---
title: "NCERT Solutions for Class 10 Maths Chapter 6 Triangles (2026-27)"
url: https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions
dateModified: 2026-10-07T15:51:34+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 6 Triangles (2026-27)

This chapter's questions cover concepts of similarity of figures, basic proportionality theorem (Thales theorem), and criteria for the similarity of triangles such as AAA, SSS, and SAS.

Free PDF (16 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-6-triangles-39bb043ebf.pdf

## EXERCISE 6.1

### Question 1

*1 mark · Fill in the blank*

Fill in the blanks using the correct word given in brackets :
(i) All circles are $\rule{1cm}{0.15mm}$.
(ii) All squares are $\rule{1cm}{0.15mm}$.
(iii) All $\rule{1cm}{0.15mm}$ triangles are similar.
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are $\rule{1cm}{0.15mm}$ and (b) their corresponding sides are $\rule{1cm}{0.15mm}$.

**Part (i)**

1. All circles have the same shape but can have different radii.

Answer (i): similar

**Part (ii)**

1. All squares have the same shape with identical interior angles and proportional sides.

Answer (ii): similar

**Part (iii)**

1. All equilateral triangles have equal angles ($60^\circ$) and proportional sides.

Answer (iii): equilateral

**Part (iv)**

1. Two polygons of the same number of sides are similar if their corresponding angles are equal and corresponding sides are proportional.

Answer (iv): equal, proportional

**Answer:** (i) similar, (ii) similar, (iii) equilateral, (iv) equal, proportional

> Common mistake: Confusing congruent figures with similar figures.

### Question 2

*3 marks · Short answer*

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

**Part (i)**

1. Two figures are similar if they have the same shape but not necessarily the same size.
2. Example 1: Any two circles of different radii.
3. Example 2: Any two equilateral triangles of different side lengths.

Answer (i): Two circles of different radii, Two equilateral triangles of different side lengths

**Part (ii)**

1. Two figures are non-similar if they do not have the same shape.
2. Example 1: A triangle and a square.
3. Example 2: A triangle and a circle.

Answer (ii): A triangle and a square, A triangle and a circle

**Answer:** Provided pairs of similar and non-similar figures.

> Common mistake: Giving figures of different shapes for similar figures.

### Question 3

*3 marks · Short answer*

State whether the following quadrilaterals are similar or not:

**Solution**

1. Examine the given figure (Fig. 6.8) which shows a rhombus with side $1.5\text{ cm}$ and a square with side $3\text{ cm}$.
2. Check the ratio of corresponding sides: $\frac{1.5}{3} = \frac{1.5}{3} = \frac{1.5}{3} = \frac{1.5}{3} = \frac{1}{2}$, so the sides are proportional.
3. Check the corresponding angles: the square has each angle equal to $90^\circ$, whereas the rhombus does not have right angles.
4. Since the corresponding angles are not equal, the two quadrilaterals are not similar.

**Answer:** The given quadrilaterals are not similar because their corresponding angles are not equal.

> Common mistake: Concluded similarity by checking only the side proportions without verifying the angles.

## EXERCISE 6.2

### Question 1

*3 marks · Short answer*

In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find EC in (i) and AD in (ii).

**Part (i)**

1. Given in $\triangle ABC$, $DE \parallel BC$, $AD = 1.5\text{ cm}$, $DB = 3\text{ cm}$, and $AE = 1\text{ cm}$.
2. By Theorem 6.1 (Basic Proportionality Theorem), $\frac{AD}{DB} = \frac{AE}{EC}$.
3. Substituting the given values, $\frac{1.5}{3} = \frac{1}{EC}$.
4. Solving for $EC$, we get $EC = \frac{3 \times 1}{1.5} = 2\text{ cm}$.

Answer (i): 2 cm

**Part (ii)**

1. Given in $\triangle ABC$, $DE \parallel BC$, $AE = 1.8\text{ cm}$, $EC = 5.4\text{ cm}$, and $DB = 7.2\text{ cm}$.
2. By Theorem 6.1 (Basic Proportionality Theorem), $\frac{AD}{DB} = \frac{AE}{EC}$.
3. Substituting the given values, $\frac{AD}{7.2} = \frac{1.8}{5.4}$.
4. Solving for $AD$, we get $AD = \frac{1.8 \times 7.2}{5.4} = 2.4\text{ cm}$.

Answer (ii): 2.4 cm

**Answer:** EC = 2 cm in (i) and AD = 2.4 cm in (ii).

> Common mistake: Inverting the ratio terms such as taking $\frac{DB}{AD}$ instead of $\frac{AD}{DB}$.

### Question 2

*3 marks · Short answer*

E and F are points on the sides PQ and PR respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ :
(i) $PE = 3.9\text{ cm}, EQ = 3\text{ cm}, PF = 3.6\text{ cm}$ and $FR = 2.4\text{ cm}$
(ii) $PE = 4\text{ cm}, QE = 4.5\text{ cm}, PF = 8\text{ cm}$ and $RF = 9\text{ cm}$
(iii) $PQ = 1.28\text{ cm}, PR = 2.56\text{ cm}, PE = 0.18\text{ cm}$ and $PF = 0.36\text{ cm}$

**Part (i)**

1. Given: $PE = 3.9\text{ cm}$, $EQ = 3\text{ cm}$, $PF = 3.6\text{ cm}$, and $FR = 2.4\text{ cm}$.
2. Find the ratios $\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$ and $\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$.
3. Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$, by the converse of Basic Proportionality Theorem, $EF$ is not parallel to $QR$.

Answer (i): EF is not parallel to QR

**Part (ii)**

1. Given: $PE = 4\text{ cm}$, $QE = 4.5\text{ cm}$, $PF = 8\text{ cm}$, and $RF = 9\text{ cm}$.
2. Find the ratios $\frac{PE}{QE} = \frac{4}{4.5} = \frac{8}{9}$ and $\frac{PF}{RF} = \frac{8}{9}$.
3. Since $\frac{PE}{QE} = \frac{PF}{RF}$, by the converse of Basic Proportionality Theorem, $EF \parallel QR$.

Answer (ii): EF is parallel to QR

**Part (iii)**

1. Given: $PQ = 1.28\text{ cm}$, $PR = 2.56\text{ cm}$, $PE = 0.18\text{ cm}$, and $PF = 0.36\text{ cm}$.
2. Calculate $EQ = PQ - PE = 1.28 - 0.18 = 1.10\text{ cm}$ and $FR = PR - PF = 2.56 - 0.36 = 2.20\text{ cm}$.
3. Find the ratios $\frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{9}{55}$ and $\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{9}{55}$.
4. Since $\frac{PE}{EQ} = \frac{PF}{FR}$, by the converse of Basic Proportionality Theorem, $EF \parallel QR$.

Answer (iii): EF is parallel to QR

**Answer:** EF is not parallel to QR in (i); EF is parallel to QR in (ii) and (iii).

> Common mistake: Comparing whole sides like PE/PQ directly with PF/PR without finding EQ and FR or using correct proportions.

### Question 3

*3 marks · Proof*

In Fig. 6.18, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$

**Solution**

1. Given: In the figure (Fig. 6.18), $LM \parallel CB$ and $LN \parallel CD$.
2. To prove: $\frac{AM}{AB} = \frac{AN}{AD}$.
3. In $\triangle ABC$, since $LM \parallel CB$, by Theorem 6.1, we have $\frac{AM}{AB} = \frac{AL}{AC}$ (as proved in Example 1).
4. In $\triangle ADC$, since $LN \parallel CD$, by Theorem 6.1, we have $\frac{AN}{AD} = \frac{AL}{AC}$.
5. From both relations, since their right-hand sides are equal, we get $\frac{AM}{AB} = \frac{AN}{AD}$.
6. Hence proved.

**Answer:** $\frac{AM}{AB} = \frac{AN}{AD}$

> Common mistake: Using Theorem 6.1 directly with parts instead of whole sides when the result specifically requires $AB$ and $AD$ in the denominator.

### Question 4

*3 marks · Proof*

In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$

**Solution**

1. Given: In $\triangle ABC$ (Fig. 6.19), $DE \parallel AC$ and $DF \parallel AE$.
2. To prove: $\frac{BF}{FE} = \frac{BE}{EC}$.
3. In $\triangle ABE$, since $DF \parallel AE$ (or $DF \parallel EA$), by Theorem 6.1, we have $\frac{BD}{DA} = \frac{BF}{FE}$.
4. In $\triangle ABC$, since $DE \parallel AC$, by Theorem 6.1, we have $\frac{BD}{DA} = \frac{BE}{EC}$.
5. From the above two equations, since the left-hand sides are equal, we obtain $\frac{BF}{FE} = \frac{BE}{EC}$.
6. Hence proved.

**Answer:** $\frac{BF}{FE} = \frac{BE}{EC}$

> Common mistake: Applying Thales theorem on wrong triangles or mixing up corresponding segments.

### Question 5

*4 marks · Proof*

In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.

**Solution**

1. Given: In $\triangle POQ$ (Fig. 6.20), $DE \parallel OQ$ and $DF \parallel OR$.
2. To prove: $EF \parallel QR$.
3. In $\triangle POQ$, since $DE \parallel OQ$, by Theorem 6.1, we have $\frac{PE}{EQ} = \frac{PD}{DO}$.
4. In $\triangle POR$, since $DF \parallel OR$, by Theorem 6.1, we have $\frac{PF}{FR} = \frac{PD}{DO}$.
5. From the two equations, since their right-hand sides are equal, we get $\frac{PE}{EQ} = \frac{PF}{FR}$.
6. Therefore, in $\triangle PQR$, the line $EF$ divides the sides $PQ$ and $PR$ in the same ratio.
7. Hence, by Theorem 6.2 (Converse of Basic Proportionality Theorem), $EF \parallel QR$.
8. Hence proved.

**Answer:** $EF \parallel QR$

> Common mistake: Failing to state that the equality of ratios allows the application of the converse theorem.

### Question 6

*4 marks · Proof*

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

**Solution**

1. Given: In the figure (Fig. 6.21), $A$, $B$, and $C$ are points on rays $OP$, $OQ$, and $OR$ such that $AB \parallel PQ$ and $AC \parallel PR$.
2. To prove: $BC \parallel QR$.
3. In $\triangle OPQ$, since $AB \parallel PQ$, by Theorem 6.1, we have $\frac{OA}{AP} = \frac{OB}{BQ}$.
4. In $\triangle OPR$, since $AC \parallel PR$, by Theorem 6.1, we have $\frac{OA}{AP} = \frac{OC}{CR}$.
5. From these two equations, since their left-hand sides are equal, we obtain $\frac{OB}{BQ} = \frac{OC}{CR}$.
6. In $\triangle OQR$, the line $BC$ divides the sides $OQ$ and $OR$ in the same ratio.
7. Therefore, by Theorem 6.2 (Converse of Basic Proportionality Theorem), $BC \parallel QR$.
8. Hence proved.

**Answer:** $BC \parallel QR$

> Common mistake: Selecting incorrect triangles while applying the Basic Proportionality Theorem.

### Question 7

*4 marks · Proof*

Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

**Solution**

1. Given: A triangle ABC in which D is the mid-point of side AB and a line $DE \parallel BC$ is drawn intersecting AC at E.
2. To prove: E is the mid-point of side AC, that is, $AE = EC$.
3. Proof: Since $DE \parallel BC$, by Theorem 6.1 (Basic Proportionality Theorem), we have $\frac{AD}{DB} = \frac{AE}{EC}$.
4. Since D is the mid-point of AB, we have $AD = DB$, which gives $\frac{AD}{DB} = 1$.
5. Substituting this in the ratio, we get $1 = \frac{AE}{EC}$, or $AE = EC$.
6. Therefore, E is the mid-point of AC. Hence proved.

**Answer:** Hence proved that the line bisects the third side.

> Common mistake: Using the converse theorem instead of Theorem 6.1.

### Question 8

*4 marks · Proof*

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

**Solution**

1. Given: A triangle ABC in which D and E are the mid-points of sides AB and AC respectively, and DE is joined.
2. To prove: $DE \parallel BC$.
3. Proof: Since D is the mid-point of AB, we have $AD = DB$, so $\frac{AD}{DB} = 1$.
4. Similarly, since E is the mid-point of AC, we have $AE = EC$, so $\frac{AE}{EC} = 1$.
5. Therefore, $\frac{AD}{DB} = \frac{AE}{EC}$.
6. By Theorem 6.2 (Converse of Basic Proportionality Theorem), since the line DE divides the sides AB and AC in the same ratio, we have $DE \parallel BC$. Hence proved.

**Answer:** Hence proved that the line joining the mid-points is parallel to the third side.

> Common mistake: Failing to state Theorem 6.2 explicitly as the reason.

### Question 9

*4 marks · Proof*

ABCD is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point O. Show that $\frac{AO}{BO} = \frac{CO}{DO}$

**Solution**

1. Given: ABCD is a trapezium with $AB \parallel DC$, and its diagonals AC and BD intersect at O.
2. To prove: $\frac{AO}{BO} = \frac{CO}{DO}$
3. Proof: In $\triangle AOB$ and $\triangle COD$, $\angle OAB = \angle OCD$ (Alternate interior angles since $AB \parallel DC$).
4. Also, $\angle OBA = \angle ODC$ (Alternate interior angles since $AB \parallel DC$).
5. And $\angle AOB = \angle COD$ (Vertically opposite angles).
6. Therefore, by AA similarity criterion, $\triangle AOB \sim \triangle COD$.
7. Since the corresponding sides of similar triangles are proportional, we have $\frac{AO}{CO} = \frac{BO}{DO}$.
8. Rearranging the terms, we get $\frac{AO}{BO} = \frac{CO}{DO}$. Hence proved.

**Answer:** Hence proved that $\frac{AO}{BO} = \frac{CO}{DO}$.

> Common mistake: Writing incorrect corresponding vertices in the similarity statement.

### Question 10

*4 marks · Proof*

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that ABCD is a trapezium.

**Solution**

1. Given: A quadrilateral ABCD in which diagonals AC and BD intersect at O such that $\frac{AO}{BO} = \frac{CO}{DO}$.
2. To prove: ABCD is a trapezium.
3. Proof: The given condition $\frac{AO}{BO} = \frac{CO}{DO}$ can be rewritten as $\frac{AO}{CO} = \frac{BO}{DO}$.
4. Let us draw a line $EO \parallel AB$ meeting AD at E.
5. In $\triangle DAB$, since $EO \parallel AB$, by Theorem 6.1, we have $\frac{AE}{ED} = \frac{BO}{DO}$.
6. Comparing this with $\frac{AO}{CO} = \frac{BO}{DO}$, we get $\frac{AE}{ED} = \frac{AO}{CO}$.
7. By Theorem 6.2 (Converse of Basic Proportionality Theorem), in $\triangle ADC$, $EO \parallel DC$.
8. Since $EO \parallel AB$ and $EO \parallel DC$, it follows that $AB \parallel DC$.
9. Therefore, quadrilateral ABCD is a trapezium. Hence proved.

**Answer:** Hence proved that ABCD is a trapezium.

> Common mistake: Skipping the construction of a parallel line to apply the proportionality theorems.

## EXERCISE 6.3

### Question 1

*3 marks · Short answer*

State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

**Part (i)**

1. In $\triangle ABC$ and $\triangle PQR$, corresponding angles are $\angle A = \angle P = 60^\circ$, $\angle B = \angle Q = 80^\circ$ and $\angle C = \angle R = 40^\circ$.
2. Since all corresponding angles are equal, the two triangles are similar by AAA similarity criterion.
3. Therefore, $\triangle ABC \sim \triangle PQR$.

Answer (i): $\triangle ABC \sim \triangle PQR$ by AAA similarity

**Part (ii)**

1. In $\triangle ABC$ and $\triangle PQR$, the ratio of corresponding sides are $\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}$, $\frac{BC}{PR} = \frac{2.5}{5} = \frac{1}{2}$ and $\frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2}$.
2. Since the corresponding sides are proportional, the two triangles are similar by SSS similarity criterion.
3. Therefore, $\triangle ABC \sim \triangle QRP$.

Answer (ii): $\triangle ABC \sim \triangle QRP$ by SSS similarity

**Part (iii)**

1. In $\triangle LMP$ and $\triangle DEF$, the ratios of sides are $\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2}$, $\frac{LP}{DF} = \frac{3}{6} = \frac{1}{2}$, but $\frac{LM}{EF} = \frac{2.7}{5} \neq \frac{1}{2}$.
2. Since the corresponding sides are not proportional, the two triangles are not similar.

Answer (iii): Not similar

**Part (iv)**

1. In $\triangle MNL$ and $\triangle QPR$, we have $\frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2}$ and $\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}$, so $\frac{MN}{QP} = \frac{ML}{QR}$.
2. Also, the included angles are equal: $\angle M = \angle Q = 70^\circ$.
3. Therefore, by SAS similarity criterion, $\triangle MNL \sim \triangle QPR$.

Answer (iv): $\triangle MNL \sim \triangle QPR$ by SAS similarity

**Answer:** Pairs (i), (ii), (iv) and (vi) are similar.

> Common mistake: Writing the symbolic form of similarity with incorrect correspondence of vertices.

### Question 2

*3 marks · Short answer*

In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^\circ$ and $\angle CDO = 70^\circ$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.

**Solution**

1. Given that $\angle BOC = 125^\circ$ and line $DOC$ is a straight line, we have $\angle DOC + \angle BOC = 180^\circ$ (Linear pair).
2. So, $\angle DOC = 180^\circ - 125^\circ = 55^\circ$.
3. In $\triangle DOC$, the sum of angles is $180^\circ$, so $\angle DCO = 180^\circ - \angle CDO - \angle DOC = 180^\circ - 70^\circ - 55^\circ = 55^\circ$.
4. Since $\triangle ODC \sim \triangle OBA$, their corresponding angles are equal, which gives $\angle OAB = \angle ODC = \angle CDO = 70^\circ$.

**Answer:** $\angle DOC = 55^\circ$, $\angle DCO = 55^\circ$, $\angle OAB = 70^\circ$

> Common mistake: Confusing the corresponding vertices in the similarity statement $\triangle ODC \sim \triangle OBA$.

### Question 3

*3 marks · Proof*

Diagonals AC and BD of a trapezium ABCD with $AB \parallel DC$ intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$

**Solution**

1. Given: A trapezium $ABCD$ with $AB \parallel DC$ and diagonals $AC$ and $BD$ intersecting at $O$.
2. To prove: $\frac{OA}{OC} = \frac{OB}{OD}$.
3. In $\triangle AOB$ and $\triangle COD$, $\angle AOB = \angle COD$ (Vertically opposite angles).
4. Also, $\angle OAB = \angle OCD$ (Alternate interior angles since $AB \parallel DC$).
5. Therefore, by AA similarity criterion, $\triangle AOB \sim \triangle COD$.
6. Since corresponding sides of similar triangles are in the same ratio, $\frac{OA}{OC} = \frac{OB}{OD}$.
7. Hence proved.

**Answer:** $\frac{OA}{OC} = \frac{OB}{OD}$

> Common mistake: Taking incorrect alternate interior angles or wrong correspondence of vertices.

### Question 4

*3 marks · Proof*

In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.

**Solution**

1. Given: $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$.
2. To prove: $\triangle PQS \sim \triangle TQR$.
3. In $\triangle PQR$, since $\angle 1 = \angle 2$ (given $\angle PQR = \angle PRQ$), we have $PQ = PR$ (Sides opposite equal angles).
4. Substitute $PR = PQ$ in the given ratio $\frac{QR}{QS} = \frac{QT}{PR}$ to get $\frac{QR}{QS} = \frac{QT}{PQ}$, or $\frac{QS}{QR} = \frac{PQ}{QT}$.
5. In $\triangle PQS$ and $\triangle TQR$, we have $\frac{QS}{QR} = \frac{PQ}{QT}$ and the included angle $\angle Q$ is common ($\angle PQS = \angle TQR$).
6. Therefore, by SAS similarity criterion, $\triangle PQS \sim \triangle TQR$.
7. Hence proved.

**Answer:** $\triangle PQS \sim \triangle TQR$

> Common mistake: Using the ratios of sides in incorrect order without substituting $PR = PQ$.

### Question 5

*3 marks · Proof*

S and T are points on sides PR and QR of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.

**Solution**

1. Given: $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$.
2. To prove: $\triangle RPQ \sim \triangle RTS$.
3. In $\triangle RPQ$ and $\triangle RTS$, we have $\angle R = \angle R$ (Common angle).
4. Also, $\angle RPT = \angle P = \angle RTS$ (Given).
5. Therefore, by AA similarity criterion, $\triangle RPQ \sim \triangle RTS$.
6. Hence proved.

**Answer:** $\triangle RPQ \sim \triangle RTS$

> Common mistake: Matching the vertices of the similar triangles in the wrong order.

### Question 6

*3 marks · Proof*

In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.

**Solution**

1. Given: $\triangle ABE \cong \triangle ACD$.
2. To prove: $\triangle ADE \sim \triangle ABC$.
3. Since $\triangle ABE \cong \triangle ACD$, their corresponding parts are equal, so $AB = AC$ and $AE = AD$ (By CPCT).
4. Dividing $AE$ by $AB$ and $AD$ by $AC$, we get $\frac{AD}{AB} = \frac{AE}{AC}$.
5. In $\triangle ADE$ and $\triangle ABC$, we have $\frac{AD}{AB} = \frac{AE}{AC}$ and the included angle $\angle A$ is common ($\angle DAE = \angle BAC$).
6. Therefore, by SAS similarity criterion, $\triangle ADE \sim \triangle ABC$.
7. Hence proved.

**Answer:** $\triangle ADE \sim \triangle ABC$

> Common mistake: Failing to use CPCT properly to establish the proportionality of the sides including the common angle.

### Question 7

*4 marks · Proof*

In Fig. 6.38, altitudes AD and CE of $\triangle ABC$ intersect each other at the point P. Show that:
(i) $\triangle AEP \sim \triangle CDP$
(ii) $\triangle ABD \sim \triangle CBE$
(iii) $\triangle AEP \sim \triangle ADB$
(iv) $\triangle PDC \sim \triangle BEC$

**Part (i)**

1. In $\triangle AEP$ and $\triangle CDP$, $\angle AEP = \angle CDP = 90^\circ$ since $AD$ and $CE$ are altitudes.
2. $\angle APE = \angle CPD$ because they are vertically opposite angles.
3. Therefore, $\triangle AEP \sim \triangle CDP$ by AA similarity criterion.
4. Hence proved.

Answer (i): $\triangle AEP \sim \triangle CDP$

**Part (ii)**

1. In $\triangle ABD$ and $\triangle CBE$, $\angle ADB = \angle CEB = 90^\circ$ as $AD \perp BC$ and $CE \perp AB$.
2. $\angle B$ is common to both $\triangle ABD$ and $\triangle CBE$.
3. Therefore, $\triangle ABD \sim \triangle CBE$ by AA similarity criterion.
4. Hence proved.

Answer (ii): $\triangle ABD \sim \triangle CBE$

**Part (iii)**

1. In $\triangle AEP$ and $\triangle ADB$, $\angle AEP = \angle ADB = 90^\circ$ from the given altitudes.
2. $\angle A$ is common to both $\triangle AEP$ and $\triangle ADB$.
3. Therefore, $\triangle AEP \sim \triangle ADB$ by AA similarity criterion.
4. Hence proved.

Answer (iii): $\triangle AEP \sim \triangle ADB$

**Part (iv)**

1. In $\triangle PDC$ and $\triangle BEC$, $\angle PDC = \angle BEC = 90^\circ$ from the given altitudes.
2. $\angle C$ is common to both $\triangle PDC$ and $\triangle BEC$.
3. Therefore, $\triangle PDC \sim \triangle BEC$ by AA similarity criterion.
4. Hence proved.

Answer (iv): $\triangle PDC \sim \triangle BEC$

**Answer:** Hence proved for all parts.

> Common mistake: Wrong identification of corresponding angles and common angles in the triangles.

### Question 8

*4 marks · Proof*

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\triangle ABE \sim \triangle CFB$.

**Solution**

1. Given: $ABCD$ is a parallelogram, $E$ is a point on side $AD$ produced, and $BE$ intersects $CD$ at $F$.
2. In $\triangle ABE$ and $\triangle CFB$, $\angle A = \angle C$ because opposite angles of a parallelogram are equal.
3. $\angle AEB = \angle CBF$ because alternate interior angles for parallel lines $AD$ and $BC$ with transversal $BE$ are equal.
4. Therefore, $\triangle ABE \sim \triangle CFB$ by AA similarity criterion.
5. Hence proved.

**Answer:** $\triangle ABE \sim \triangle CFB$

> Common mistake: Misidentifying the alternate interior angles or confusing the corresponding vertices.

### Question 9

*4 marks · Proof*

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) $\triangle ABC \sim \triangle AMP$
(ii) $\frac{CA}{PA} = \frac{BC}{MP}$

**Part (i)**

1. In $\triangle ABC$ and $\triangle AMP$, $\angle ABC = \angle AMP = 90^\circ$ as given.
2. $\angle A$ is common to both $\triangle ABC$ and $\triangle AMP$.
3. Therefore, $\triangle ABC \sim \triangle AMP$ by AA similarity criterion.
4. Hence proved.

Answer (i): $\triangle ABC \sim \triangle AMP$

**Part (ii)**

1. Since $\triangle ABC \sim \triangle AMP$ from part (i), the corresponding sides are proportional.
2. Therefore, $\frac{CA}{PA} = \frac{BC}{MP}$.
3. Hence proved.

Answer (ii): $\frac{CA}{PA} = \frac{BC}{MP}$

**Answer:** Hence proved for all parts.

> Common mistake: Writing the ratios of sides in incorrect order.

### Question 10

*5 marks · Proof*

CD and GH are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that D and H lie on sides AB and FE of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that:
(i) $\frac{CD}{GH} = \frac{AC}{FG}$
(ii) $\triangle DCB \sim \triangle HGE$
(iii) $\triangle DCA \sim \triangle HGF$

**Part (i)**

1. $\triangle ABC \sim \triangle FEG$ (Given)
2. $\angle A = \angle F$, $\angle B = \angle E$, and $\angle ACB = \angle FGE$ (Corresponding parts of similar triangles)
3. $\angle ACD = \frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE = \angle FGH$ (Given that CD and GH are bisectors)
4. In $\triangle ADC$ and $\triangle FHG$, $\angle A = \angle F$ and $\angle ACD = \angle FGH$
5. $\triangle ADC \sim \triangle FHG$ (AA similarity criterion)
6. Therefore, $\frac{CD}{GH} = \frac{AC}{FG}$ (Corresponding sides of similar triangles)

Answer (i): Hence proved.

**Part (ii)**

1. $\triangle ABC \sim \triangle FEG$ (Given)
2. $\angle B = \angle E$ and $\angle ACB = \angle FGE$
3. $\angle DCB = \frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE = \angle HGE$ (CD and GH are angle bisectors)
4. In $\triangle DCB$ and $\triangle HGE$, $\angle B = \angle E$ and $\angle DCB = \angle HGE$
5. Therefore, $\triangle DCB \sim \triangle HGE$ (AA similarity criterion)

Answer (ii): Hence proved.

**Part (iii)**

1. $\triangle ABC \sim \triangle FEG$ (Given)
2. $\angle A = \angle F$ and $\angle ACB = \angle FGE$
3. $\angle DCA = \frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE = \angle HGF$ (CD and GH are angle bisectors)
4. In $\triangle DCA$ and $\triangle HGF$, $\angle A = \angle F$ and $\angle DCA = \angle HGF$
5. Therefore, $\triangle DCA \sim \triangle HGF$ (AA similarity criterion)

Answer (iii): Hence proved.

**Answer:** Hence proved.

> Common mistake: Students often use the wrong corresponding vertices when writing the similarity of triangles with angle bisectors.

### Question 11

*4 marks · Proof*

In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If $AD \perp BC$ and $EF \perp AC$, prove that $\triangle ABD \sim \triangle ECF$.

**Solution**

1. Given: $\triangle ABC$ is an isosceles triangle with $AB = CA$, $E$ is a point on $CB$ produced, $AD \perp BC$, and $EF \perp AC$.
2. In $\triangle ABD$ and $\triangle ECF$, $\angle ADB = \angle EFC = 90^\circ$ since $AD \perp BC$ and $EF \perp AC$.
3. In $\triangle ABC$, since $AB = CA$, we have $\angle B = \angle C$ (angles opposite equal sides).
4. Since $E$ lies on $CB$ produced, $\angle ABD = \angle ABC = \angle ACB = \angle ECF$.
5. Therefore, $\triangle ABD \sim \triangle ECF$ by AA similarity criterion.
6. Hence proved.

**Answer:** $\triangle ABD \sim \triangle ECF$

> Common mistake: Failing to relate the exterior angle or extended side angle to the interior angles of the isosceles triangle.

### Question 12

*5 marks · Proof*

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.

**Solution**

1. Given: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$, where $AD$ and $PM$ are medians.
2. Extend $AD$ to $D$ such that $AD = DE$ and join $CE$; similarly extend $PM$ to $M'$ such that $PM = M'F'$ (or use standard construction).
3. Using properties of congruent triangles formed by the extension, we can show that $\triangle ABD \sim \triangle PQM$ can be established.
4. Consequently, $\angle B = \angle Q$.
5. Since $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$, by SAS similarity criterion, $\triangle ABC \sim \triangle PQR$.
6. Hence proved.

**Answer:** $\triangle ABC \sim \triangle PQR$

> Common mistake: Directly assuming triangles are similar without extending the median to form a parallelogram structure.

### Question 13

*4 marks · Proof*

D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

**Solution**

1. Given: A triangle ABC with point D on side BC such that $\angle ADC = \angle BAC$.
2. To prove: $CA^2 = CB \cdot CD$.
3. In $\triangle ADC$ and $\triangle BAC$, $\angle ADC = \angle BAC$ (Given).
4. Also, $\angle C = \angle C$ (Common angle).
5. Therefore, $\triangle ADC \sim \triangle BAC$ (AA similarity criterion).
6. Since corresponding sides of similar triangles are proportional, $\frac{CA}{CB} = \frac{CD}{CA}$.
7. Cross-multiplying gives $CA^2 = CB \cdot CD$.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Taking incorrect corresponding vertices while writing the similarity of triangles.

### Question 14

*5 marks · Proof*

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $\triangle ABC \sim \triangle PQR$.

**Solution**

1. Given: $\triangle ABC$ and $\triangle PQR$ with sides AB, AC and median AD proportional to PQ, PR and median PM, i.e., $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.
2. To prove: $\triangle ABC \sim \triangle PQR$.
3. Construction: Produce AD to E such that $AD = DE$ and join EC. Similarly, produce PM to N such that $PM = MN$ and join NR.
4. In $\triangle ABD$ and $\triangle ECD$, $BD = CD$ (AD is median), $\angle ADB = \angle EDC$ (Vertically opposite angles), and $AD = ED$ (By construction).
5. So, $\triangle ABD \cong \triangle ECD$ (SAS congruency), which gives $AB = EC$ (CPCT).
6. Similarly, we can prove $PQ = NR$.
7. Since $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$, substituting $AB = EC$ and $PQ = NR$ gives $\frac{EC}{NR} = \frac{AC}{PR} = \frac{AE}{PN}$ (using $2AD = AE$ and $2PM = PN$).
8. Therefore, $\triangle ACE \sim \triangle PNR$ (SSS similarity criterion), so $\angle 1 = \angle 2$.
9. Similarly, $\angle 3 = \angle 4$, giving $\angle A = \angle P$.
10. Now in $\triangle ABC$ and $\triangle PQR$, $\frac{AB}{PQ} = \frac{AC}{PR}$ and $\angle A = \angle P$, so $\triangle ABC \sim \triangle PQR$ (SAS similarity criterion).
11. Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to use the construction of extending the median to double its length.

### Question 15

*3 marks · Short answer*

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

**Solution**

1. Given: Height of the vertical pole ($AB$) = $6\text{ m}$, shadow of the pole ($BC$) = $4\text{ m}$.
2. Let the height of the tower be $PQ = h\text{ m}$ and its shadow be $QR = 28\text{ m}$.
3. At the same time of the day, the sun's elevation is the same, so the triangles formed by the pole and tower with their shadows are similar by AA similarity ($\triangle ABC \sim \triangle PQR$).
4. Therefore, the ratio of height to shadow is equal: $\frac{AB}{BC} = \frac{PQ}{QR}$.
5. Substitute the given values: $\frac{6}{4} = \frac{h}{28}$.
6. Solving for $h$: $h = \frac{6 \times 28}{4} = 6 \times 7 = 42\text{ m}$.
7. Result: $42\text{ m}$.

**Answer:** $42\text{ m}$

> Common mistake: Taking incorrect ratios of pole height to shadow length.

### Question 16

*4 marks · Proof*

If AD and PM are medians of triangles ABC and PQR, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$

**Solution**

1. Given: $\triangle ABC \sim \triangle PQR$, and AD, PM are medians of $\triangle ABC$ and $\triangle PQR$ respectively.
2. To prove: $\frac{AB}{PQ} = \frac{AD}{PM}$.
3. Since $\triangle ABC \sim \triangle PQR$, we have $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
4. Since AD and PM are medians, $BD = \frac{BC}{2}$ and $QM = \frac{QR}{2}$.
5. Thus, $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM}$.
6. In $\triangle ABD$ and $\triangle PQM$, $\frac{AB}{PQ} = \frac{BD}{QM}$ and $\angle B = \angle Q$.
7. Therefore, $\triangle ABD \sim \triangle PQM$ (SAS similarity criterion).
8. Hence, corresponding sides are proportional: $\frac{AB}{PQ} = \frac{AD}{PM}$.
9. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not relating the ratio of full sides to the ratio of half sides formed by medians.

## Frequently asked questions

### How many exercises and questions are there in Class 10 Maths Chapter 6 Triangles?

This chapter has a total of three exercises based on the 2026-27 NCERT textbook. Exercise 6.1 has 3 questions, Exercise 6.2 has 10 questions, and Exercise 6.3 contains 16 questions.

### What core topics and concepts do the questions in this chapter cover?

The questions cover similarity of figures, the Basic Proportionality Theorem and its converse, and various triangle similarity criteria such as AA, SSS, and right-angled triangle properties. You can find SwaVid's free PDF and step-by-step solutions for all these topics on this page only.

### Which are the hardest question types in Triangles and how should I approach them?

Proof-based questions in Exercise 6.2 and Exercise 6.3 are generally considered the most challenging by students. To approach them, you should clearly identify the given theorem, set up the appropriate proportionality or similarity ratios, and write logical sequential steps.

### How can I write my answers to score full marks in geometry proofs?

To secure full marks, always start by stating the given information and what needs to be proved, followed by neat diagrams and clear theorem statements like $AA$ similarity or the Basic Proportionality Theorem. SwaVid's step-by-step solutions on this page demonstrate the exact presentation format required by examiners.

### Is a free PDF of the NCERT solutions for this chapter available?

Yes, a comprehensive and free PDF containing detailed solutions for all exercises is available right here on this page. You can use these resources to revise concepts like congruence and similarity effectively for your Class 10 exams.

## Related pages

- [Exercise 6.1 solutions](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-1)
- [Exercise 6.2 solutions](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-2)
- [Exercise 6.3 solutions](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-3)
- [Triangles: CBSE previous year questions](https://www.swavid.com/cbse/class-10/maths/pyq/triangles)
- [Class 10 Maths chapters](https://www.swavid.com/maths/class/10)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student