Class 10 NCERT Solutions Chapter 6 Triangles Exercise 6.2 (original) (raw)

Last Updated : 23 Jul, 2025

NCERT Solutions for Class 10 Maths Chapter 6 - Triangles (Exercise 6.2)

In this article, we provide detailed solutions to the questions in Exercise 6.2 of Chapter 6: Triangles from the NCERT Class 10 Mathematics textbook. The chapter focuses on various theorems related to triangles, including the Basic Proportionality Theorem (Theorem 6.1) and its converse (Theorem 6.2). These theorems help in understanding the properties of similar triangles and parallel lines.

**Theorem 6.1 :

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

**Theorem 6.2 :

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Question 1. In Figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

**Solution:

****(i)** Here, In △ ABC,

DE || BC

So, according to Theorem 6.1

\frac{AD}{DB} = \frac{AE}{EC}

⇒\frac{1.5}{3} = \frac{1}{EC}

⇒EC =\frac{3×1}{1.5}

EC = 2 cm

**Hence, EC = 2 cm.

****(ii)** Here, In △ ABC,

So, according to Theorem 6.1 , if DE || BC

\frac{AD}{DB} = \frac{AE}{EC}

⇒\frac{AD}{7.2} = \frac{1.8}{5.4}

⇒AD =\frac{1.8×7.2}{5.4}

AD = 2.4 cm

**Hence, AD = 2.4 cm.

Question 2. E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF || QR :

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

**Solution:

According to the Theorem 6.2,

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

So, lets check the ratios

Here, In △ PQR,

\frac{PE}{EQ} = \frac{3.9}{3} = 1.3 ...........................(i)

\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5 ...........................(ii)

As,\frac{PE}{EQ} ≠ \frac{PF}{FR}

Hence, EF is **not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

**Solution:

According to the Theorem 6.2,

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

So, lets check the ratios

Here, In △ PQR,

\frac{PE}{EQ} = \frac{4}{4.5} = \frac{8}{9} ...........................(i)

\frac{PF}{FR} = \frac{8}{9} ...........................(ii)

As,\frac{PE}{EQ} = \frac{PF}{FR}

Hence, EF is **parallel to QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

**Solution:

EQ = PQ - PE = 1.28 - 0.18 = 1.1

and, FR = PR - PF = 2.56 - 0.36 = 2.2

According to the Theorem 6.2,

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

So, lets check the ratios

Here, In △ PQR,

\frac{PE}{EQ} = \frac{0.18}{1.1} = \frac{9}{55} ...........................(i)

\frac{PF}{FR} = \frac{0.36}{2.2} = \frac{9}{55} ...........................(ii)

As,\frac{PE}{EQ} = \frac{PF}{FR}

Hence, EF is **parallel to QR.

Question 3. In Figure, if LM || CB and LN || CD, prove that

\frac{AM}{AB} = \frac{AN}{AD}

**Solution:

Here, In △ ABC,

According to Theorem 6.1, if LM || CB

then,\frac{AM}{AB} = \frac{AL}{AC} ..................................(I)

and, In △ ADC,

According to Theorem 6.1, if LN || CD

then,\frac{AN}{AD} = \frac{AL}{AC} ..................................(II)

From (I) and (II), we conclude that

\frac{AM}{AB} = \frac{AN}{AD}

Hence Proved !!

Question 4. In Figure, DE || AC and DF || AE. Prove that

\frac{BF}{FE} = \frac{BE}{EC}

**Solution:

Here, In △ ABC,

According to Theorem 6.1, if DE || AC

then,\frac{BD}{DA} = \frac{BE}{EC} ..................................(I)

and, In △ ABE,

According to Theorem 6.1, if DF || AE

then,\frac{BD}{DA} = \frac{BF}{FE} ..................................(II)

From (I) and (II), we conclude that

\frac{BF}{FE} = \frac{BE}{EC}

Hence Proved !!

Question 5. In Figure, DE || OQ and DF || OR. Show that EF || QR.

**Solution:

Here, In △ POQ,

According to Theorem 6.1, if DE || OQ

then,\frac{PE}{EQ} = \frac{PD}{DO} ..................................(I)

and, In △ POR,

According to Theorem 6.1, if DF || OR

then,\frac{PD}{DO} = \frac{PF}{FR} ..................................(II)

From (I) and (II), we conclude that

\frac{PE}{EQ} = \frac{PF}{FR} ....................................(III)

According to Theorem 6.2 and eqn. (III)

**EF || QR, in △ PQR

Hence Proved !!

Question 6. In Figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

**Solution:

Here, In △ POQ,

According to Theorem 6.1, if AB || PQ

then,\frac{OA}{AP} = \frac{OB}{BQ} ..................................(I)

and, In △ POR,

According to Theorem 6.1, if AC || PR

then,\frac{OA}{AP} = \frac{OC}{CR} ..................................(II)

From (I) and (II), we conclude that

\frac{OB}{BQ} = \frac{OC}{CR} ....................................(III)

According to Theorem 6.2 and eqn. (III)

**BC || QR, in △ OQR

Hence Proved !!

Question 7. 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.

**Solution:

Given, in ΔABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E

So, DE || BC.

We have to prove that E is the mid point of AC.

As, AD=DB

⇒\frac{AD}{DB} = 1 …………………………. (I)

Here, In △ ABC,

According to Theorem 6.1, if DE || BC

then,\frac{AD}{DB} = \frac{AE}{EC} ..................................(II)

From (I) and (II), we conclude that

\frac{AD}{DB} = \frac{AE}{EC} = 1

\frac{AE}{EC} = 1

**AE = EC

**E is the midpoint of AC.

Hence proved !!

Question 8. 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.

**Solution:

Given, in ΔABC, D and E are the mid points of AB and AC respectively

AD=BD and AE=EC.

We have to prove that: DE || BC.

As, AD=DB

⇒\frac{AD}{DB} = 1 …………………………. (I)

and, AE=EC

⇒\frac{AE}{EC} = 1 …………………………. (II)

From (I) and (II), we conclude that

\frac{AD}{DB} = \frac{AE}{EC} = 1 ...................(III)

According to Theorem 6.2 and eqn. (III)

**DE || BC, in △ ABC

Hence Proved !!

Question 9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that

\frac{AO}{BO} = \frac{CO}{DO}

**Solution:

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

Here, In △ ADB,

According to Theorem 6.1, if AB || EO

then,\frac{AE}{ED} = \frac{OB}{OD} ..................................(I)

and, In △ ADC,

According to Theorem 6.1, if AC || PR

then,\frac{AE}{ED} = \frac{AO}{OC} ..................................(II)

From (I) and (II), we conclude that

\frac{AO}{OC} = \frac{OB}{OD}

After rearranging, we get

\frac{AO}{OB} = \frac{OC}{OD}

Hence Proved !!

Question 10. 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:

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

Here, In △ ADB,

According to Theorem 6.1, if AB || EO

then,\frac{AE}{ED} = \frac{OB}{OD} ..................................(I)

\frac{AO}{OB} = \frac{OC}{OD} (Given)

\frac{AO}{OC} = \frac{OB}{OD} (After rearranging) ......................................(II)

From (I) and (II), we conclude that

\frac{AE}{ED} = \frac{AO}{OC} ......................................(III)

According to Theorem 6.2 and eqn. (III)

EO || DC and also EO || AB

**⇒ AB || DC.

Hence, **quadrilateral ABCD is a trapezium with AB || CD.

Summary

Chapter 6 of Class 10 Mathematics delves into the properties of triangles, particularly focusing on theorems that deal with parallel lines and proportional segments. The exercises help solidify the understanding of these concepts by applying theorems to solve various geometric problems.