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NCERT Solutions · Class 10 Maths · Chapter 6

Chapter 6: Triangles (Geometry)

Step-by-step NCERT solutions for Class 10 Maths Chapter 6, Triangles (2026-27 reprint): Exercise 6.1 similar figures, Exercise 6.2 the Basic Proportionality Theorem and its converse, and Exercise 6.3 the AA, SSS and SAS similarity criteria with board-style proofs. All 29 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-6-triangles

Basic Proportionality Theorem (BPT, Theorem 6.1): a line parallel to one side of a triangle divides the other two sides in the same ratio. Converse (Theorem 6.2): a line dividing two sides in the same ratio is parallel to the third side. Triangles are similar by AA (two angles equal), SSS (all three sides in the same ratio) or SAS (one angle equal and the sides including it in the same ratio).

Exercise 6.1

Fill in the blanks: (i) All circles are ____ (congruent, similar). (ii) All squares are ____ (similar, congruent). (iii) All ____ triangles are similar (isosceles, equilateral). (iv) Two polygons with the same number of sides are similar if their corresponding angles are ____ and their corresponding sides are ____ (equal, proportional).

Solution

(i) similar (ii) similar (iii) equilateral (iv) equal; proportional

Give two different examples of pairs of (i) similar figures (ii) non-similar figures.

Solution

(i) Similar: two circles of radii 2 cm and 5 cm; two squares of sides 3 cm and 7 cm; two equilateral triangles of different sizes; a photograph and its enlargement.

(ii) Non-similar: a square and a rhombus that is not a square; a square and a rectangle of sides 2 cm and 6 cm; an equilateral triangle and a right triangle; a circle and an ellipse.

Similar: any two circles, any two squares. Non-similar: a square and a non-square rhombus, an equilateral triangle and a right triangle.

State whether the quadrilaterals in Fig. 6.8 (the rhombus PQRS of side 1.5 cm and the square ABCD of side 3 cm) are similar.

Solution

The sides are in the same ratio (1.5 : 3 = 1 : 2), but the angles are not equal: ABCD has four right angles, while the rhombus PQRS does not.

Not similar: the corresponding angles are not equal.

Exercise 6.2

In Fig. 6.17, DE ∥ BC. Find EC in (i) and AD in (ii). [(i) AD = 1.5 cm, DB = 3 cm, AE = 1 cm; (ii) DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm]

Solution

By BPT, $\frac{A D}{D B} = \frac{A E}{E C}$ .

$\frac{1.5}{3} = \frac{1}{E C} \Rightarrow E C = 2$ cm
$\frac{A D}{7.2} = \frac{1.8}{5.4} \Rightarrow A D = \frac{7.2}{3} = 2.4$ cm

(i) EC = 2 cm (ii) AD = 2.4 cm

E and F are points on PQ and PR of △PQR. Is EF ∥ QR? (i) PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4 (ii) PE = 4, QE = 4.5, PF = 8, RF = 9 (iii) PQ = 1.28, PR = 2.56, PE = 0.18, PF = 0.36 (all in cm)

Solution

By the converse of BPT, EF ∥ QR exactly when $\frac{P E}{E Q} = \frac{P F}{F R}$ .

$\frac{3.9}{3} = 1.3$ , $\frac{3.6}{2.4} = 1.5$ . Unequal, so not parallel.
$\frac{4}{4.5} = \frac{8}{9}$ , $\frac{8}{9}$ . Equal, so parallel.
$E Q = 1.10$ , $F R = 2.20$ : $\frac{0.18}{1.10} = \frac{9}{55}$ , $\frac{0.36}{2.20} = \frac{9}{55}$ . Equal, so parallel.

(i) No (ii) Yes (iii) Yes

In Fig. 6.18, LM ∥ CB and LN ∥ CD. Prove that AM/AB = AN/AD.

Solution

In △ABC, LM ∥ CB, so by BPT $\frac{A M}{A B} = \frac{A L}{A C}$ … (1)

In △ADC, LN ∥ CD, so by BPT $\frac{A N}{A D} = \frac{A L}{A C}$ … (2)

From (1) and (2), $\frac{A M}{A B} = \frac{A N}{A D}$ .

Both ratios equal AL/AC (by BPT in △ABC and △ADC), so AM/AB = AN/AD.

In Fig. 6.19, DE ∥ AC and DF ∥ AE. Prove that BF/FE = BE/EC.

Solution

In △ABC, DE ∥ AC, so $\frac{B D}{D A} = \frac{B E}{E C}$ … (1)

In △ABE, DF ∥ AE, so $\frac{B D}{D A} = \frac{B F}{F E}$ … (2)

From (1) and (2), $\frac{B F}{F E} = \frac{B E}{E C}$ .

Both ratios equal BD/DA, so BF/FE = BE/EC.

In Fig. 6.20, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR.

Solution

In △POQ, DE ∥ OQ, so $\frac{P E}{E Q} = \frac{P D}{D O}$ … (1)

In △POR, DF ∥ OR, so $\frac{P F}{F R} = \frac{P D}{D O}$ … (2)

So $\frac{P E}{E Q} = \frac{P F}{F R}$ , and by the converse of BPT in △PQR, EF ∥ QR.

PE/EQ = PF/FR (both equal PD/DO), so EF ∥ QR by the converse of BPT.

In Fig. 6.21, A, B and C are points on OP, OQ and OR such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

Solution

In △OPQ, AB ∥ PQ, so $\frac{O A}{A P} = \frac{O B}{B Q}$ … (1)

In △OPR, AC ∥ PR, so $\frac{O A}{A P} = \frac{O C}{C R}$ … (2)

So $\frac{O B}{B Q} = \frac{O C}{C R}$ , and by the converse of BPT in △OQR, BC ∥ QR.

OB/BQ = OC/CR (both equal OA/AP), so BC ∥ QR.

Using Theorem 6.1, prove that a line through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

Solution

In △ABC, let D be the mid-point of AB and let the line through D parallel to BC meet AC at E.

By BPT, $\frac{A D}{D B} = \frac{A E}{E C}$ . Since $A D = D B$ , $\frac{A D}{D B} = 1$ , so $\frac{A E}{E C} = 1$ , i.e. $A E = E C$ .

AE/EC = AD/DB = 1, so E is the mid-point of AC.

Using Theorem 6.2, prove that the line joining the mid-points of two sides of a triangle is parallel to the third side.

Solution

In △ABC, let D and E be the mid-points of AB and AC. Then $A D = D B$ and $A E = E C$ , so

$\frac{A D}{D B} = 1 = \frac{A E}{E C}$

By the converse of BPT, DE ∥ BC.

AD/DB = AE/EC = 1, so DE ∥ BC.

ABCD is a trapezium with AB ∥ DC, and its diagonals meet at O. Show that AO/BO = CO/DO.

Solution

Through O, draw EF ∥ DC (and so ∥ AB), meeting AD at E and BC at F.

In △ADC, EO ∥ DC, so $\frac{A E}{E D} = \frac{A O}{O C}$ … (1)

In △DAB, EO ∥ AB, so $\frac{A E}{E D} = \frac{B O}{O D}$ … (2)

From (1) and (2), $\frac{A O}{O C} = \frac{B O}{O D}$ , i.e. $\frac{A O}{B O} = \frac{C O}{D O}$ .

(Alternatively: △AOB ~ △COD by AA, since ∠OAB = ∠OCD and ∠OBA = ∠ODC are alternate angles.)

Drawing EO ∥ AB ∥ DC and applying BPT twice gives AO/OC = BO/OD, i.e. AO/BO = CO/DO.

The diagonals of a quadrilateral ABCD meet at O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Solution

Through O, draw OE ∥ AB, meeting AD at E.

In △DAB, OE ∥ AB, so $\frac{D E}{E A} = \frac{D O}{O B}$ … (1)

Given $\frac{A O}{B O} = \frac{C O}{D O}$ , i.e. $\frac{D O}{O B} = \frac{C O}{O A}$ … (2)

From (1) and (2), $\frac{D E}{E A} = \frac{C O}{O A}$ , so in △ADC, by the converse of BPT, EO ∥ DC.

So AB ∥ EO ∥ DC, i.e. AB ∥ DC, and ABCD is a trapezium.

Drawing OE ∥ AB and using BPT and its converse gives AB ∥ DC, so ABCD is a trapezium.

Exercise 6.3

State which pairs of triangles in Fig. 6.34 are similar, giving the criterion and the similarity in symbolic form.

Solution

∠A = ∠P = 60°, ∠B = ∠Q = 80°, ∠C = ∠R = 40°: similar by AAA, △ABC ~ △PQR.
$\frac{A B}{QR} = \frac{2}{4}$ , $\frac{B C}{R P} = \frac{2.5}{5}$ , $\frac{C A}{P Q} = \frac{3}{6}$ , all $\frac{1}{2}$ : similar by SSS, △ABC ~ △QRP.
$\frac{M P}{D E} = \frac{2}{4}$ , $\frac{L P}{D F} = \frac{3}{6}$ , $\frac{L M}{E F} = \frac{2.7}{5}$ : not all equal, so not similar.
$\frac{M N}{QP} = \frac{2.5}{5}$ , $\frac{M L}{QR} = \frac{5}{10}$ and the included angles ∠M = ∠Q = 70°: similar by SAS, △MNL ~ △QPR.
We know AB, BC and ∠A, but ∠A is not between the known sides (∠B is), and only two sides of △DEF are given: the conditions are not enough, so not similar (on the given data).
In △DEF, ∠F = 180° – 70° – 80° = 30°; in △PQR, ∠P = 180° – 80° – 30° = 70°. So ∠D = ∠P, ∠E = ∠Q, ∠F = ∠R: similar by AA, △DEF ~ △PQR.

(i) AAA, △ABC ~ △PQR (ii) SSS, △ABC ~ △QRP (iii) Not similar (iv) SAS, △MNL ~ △QPR (v) Not similar (vi) AA, △DEF ~ △PQR

In Fig. 6.35, △ODC ~ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution

DOB is a straight line, so ∠DOC = 180° – 125° = 55°.

In △ODC, ∠DCO = 180° – 70° – 55° = 55°.

Since △ODC ~ △OBA, corresponding angles are equal: ∠OAB = ∠OCD = 55°.

∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°

Diagonals AC and BD of a trapezium ABCD with AB ∥ DC meet at O. Using a similarity criterion, show that OA/OC = OB/OD.

Solution

In △AOB and △COD:

∠OAB = ∠OCD (alternate angles, AB ∥ DC)
∠OBA = ∠ODC (alternate angles)

So △AOB ~ △COD (AA), and the corresponding sides are proportional: $\frac{O A}{O C} = \frac{O B}{O D}$ .

△AOB ~ △COD by AA (alternate angles), so OA/OC = OB/OD.

In Fig. 6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that △PQS ~ △TQR.

Solution

In △PQR, ∠1 = ∠2 (i.e. ∠PQR = ∠PRQ), so $P Q = P R$ .

Then $\frac{QR}{QS} = \frac{QT}{P R} = \frac{QT}{QP}$ .

In △PQS and △TQR, ∠Q is common, and the sides including it are proportional: $\frac{QS}{QR} = \frac{QP}{QT}$ .

So △PQS ~ △TQR (SAS).

Since ∠1 = ∠2, PQ = PR, so QR/QS = QT/QP; with the common angle Q, △PQS ~ △TQR by SAS.

S and T are points on PR and QR of △PQR such that ∠P = ∠RTS. Show that △RPQ ~ △RTS.

Solution

In △RPQ and △RTS:

∠RPQ = ∠RTS (given)
∠R is common

So △RPQ ~ △RTS (AA).

∠P = ∠RTS and ∠R is common, so △RPQ ~ △RTS by AA.

In Fig. 6.37, △ABE ≅ △ACD. Show that △ADE ~ △ABC.

Solution

Since △ABE ≅ △ACD (CPCT): $A B = A C$ and $A E = A D$ .

So $\frac{A D}{A B} = \frac{A E}{A C}$ (both sides of each fraction are equal pairs).

In △ADE and △ABC, ∠A is common and the sides including it are proportional, so △ADE ~ △ABC (SAS).

AB = AC and AD = AE give AD/AB = AE/AC; with the common angle A, △ADE ~ △ABC by SAS.

In Fig. 6.38, altitudes AD and CE of △ABC meet at P. Show that (i) △AEP ~ △CDP (ii) △ABD ~ △CBE (iii) △AEP ~ △ADB (iv) △PDC ~ △BEC.

Solution

∠AEP = ∠CDP = 90°, and ∠APE = ∠CPD (vertically opposite). So △AEP ~ △CDP (AA).
∠ADB = ∠CEB = 90°, and ∠B is common. So △ABD ~ △CBE (AA).
∠AEP = ∠ADB = 90°, and ∠PAE = ∠BAD (common angle at A). So △AEP ~ △ADB (AA).
∠PDC = ∠BEC = 90°, and ∠PCD = ∠BCE (common angle at C). So △PDC ~ △BEC (AA).

In each pair there is a right angle and a common or vertically opposite angle, so each similarity follows by AA.

E is a point on side AD produced of a parallelogram ABCD, and BE meets CD at F. Show that △ABE ~ △CFB.

Solution

In △ABE and △CFB:

∠A = ∠C (opposite angles of a parallelogram)
∠AEB = ∠CBF (alternate angles, since AE ∥ BC)

So △ABE ~ △CFB (AA).

∠A = ∠C and ∠AEB = ∠CBF (AD ∥ BC), so △ABE ~ △CFB by AA.

In Fig. 6.39, ABC and AMP are right triangles, right-angled at B and M. Prove that (i) △ABC ~ △AMP (ii) CA/PA = BC/MP.

Solution

(i) ∠ABC = ∠AMP = 90° and ∠A is common, so △ABC ~ △AMP (AA).

(ii) Corresponding sides of similar triangles are proportional, so $\frac{C A}{P A} = \frac{B C}{M P}$ .

(i) Right angle and common ∠A give AA similarity (ii) hence CA/PA = BC/MP.

CD and GH bisect ∠ACB and ∠EGF, with D on AB and H on FE. If △ABC ~ △FEG, show that (i) CD/GH = AC/FG (ii) △DCB ~ △HGE (iii) △DCA ~ △HGF.

Solution

From △ABC ~ △FEG: ∠A = ∠F, ∠B = ∠E, ∠ACB = ∠FGE. Halving the last: ∠ACD = ∠FGH and ∠DCB = ∠HGE.

(iii) In △DCA and △HGF: ∠A = ∠F and ∠ACD = ∠FGH, so △DCA ~ △HGF (AA).

(i) From (iii), $\frac{C D}{G H} = \frac{A C}{F G}$ .

(ii) In △DCB and △HGE: ∠B = ∠E and ∠DCB = ∠HGE, so △DCB ~ △HGE (AA).

Halving the equal angles C and G, AA gives △DCA ~ △HGF and △DCB ~ △HGE, and so CD/GH = AC/FG.

In Fig. 6.40, E is a point on side CB produced of an isosceles △ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that △ABD ~ △ECF.

Solution

AB = AC, so ∠ABC = ∠ACB, i.e. ∠ABD = ∠ECF.

In △ABD and △ECF:

∠ADB = ∠EFC = 90°
∠ABD = ∠ECF

So △ABD ~ △ECF (AA).

∠ABD = ∠ECF (base angles of the isosceles triangle) and both have a right angle, so △ABD ~ △ECF by AA.

Sides AB, BC and median AD of △ABC are proportional to sides PQ, QR and median PM of △PQR. Show that △ABC ~ △PQR.

Solution

Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ . Since D and M are mid-points, $B D = \frac{B C}{2}$ and $QM = \frac{QR}{2}$ , so $\frac{B C}{QR} = \frac{B D}{QM}$ .

So $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ , and △ABD ~ △PQM (SSS). Hence ∠B = ∠Q.

In △ABC and △PQR: $\frac{A B}{P Q} = \frac{B C}{QR}$ and ∠B = ∠Q, so △ABC ~ △PQR (SAS).

△ABD ~ △PQM by SSS gives ∠B = ∠Q; with AB/PQ = BC/QR, △ABC ~ △PQR by SAS.

D is a point on side BC of △ABC such that ∠ADC = ∠BAC. Show that CA² = CB·CD.

Solution

In △ADC and △BAC:

∠ADC = ∠BAC (given)
∠C is common

So △ADC ~ △BAC (AA), and $\frac{C A}{C B} = \frac{C D}{C A}$ , i.e. $C A^{2} = C B \cdot C D$ .

△ADC ~ △BAC by AA, so CA/CB = CD/CA, giving CA² = CB·CD.

Sides AB, AC and median AD of △ABC are proportional to sides PQ, PR and median PM of △PQR. Show that △ABC ~ △PQR.

Solution

Produce AD to E with $D E = A D$ , and PM to N with $M N = P M$ . Join EC and NR.

ABEC is a parallelogram (its diagonals bisect each other), so $C E = A B$ ; similarly $R N = P Q$ .

Given $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ , so $\frac{C E}{R N} = \frac{A C}{P R} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ , and △AEC ~ △PNR (SSS). So ∠CAE = ∠RPN, i.e. ∠CAD = ∠RPM.

In the same way (using parallelogram ABEC again, with ∠BAE = ∠AEC), ∠BAD = ∠QPM.

Adding, ∠BAC = ∠QPR. With $\frac{A B}{P Q} = \frac{A C}{P R}$ , △ABC ~ △PQR (SAS).

Doubling the medians to form parallelograms gives △AEC ~ △PNR (SSS), so ∠BAC = ∠QPR; then △ABC ~ △PQR by SAS.

A vertical pole 6 m long casts a shadow 4 m long, and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution

The sun's rays make the same angle with the ground for both, so the pole–shadow and tower–shadow triangles are similar (AA: each has a right angle and the same angle of elevation).

$\frac{h}{28} = \frac{6}{4} \Rightarrow h = 42$

The tower is 42 m high.

AD and PM are medians of △ABC and △PQR, with △ABC ~ △PQR. Prove that AB/PQ = AD/PM.

Solution

From △ABC ~ △PQR: ∠B = ∠Q and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{2 B D}{2 QM} = \frac{B D}{QM}$ .

In △ABD and △PQM: ∠B = ∠Q and the including sides are proportional, so △ABD ~ △PQM (SAS).

Hence $\frac{A B}{P Q} = \frac{A D}{P M}$ .

△ABD ~ △PQM by SAS (∠B = ∠Q, AB/PQ = BD/QM), so AB/PQ = AD/PM.

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