Triangles — Detailed Notes

Chapter 6 | CBSE Class 10 Mathematics (NCERT)

Study Diagrams (Visual Revision)

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Triangles — Decision Map

Triangles — Must-Write Strip

Basic Proportionality Theorem

6.1 Introduction

Two figures are congruent if they have the same shape and the same size. This chapter studies figures with the same shape but not necessarily the same size — similar figures. Similarity is the basis of many indirect measurements (heights, distant lengths) used later in trigonometry chapters.

6.2 Similar Figures

Idea

All circles are similar (same shape).
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 not similar (different shapes).

Definition (polygons)

Two polygons with the same number of sides are similar if:

Their corresponding angles are equal, and
Their corresponding sides are in the same ratio (proportion).

That common ratio is the scale factor (or representative fraction), as in maps and photo enlargements.

Key insight: Either condition alone is not enough.

Case	Angles	Sides	Similar?
Square & rectangle	Equal	Not proportional	No
Square & rhombus	Not equal	Proportional	No

Similarity is transitive: if $P \sim Q$ and $Q \sim R$ , then $P \sim R$ .

6.3 Similarity of Triangles

Two triangles are similar if corresponding angles are equal and corresponding sides are proportional. Triangles with equal corresponding angles are called equiangular. Thales linked equiangular triangles to proportional sides via the Basic Proportionality Theorem.

Theorem 6.1 (Basic Proportionality Theorem / Thales)

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.

Given: In $△ A B C$ , $D E ∥ B C$ with $D$ on $A B$ , $E$ on $A C$ .
Prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .

Proof outline (NCERT): Join $B E$ , $C D$ ; draw $D M ⊥ A C$ , $E N ⊥ A B$ . Compare areas:

$\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}, \frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C} .$

$△ B D E$ and $△ D E C$ have equal areas (same base $D E$ , between same parallels $D E$ and $B C$ ), so the ratios 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.

Proof idea: Assume $D E$ meets $A B$ , $A C$ with $\frac{A D}{D B} = \frac{A E}{E C}$ . Draw $D E^{'} ∥ B C$ . By Theorem 6.1, $\frac{A D}{D B} = \frac{A E ^{'}}{E ^{'} C}$ . Equality of ratios forces $E$ and $E^{'}$ to coincide, so $D E ∥ B C$ .

Useful corollary form (Example 1)

If $D E ∥ B C$ , then also $\frac{A D}{A B} = \frac{A E}{A C} .$ (From BPT, take reciprocals and add 1.)

Worked ideas

Trapezium (Example 2 style): $A B C D$ with $A B ∥ D C$ ; $E$ , $F$ on non-parallel sides with $E F ∥ A B$ . Join $A C$ meeting $E F$ at $G$ . Apply BPT in $△ A D C$ and $△ C A B$ to get $\frac{A E}{E D} = \frac{B F}{F C}$ .

Isosceles via converse (Example 3 style): If $\frac{P S}{S Q} = \frac{P T}{T R}$ and $∠ P S T = ∠ P R Q$ , then $S T ∥ QR$ (Thm 6.2), corresponding angles give $∠ P QR = ∠ P R Q$ , so $P Q = P R$ .

Better Method / Strategy: For parallel lines inside a triangle, default to BPT. To prove a line is parallel, aim for equal ratios then quote Theorem 6.2. In trapeziums, draw a diagonal to create two similar or BPT-friendly triangles.

6.4 Criteria for Similarity of Triangles

Write similarity with correct correspondence: $△ A B C \sim △ D E F \Rightarrow ∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F, \frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D} .$ You cannot write $△ A B C \sim △ E D F$ for the same pairing.

Theorem 6.3 — AAA similarity

If in two triangles corresponding angles are equal, then corresponding sides are proportional and the triangles are similar.

Remark (AA): If two angles of one triangle equal two angles of another, the third angles match (angle sum). So AA is enough and is the form used most in exams.

Theorem 6.4 — SSS similarity

If sides of one triangle are proportional to sides of the other, then corresponding angles are equal and the triangles are similar.

Theorem 6.5 — SAS similarity

If one angle of a triangle equals one angle of the other and the including sides are proportional, then the triangles are similar.

Better Method / Strategy:

Prefer AA when parallels/VOA/corresponding angles are visible.

Use SSS when three side ratios are given (or easy to compute).

Use SAS when one included angle and the two adjacent side ratios match.

Always list equal angles before claiming a criterion.

Worked ideas

AA with parallels (Example 4): $P Q ∥ R S$ $\Rightarrow$ alternate angles equal; vertically opposite angles at $O$ equal $\Rightarrow$ $△ P O Q \sim △ S O R$ (AAA/AA).

SSS to find an angle (Example 5): If $\frac{A B}{R Q} = \frac{B C}{QP} = \frac{C A}{P R},$ then $△ A B C \sim △ R QP$ , so corresponding angles equal. Use angle sum in the known triangle to find the unknown angle.

SAS from products (Example 6): $O A \cdot O B = O C \cdot O D$ $\Rightarrow$ $\frac{O A}{O C} = \frac{O D}{O B}$ , with vertically opposite $∠ A O D = ∠ C O B$ $\Rightarrow$ $△ A O D \sim △ C O B$ (SAS).

Application — lamp-post shadow (Example 7): Girl height $0.9$ m, lamp $3.6$ m, walks $4.8$ m in $4$ s. By AA ( $△ A B E \sim △ C D E$ ), $\frac{4.8 + x}{x} = \frac{3.6}{0.9} = 4 \Rightarrow x = 1.6 m .$

Medians in similar triangles (Example 8): If $△ A B C \sim △ P QR$ and $C M$ , $R N$ are medians, prove $△ A M C \sim △ P N R$ (SAS), $\frac{C M}{R N} = \frac{A B}{P Q}$ , and $△ C M B \sim △ R N Q$ (SSS).

Note to the reader (RHS similarity)

If in two right triangles the hypotenuse and one side are proportional to the hypotenuse and one side of the other, the triangles are similar (RHS similarity). Useful later in trigonometry proofs.

Extra Board-Style Patterns

Pattern A — Parallel lines stack

If $D E ∥ B C$ and another line through $D$ or $E$ is also parallel to a side, apply BPT twice (once in each small triangle) and equate ratios. Classic NCERT figures with $L M ∥ C B$ and $L N ∥ C D$ reduce to $\frac{A M}{A B} = \frac{A N}{A D}$ .

Pattern B — “Show similar, then transfer sides”

Establish AA/SSS/SAS.
Write the three side ratios in matching order.
Cross-multiply only after correspondence is fixed.

Pattern C — Altitudes intersecting

When altitudes $A D$ and $C E$ meet at $P$ , right angles create several AA pairs: $△ A E P \sim △ C D P$ , $△ A B D \sim △ C B E$ , etc. Look for shared angles and vertical angles.

Pattern D — Median + similarity

If $△ A B C \sim △ P QR$ , medians to corresponding sides are in the same scale factor: $\frac{A D}{P M} = \frac{A B}{P Q} .$ Proof idea: mid-points halve corresponding sides, so SAS/SSS applies to half-triangles.

Pattern E — Power-of-point flavour (Ex. 6.3 style)

If $∠ A D C = ∠ B A C$ with $D$ on $B C$ , then $△ A D C \sim △ B A C$ (AA) and $C A^{2} = C B \cdot C D .$

How to Write a Full Similarity Proof (Template)

Given: …
To prove: △___ ~ △___   OR   a length/ratio relation

Proof:
1. List equal angles with reasons
   (common / corresponding / alternate / VOA / given).
2. Name criterion: AA / SSS / SAS.
3. Write △ABC ~ △XYZ with correct order.
4. If needed, write AB/XY = BC/YZ = CA/ZX and conclude.

Micro-example (AA)

$P Q ∥ R S$ , lines cross at $O$ .
$∠ O P Q = ∠ O S R$ (alternate), $∠ O QP = ∠ O R S$ (alternate)
$\Rightarrow$ $△ P O Q \sim △ S O R$ (AA).

Micro-example (SSS)

Sides $3.8, 6, 33$ proportional to $7.6, 12, 63$ (ratio $\frac{1}{2}$ )
$\Rightarrow$ SSS $\Rightarrow$ corresponding angles equal.

Micro-example (SAS)

$O A \cdot O B = O C \cdot O D$ $\Rightarrow$ $\frac{O A}{O C} = \frac{O D}{O B}$ , and $∠ A O D = ∠ C O B$ (VOA)
$\Rightarrow$ SAS similarity of $△ A O D$ and $△ C O B$ .

Linking to Later Chapters

Chapter 7: Section formula derivation uses AA similarity of right triangles.
Chapters 8–9: Heights/distances and trig ratios sit on similar right triangles (often AA).
RHS note: Speeds up some right-triangle similarity arguments in trigonometry.

Exam-Oriented Analysis

Aspect	Detail
Weight	Geometry block; often 3–6 marks (sometimes more across paper)
Question types	Fill blanks; “which pairs similar?”; BPT find length; short proofs; application numericals
1-mark	Definition / fill (all circles similar, etc.)
2-mark	BPT numerical or name the criterion
3–4 mark	Prove similarity or a ratio using AA/BPT
Common mistakes	Wrong vertex order; using only angles or only sides for polygons; confusing SAS similarity with SAS congruence; mixing $A D / D B$ with $A D / A B$
Scoring tips	State theorem/criterion by name; mark equal angles on figure; write ratios in matching correspondence order

Scoring tip: For a 3-mark proof, structure as: Given $\to$ equal angles/ratios with reasons $\to$ named criterion $\to$ required conclusion.

Quick self-test before the exam

State BPT and converse in under 20 seconds.
Given three equal side ratios, which criterion?
Why AA is enough.
Write one correct and one incorrect similarity statement for the same triangles.