---
title: "NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2"
url: https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-2
dateModified: 2026-10-07T15:51:34+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2

Chapter 6: Triangles. Every question from Exercise 6.2, with full working and the final answer.

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## 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.

## Related pages

- [All Chapter 6 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions)
- [Exercise 6.1](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-1)
- [Exercise 6.3](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-3)

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