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Chapter 6 · CBSE Class 10 Maths

Triangles

Two ways to learn — a crisp formula reference, or a story that makes similarity criteria feel obvious.

Basic Proportionality TheoremAA similaritySSS similaritySAS similarity
CBSEClass 10MathematicsChapter 68 min crisp · 12 min story

Four scenes that build the ideas from scratch, step by step.

How this works

Four scenes, one thread: same-shape triangles hide everywhere — shadows, intersecting lines, parallel cuts — and a handful of shortcuts let you prove similarity from just 2-3 facts instead of checking every angle and side. Read the story, then tap "The Maths" for the formal result.

1

The Girl and the Lamp-post

A girl 90 cm tall walks away from a 3.6 m lamp-post at 1.2 m/s. After 4 seconds, she has walked 4.8 m. How long is her shadow at that moment?

Student

The lamp-post, the girl, and the shadow's tip form some kind of triangle picture — but how does that actually give me the shadow length?

Both the lamp-post and the girl stand straight up (each making a 90° angle with the ground), and both share the exact same angle at the tip of the shadow — the sun's ray hits both at the same angle. Two matching angles are enough: the big triangle (lamp-post to shadow tip) and the small triangle (girl to shadow tip) are similar.

Let shadow length = x metres. Total distance from lamp-post base to shadow tip = 4.8 + x

(lamp-post height)/(girl height) = (lamp-post's shadow-distance)/(girl's shadow length)

3.6/0.9 = (4.8+x)/x → 4 = (4.8+x)/x → 4x = 4.8+x → 3x = 4.8

The shadow is 1.6 m long — found without measuring anything directly, just from the fact that two right-angled triangles sharing an angle must be the same shape, scaled.

2

Thales' Trick

Draw any triangle ABC. Draw a line DE parallel to side BC, cutting the other two sides at D and E. Measure AD, DB, AE, EC. No matter how the triangle is drawn, AD/DB always equals AE/EC exactly.

Thales

A line parallel to one side of a triangle always splits the other two sides in the same ratio — try it with any triangle, any parallel line, and you'll get the same result every time.

DE || BC ⟹ AD/DB = AE/EC

This is the Basic Proportionality Theorem (BPT), also called Thales' Theorem

The reverse is equally powerful: if you already know a ratio holds — AD/DB = AE/EC — you can conclude DE must be parallel to BC, without measuring any angles at all.

Real use case: PS/SQ = PT/TR is given, along with ∠PST = ∠PRQ. The ratio alone (converse BPT) proves ST || QR, which then makes ∠PST = ∠PQR (corresponding angles). Combined with the given ∠PST = ∠PRQ, this forces ∠PRQ = ∠PQR — so △PQR turns out to be isosceles, entirely from one ratio fact.
3

Checking Similarity Without Measuring Everything

Two triangles have 6 total pieces of correspondence to check (3 angles + 3 sides) if you wanted to prove similarity the "hard way." Two lines PQ and RS cross at O, with PQ || RS. Is △POQ similar to △SOR?

Teacher

You don't need all 6 pieces. Just find two angle matches, and the third follows automatically.

∠P = ∠S (alternate angles, since PQ || RS)

∠Q = ∠R (alternate angles, same reason)

Two angle matches already found — that's enough

The third angle pair (∠POQ and ∠SOR) turns out to also be equal — they're vertically opposite angles at the crossing point — but that wasn't even necessary to prove similarity. Two angle matches alone guarantee the third, by the angle-sum property of a triangle.

4

SSS and SAS — Similarity's Other Shortcuts

Sometimes angles aren't given directly, but side lengths (or their ratios) are. Take △ABC and △RQP where AB/RQ = BC/QP = CA/PR — all three ratios equal. If ∠A=80° and ∠B=60°, what is ∠P?

Student

All three side ratios match, but I don't see any angle information linking the two triangles directly.

All 3 corresponding side ratios equal ⟹ △ABC ~ △RQP (SSS similarity)

So ∠C = ∠P (corresponding angles of similar triangles)

∠C = 180° − 80° − 60° = 40°

Now a different setup: two chords intersect such that OA·OB = OC·OD. This can be rearranged into a side-ratio: OA/OC = OD/OB. Combined with the vertically-opposite angle ∠AOD = ∠COB right at the intersection point — one angle, plus the two sides that INCLUDE it — that's the SAS pattern.

The Big Picture

Every idea in this chapter branches from one fact: similar triangles are the same shape at different scales. A parallel line inside a triangle always creates a fixed ratio (BPT); and three shortcuts — AA, SSS, SAS — each prove full similarity from a small subset of the 6 possible correspondences.

Q

Line parallel to one side of a triangle?

BPT: AD/DB = AE/EC

Q

Know a side ratio, want to prove parallel?

Converse BPT: ratio ⟹ parallel

Q

Only angles known, no sides?

AA: 2 angle matches ⟹ similar

Q

Only side ratios known, no angles?

SSS: all 3 side ratios equal ⟹ similar

Q

One angle + 2 sides around it?

SAS: included angle + 2 sides ⟹ similar

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