# Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.2

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  ⇒EC = EC = 2 cm

Hence, EC = 2 cm. (ii) Here, In △ ABC,

So, according to Theorem 6.1 , if DE || BC  ⇒AD = ### 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, = 1.3 ………………………(i) = 1.5 ………………………(ii)

As, 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, ………………………(i) ………………………(ii)

As, 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, ………………………(i) ………………………(ii)

As, Hence, EF is parallel to QR.

###  Solution:

Here, In △ ABC,

According to Theorem 6.1, if LM || CB

then, …………………………….(I)

According to Theorem 6.1, if LN || CD

then, …………………………….(II)

From (I) and (II), we conclude that Hence Proved !!

###  Solution:

Here, In △ ABC,

According to Theorem 6.1, if DE || AC

then, …………………………….(I)

and, In △ ABE,

According to Theorem 6.1, if DF || AE

then, …………………………….(II)

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

and, In △ POR,

According to Theorem 6.1, if DF || OR

then, …………………………….(II)

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

and, In △ POR,

According to Theorem 6.1, if AC || PR

then, …………………………….(II)

From (I) and (II), we conclude that ………………………………(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.  = 1 …………………………. (I)

Here, In △ ABC,

According to Theorem 6.1, if DE || BC

then, …………………………….(II)

From (I) and (II), we conclude that = 1 = 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

We have to prove that: DE || BC.  = 1 …………………………. (I)

and, AE=EC = 1 …………………………. (II)

From (I) and (II), we conclude that = 1 ……………….(III)

According to Theorem 6.2 and eqn. (III)

DE || BC, in △ ABC

Hence Proved !!

### Solution:

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

EO || DC || AB According to Theorem 6.1, if AB || EO

then, …………………………….(I)

According to Theorem 6.1, if AC || PR

then, …………………………….(II)

From (I) and (II), we conclude that After rearranging, we get Hence Proved !!

### . 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 According to Theorem 6.1, if AB || EO

then, …………………………….(I) (Given) (After rearranging) ………………………………..(II)

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

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