Triangles

Why This Chapter Matters

This is one of the largest chapters in Class 10 — and one of the highest-mark chapters in board exams (8-12 marks typically). Basic Proportionality Theorem and its converse, similarity criteria, and Pythagoras theorem are tested every year. Proofs are asked in Long Answer questions.

Prerequisites

Core Concepts

1. 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 in the same ratio.

If DE ∥ BC in △ABC, then:

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

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

Memory: BPT → the line parallel to one side creates equal ratios on the other two sides.

2. Similarity of Triangles

Two triangles are similar if:

Notation: △ABC ~ △DEF means:

3. Criteria for Similarity

#### AA (Angle-Angle) Similarity

If two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

Since angle sum = 180°, if two pairs of angles are equal, the third pair is automatically equal.

#### SSS (Side-Side-Side) Similarity

If the three pairs of corresponding sides are proportional:

AB/DE = BC/EF = CA/FD → △ABC ~ △DEF

#### SAS (Side-Angle-Side) Similarity

If one pair of corresponding angles is equal AND the sides including those angles are proportional:

AB/DE = AC/DF and ∠A = ∠D → △ABC ~ △DEF

4. Ratio of Areas of Similar Triangles

$$\frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle DEF} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{BC}{EF}\right)^2 = \left(\frac{CA}{FD}\right)^2$$

Key point: The ratio of areas of similar triangles = square of the ratio of corresponding sides (NOT the ratio itself!)

5. Pythagoras Theorem

Statement: In a right-angled triangle, the square of the hypotenuse = sum of squares of the other two sides.

If ∠B = 90° in △ABC:

$$AC^2 = AB^2 + BC^2$$

Converse (equally important!): If in a triangle, the square of one side equals the sum of squares of the other two sides, then the angle opposite the first side is 90°.

Proof of Pythagoras (using similarity — must know for board):

Solved Examples

Example 1 — BPT

Q: In △ABC, DE ∥ BC. If AD = 4 cm, DB = 3 cm, AE = 6 cm, find EC.

By BPT: AD/DB = AE/EC

4/3 = 6/EC → EC = 4.5 cm

Example 2 — Similarity (AA)

Q: In △ABC, ∠A = 60°, ∠B = 80°. In △PQR, ∠Q = 80°, ∠R = 40°. Are they similar?

△ABC: ∠A = 60°, ∠B = 80°, ∠C = 40°

△PQR: ∠P = 60°, ∠Q = 80°, ∠R = 40°

All three angles match → △ABC ~ △PQR (by AAA/AA)

Example 3 — Areas

Q: Areas of two similar triangles are 81 cm² and 49 cm². If altitude of first = 4.5 cm, find altitude of second.

(h₁/h₂)² = 81/49 → h₁/h₂ = 9/7 → 4.5/h₂ = 9/7 → h₂ = 3.5 cm

PYQs

2023

Q: In △PQR, QM ⊥ PR and PM × MR = QM². Prove that ∠PQR = 90°.

Using the given condition and similarity of triangles △PQM ~ △QRM

2022

Q: BL and CM are medians of right-angled △ABC (∠A = 90°). Prove: 4(BL² + CM²) = 5BC²

Using Pythagoras in △ABL, △ACM, and △ABC

2021

Q: State and prove the Basic Proportionality Theorem.

(Standard proof required — draw diagram, use area of triangles)

2020

Q: Sides of two similar triangles are in ratio 3:7. Ratio of their areas:

9:49 ✓ (square of side ratio)

MCQ Practice

Q1. In similar triangles, ratio of corresponding sides = 2:3. Ratio of areas:

(A) 2:3 (B) 4:9 ✓ (C) 8:27 (D) 3:2

Q2. In △ABC, D divides AB such that AD/DB = 3/5. If BC = 4 cm, find DE ∥ BC:

DE = (AD/(AD+DB)) × BC = 3/8 × 4 = 1.5 cm ✓

Q3 (Hard). ABC and BDE are two equilateral triangles where D is midpoint of BC. Ratio Area(ABC):Area(BDE):

BD = BC/2, so ratio of sides = 2:1 → ratio of areas = 4:1

Revision Notes

BPT: DE ∥ BC → AD/DB = AE/EC

Similarity Criteria:
  AA  → Two angles equal
  SSS → Three sides proportional
  SAS → One angle equal + including sides proportional

Area Ratio = (Side Ratio)²

Pythagoras Theorem:
  ∠B = 90° → AC² = AB² + BC²
  Converse: if AC² = AB² + BC², then ∠B = 90°

Common Pythagorean Triplets: (3,4,5), (5,12,13), (8,15,17), (7,24,25)

Common Mistakes:

❌ Ratio of areas = ratio of sides (wrong! it's the SQUARE of the ratio)

❌ Writing SSS similarity condition wrong (must be PROPORTIONAL, not equal)

❌ Not using the converse of BPT when proving lines are parallel

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