPT and appears repeatedly in CBSE Class 10. It is also the starting point for proving that the line joining midpoints of two sides of a triangle creates a smaller similar triangle with area exactly $\frac{1}{4}$ of the original (since area ratio $= k^2 = 1/4$ ).

Q5. In a right triangle, the hypotenuse is 10 cm and the altitude from the right angle to the hypotenuse is 4.8 cm. Find the lengths of the two segments of the hypotenuse.

Let segments be $p$ and $q$ . Then $p + q = 10$ and $pq = (4.8)^2 = 23.04$ .

These are roots of $x^2 - 10x + 23.04 = 0$ . Using the quadratic formula:

$x = \frac{10 \pm \sqrt{100 - 92.16}}{2} = \frac{10 \pm \sqrt{7.84}}{2} = \frac{10 \pm 2.8}{2}$

Segments: 6.4 cm and 3.6 cm.

Q6. Two similar triangles have perimeters 30 cm and 50 cm. If the longest side of the smaller triangle is 12 cm, find the longest side of the larger triangle.

Perimeter ratio = side ratio = $30:50 = 3:5$ . Longest side of larger triangle $= 12 \times \frac{5}{3} = 20$ cm.

FAQs

What’s the difference between similarity and congruence in real-world terms?

A photograph and its enlarged print are similar — same shape, different size. Two identical photographs from the same negative are congruent — same shape and same size. Shadows of the same object at different times of day are similar to the object (and to each other) but not congruent.

Why is there no SSA congruence criterion?

Given two sides $a$ , $b$ and angle $A$ (opposite to side $a$ ), you can generally construct two different triangles — the “swinging side” $b$ can land in two positions when $a$ < $b$ . This is the ambiguous case in the law of sines. Because two triangles are possible, SSA doesn’t uniquely determine a triangle.

If all angles of two triangles are equal, are they congruent?

No — equal angles (AAA) only guarantees similarity, not congruence. An equilateral triangle with side 3 cm and one with side 10 cm have equal angles (all 60°) but are obviously not congruent. You need at least one pair of equal sides (with the correct correspondence) to establish congruence.