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CBSE Class 10 · Maths · board exam 2027

Chapter 6: Triangles — notes

The chapter’s 7 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.

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  1. 1. Similar figures

    The board.

    • • Similar figures: same shape, size may differ
    • • Congruent: same shape and same size
    • • Every congruent pair is similar.
    • • A similar pair need not be congruent.
    • • Always similar: all circles, all squares, all equilateral triangles
    • • Two polygons are similar if:
    • • (i) corresponding angles are equal, and
    • • (ii) corresponding sides are in the same ratio (scale factor).
    • • Square 2 cm × 2 cm, rectangle 2 cm × 1 cm:
    • • angles equal, but 2/2 ≠ 2/1 ⇒ not similar
    • • Square and rhombus: sides in the same ratio,
    • • angles not equal ⇒ not similar

  2. 2. Similar triangles and their correspondence

    A triangle is also a polygon.

    • • △ABC ~ △DEF ( ~ : is similar to )
    • • ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
    • • A ↔ D, B ↔ E, C ↔ F
    • • AB/DE = BC/EF = CA/FD
    • • Equiangular triangles: corresponding angles equal
    • • ⇒ corresponding sides in the same ratio
    • • Example: △ABC ~ △PQR, AB = 6, BC = 4, AC = 8, PR = 6 cm. Find PQ + QR.
    • • AB ↔ PQ, BC ↔ QR, AC ↔ PR
    • • AB/PQ = BC/QR = AC/PR
    • • 6/PQ = 8/6
    • • PQ = (6 × 6)/8 = 36/8 = 4.5
    • • 4/QR = 8/6
    • • QR = (4 × 6)/8 = 24/8 = 3
    • • PQ + QR = 4.5 + 3 = 7.5 cm
    • • △ABC ~ △QRP ⇒ A ↔ Q, B ↔ R, C ↔ P
    • • So AB ↔ QR, BC ↔ RP, CA ↔ PQ

  3. 3. The Basic Proportionality Theorem

    This △ABC. A line DE is drawn parallel to BC. It cuts AB at D and AC at E. The Greek mathematician Thales found something beautiful here.

    • • BPT: If a line is drawn parallel to one side of a triangle
    • • to meet the other two sides in distinct points,
    • • then the other two sides are divided in the same ratio.
    • • In △ABC, DE ∥ BC ⇒ AD/DB = AE/EC
    • • Given: In △ABC, DE ∥ BC, D on AB, E on AC.
    • • To prove: AD/DB = AE/EC
    • • Construction: Join BE and CD.
    • • Draw DM ⊥ AC and EN ⊥ AB.
    • • ar(ADE) = 1/2 × AD × EN
    • • ar(BDE) = 1/2 × DB × EN
    • • ar(ADE)/ar(BDE) = (AD × EN)/(DB × EN) = AD/DB ...(1)
    • • ar(ADE) = 1/2 × AE × DM
    • • ar(DEC) = 1/2 × EC × DM
    • • ar(ADE)/ar(DEC) = (AE × DM)/(EC × DM) = AE/EC ...(2)
    • • △BDE and △DEC are on the same base DE
    • • and between the same parallels DE and BC.
    • • So ar(BDE) = ar(DEC) ...(3)
    • • From (1), (2) and (3):
    • • AD/DB = AE/EC
    • • Hence proved.
    • • DE ∥ BC ⇒ AD/DB = AE/EC
    • • and also AD/AB = AE/AC
    • • Example: AD = x, DB = x − 2, AE = x + 2, EC = x − 1. Find x.
    • • AD/DB = AE/EC
    • • x/(x − 2) = (x + 2)/(x − 1)
    • • x(x − 1) = (x + 2)(x − 2)
    • • x² − x = x² − 4
    • • −x = −4, so x = 4
    • • Check: 4/2 = 2 and 6/3 = 2 ✓

  4. 4. The converse of BPT

    Turn the theorem around.

    • • 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.
    • • AD = DB ⇒ AD/DB = 1
    • • AE = EC ⇒ AE/EC = 1
    • • So DE ∥ BC (converse of BPT)
    • • Example: PS/SQ = PT/TR and ∠PST = ∠PRQ. Prove △PQR is isosceles.
    • • PS/SQ = PT/TR ⇒ ST ∥ QR (converse of BPT)
    • • ∠PST = ∠PQR (corresponding angles)
    • • ∠PST = ∠PRQ (given)
    • • So ∠PQR = ∠PRQ
    • • So PR = PQ: △PQR is isosceles. Hence proved.

  5. 5. The AA criterion of similarity

    Checking three angles and three sides every time is a lot of work.

    • • AAA: If the corresponding angles of two triangles are equal,
    • • then the triangles are similar.
    • • ∠A + ∠B + ∠C = 180°
    • • Two angles equal ⇒ the third is equal
    • • AA: two angles equal ⇒ triangles similar
    • • Equal angles come from:
    • • - a common angle • vertically opposite angles
    • • - parallel lines (alternate, corresponding) • right angles
    • • Example: girl CD = 0.9 m, lamp AB = 3.6 m, speed 1.2 m/s, time 4 s.
    • • Let the shadow DE = x m
    • • BD = 1.2 × 4 = 4.8 m
    • • ∠B = ∠D = 90°, ∠E = ∠E
    • • So △ABE ~ △CDE (AA)
    • • CD/AB = DE/BE
    • • 0.9/3.6 = x/(4.8 + x)
    • • 0.9/3.6 = 1/4
    • • 4x = 4.8 + x
    • • 3x = 4.8
    • • x = 4.8/3 = 1.6
    • • Her shadow = 1.6 m
    • • Example: ∠ADC = ∠BAC. Prove CA² = CB × CD.
    • • In △BAC and △ADC: ∠BAC = ∠ADC (given), ∠C = ∠C (common)
    • • So △BAC ~ △ADC (AA)
    • • CB/CA = CA/CD
    • • CA² = CB × CD. Hence proved.

  6. 6. The SSS criterion of similarity

    The second test uses only sides.

    • • SSS: If the corresponding sides of two triangles are in the same ratio,
    • • then the triangles are similar.
    • • Sides 3, 4, 5 cm and 6, 8, 10 cm
    • • 3/6 = 4/8 = 5/10 = 1/2
    • • So the triangles are similar (SSS)
    • • Sides 2, 3, 4 and 4, 6, 9:
    • • 2/4 = 3/6 = 1/2, but 4/9 ≠ 1/2
    • • So not similar
    • • AB/RQ = BC/QP = CA/PR ⇒ △ABC ~ △RQP
    • • So ∠C = ∠P

  7. 7. The SAS criterion of similarity

    The third test mixes both.

    • • SAS: If one angle of a triangle is equal to one angle of the other,
    • • and the sides including these angles are proportional,
    • • then the triangles are similar.
    • • ∠A = ∠P
    • • AB/PQ = 4/2 = 2, AC/PR = 6/3 = 2
    • • So △ABC ~ △PQR (SAS)
    • • The equal angle must be the included angle.
    • • Example: OA × OB = OC × OD. Prove △AOD ~ △COB.
    • • OA × OB = OC × OD ⇒ OA/OC = OD/OB
    • • ∠AOD = ∠COB (vertically opposite angles)
    • • So △AOD ~ △COB (SAS)
    • • So ∠A = ∠C

Summary

Very good! Similar figures have the same shape, and two polygons are similar when the angles are equal and the sides are in the same ratio, both together. Then we proved the Basic Proportionality Theorem, step by step, and met its converse. Then came the three tests of similarity. AA, two angles. SSS, three sides in the same ratio. And SAS, one angle with the two sides around it. And remember the habit of this chapter: always read the order of the letters to match the vertices. Write the proof of BPT once in your notebook, without looking, with Given, To prove, Construction and Proof. Today's practice questions are on your screen. If anything is not clear, press the Ask button and ask me. Done, you worked very well today!

AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 6. Written from the textbook, checked twice.

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Where are these notes written from?
From the chapter’s own textbook text, part by part, in the book’s own terms; a second, independent model then checked every part against the book. The full lesson, a check question for each part and the practice are in the AskMedha app.

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