Triangles – Class 10 Maths Chapter 6 Notes & NCERT PDF

Class 10MathematicsChapter 6NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 6 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 6: Triangles

Introduction

In Class IX you studied congruent figures — same shape and same size. This chapter studies similar figures — same shape but not necessarily the same size. The book notes that indirect measurements, like the height of a mountain or the distance to the Moon, are based on the principle of similarity.

Key Concepts & Theorems

Similar figures

All circles are similar; all squares are similar; all equilateral triangles are similar.

All congruent figures are similar, but similar figures need not be congruent.

A circle and a square are never similar; a triangle and a square are never similar.

Two polygons with the same number of sides are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (proportion).

Both conditions are needed for polygons: a square and a rectangle have equal angles but sides not in proportion; a square and a rhombus have sides in proportion but angles not equal.

Similar triangles

Triangles ABC and DEF are similar (written △ABC ~ △DEF) if ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB/DE = BC/EF = CA/FD. The order of letters tells which parts correspond.

Theorem 6.1 — Basic Proportionality Theorem (BPT, Thales' theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

In △ABC with DE ∥ BC (D on AB, E on AC): AD/DB = AE/EC. The proof uses areas of triangles with equal heights: triangles BDE and DEC lie on the same base DE between the same parallels, so their areas are equal.

Theorem 6.2 — Converse of BPT

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Criteria for similarity

AAA (Theorem 6.3): if corresponding angles are equal, the triangles are similar. Since the angle sum is 180°, two equal angles are enough — the AA criterion.

SSS (Theorem 6.4): if the sides of one triangle are proportional to the sides of the other, the triangles are similar.

SAS (Theorem 6.5): if one angle is equal and the sides including that angle are proportional, the triangles are similar.

Methods Step by Step

To prove two triangles similar: name the triangles in corresponding order, list the equal angles or proportional sides with reasons (vertically opposite angles, alternate angles with parallel lines, common angle, given), then name the criterion.

To find a length: prove similarity first, then write the ratio of corresponding sides and solve.

BPT questions: check that DE ∥ BC is given before writing AD/DB = AE/EC.

Worked Examples (from the chapter)

Example 4: If PQ ∥ RS, then △POQ ~ △SOR — alternate angles ∠P = ∠S and ∠Q = ∠R, and vertically opposite angles at O (AA).

Example 5: AB/RQ = BC/QP = CA/PR = 1/2, so △ABC ~ △RQP (SSS). Then ∠P = ∠C = 180° − 80° − 60° = 40°.

Example 6: OA · OB = OC · OD gives OA/OC = OD/OB; with vertically opposite angles, △AOD ~ △COB (SAS), so ∠A = ∠C and ∠B = ∠D.

Example 7 (shadow problem): A 90 cm girl walks away from a 3.6 m lamp-post at 1.2 m/s. After 4 s she is 4.8 m away. With shadow x: 3.6/0.9 = (4.8 + x)/x → 4x = 4.8 + x → x = 1.6 m.

Example 2: In trapezium ABCD with AB ∥ DC and EF ∥ AB, the BPT applied twice (through a diagonal) gives AE/ED = BF/FC.

Proof Outlines

BPT (Theorem 6.1)

Given: △ABC, DE ∥ BC, D on AB, E on AC. To prove: AD/DB = AE/EC.

Join BE and CD; draw DM ⊥ AC and EN ⊥ AB.

ar(ADE) = ½ × AD × EN and ar(BDE) = ½ × DB × EN, so ar(ADE)/ar(BDE) = AD/DB.

Similarly ar(ADE)/ar(DEC) = AE/EC.

△BDE and △DEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).

Therefore AD/DB = AE/EC.

Converse (Theorem 6.2)

Assume AD/DB = AE/EC but DE is not parallel to BC.

Draw DE′ ∥ BC; by BPT, AD/DB = AE′/E′C, so AE/EC = AE′/E′C.

Adding 1 to both sides gives AC/EC = AC/E′C, so EC = E′C, meaning E and E′ coincide — a contradiction. Hence DE ∥ BC.

Why AA is enough

If ∠A = ∠D and ∠B = ∠E, then ∠C = 180° − ∠A − ∠B = 180° − ∠D − ∠E = ∠F. So two pairs of equal angles already give all three pairs (AAA).

Similarity vs Congruence

Congruence criteria (SSS, SAS, ASA, RHS) need equal sides; similarity criteria (SSS, SAS, AA) need only proportional sides.

Congruent triangles are similar with ratio 1 : 1.

Similar triangles let us measure things we cannot reach — heights of towers, widths of rivers, the distance to the Moon — by comparing with a small triangle we can measure.

Board Exam Focus

Statement and proof of BPT is a frequently asked theorem.

Application problems: shadows (lamp-post, tower), trapezium ratios, finding sides with DE ∥ BC.

Always write the correspondence correctly (△ABC ~ △RQP, not △PQR, in Example 5).

Formula / Fact Sheet

BPT: DE ∥ BC ⇒ AD/DB = AE/EC (also AD/AB = AE/AC)

Converse: AD/DB = AE/EC ⇒ DE ∥ BC

Similarity criteria: AAA / AA, SSS, SAS

Similar triangles ⇒ corresponding angles equal and corresponding sides proportional

Common Mistakes

Writing triangles in the wrong vertex order, leading to wrong ratios.

Applying BPT without the parallel-line condition.

Treating SSA as a similarity criterion (it is not).

Thinking all rectangles are similar (only sides in proportion makes them similar).

In the shadow problem, forgetting to convert 90 cm to 0.9 m.

💡 Key Learning Points

✓Similar figures have the same shape but not necessarily the same size; congruent figures are always similar.
✓Polygons are similar when corresponding angles are equal and corresponding sides are proportional.
✓Basic Proportionality Theorem: a line parallel to one side divides the other two sides in the same ratio; its converse also holds.
✓Triangles are similar by AAA (or AA), SSS or SAS criteria.
✓Similarity is used for indirect measurement, such as finding the length of a shadow or the height of a tower.

👨‍🏫 Teaching Tips

→Show photographs enlarged and reduced from the same negative to introduce 'same shape, different size'.
→Use the book's square vs rectangle and square vs rhombus examples to show why both conditions are needed for polygons.
→Let students draw a triangle, draw a line parallel to the base, and measure AD, DB, AE, EC to discover BPT before the proof.
→Present the BPT proof as an area argument, stressing why triangles BDE and DEC have equal areas.
→Insist on writing the correspondence (△ABC ~ △RQP) before any ratio; give a quick drill on matching vertices.
→Take students outside at a fixed time to measure a stick's shadow and a pole's shadow and find the pole's height by similarity.
→Compare congruence criteria (SSS, SAS, ASA) with similarity criteria (SSS, SAS, AA) side by side.

📋 Assessment Questions

State and prove the Basic Proportionality Theorem.
Are all squares similar? Are all rectangles similar? Give reasons.
In △ABC, DE ∥ BC with AD = 4, DB = 6, AE = 6. Find EC.
Prove that if PQ ∥ RS then △POQ ~ △SOR.
A 90 cm tall girl is 4.8 m from a 3.6 m lamp-post; find her shadow length.

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