# Triangles : Exercise - 6.2 (Mathematics NCERT Class 10th)

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**Q.1Â Â Â Â Â In figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).**

**Sol. **

(i) In fig. (i),

since DE || BC,

2 cm.

(ii) In fig. (ii),

since DE || BC,

2.4 cm.

**Q.2Â Â Â Â Â Â E and F are points on the sides PQ and PB respectively of a PQR. For each of the following cases, state whether EF || QR :**

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) PE = 3.9 cm, EQ = 4 cm, PF = 3.6 cm and FR = 2.4 cm.**

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

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

**Sol. **

(i) We have,

Â Â Â Â Â PE = 3.9 cm, EQ = 4 cm,

Â Â Â Â Â PF = 3.6 cm and FR = 2.4 cm

Â Â Â Â Â Now,Â cm

Â Â Â Â Â and, cm

Â Â Â Â Â

Â Â Â Â Â EF does not divide the sides PQ and PR of PQR in the same ratio.

Â Â Â Â Â Therefore, EF is not parallel to QR.

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

Â Â Â Â Â Â Now,

Â Â Â Â Â Â and,

Â Â Â Â Â Â

Â Â Â Â Â Â Thus, EF divides sides PQ and PR of PQR in the same ratio.

Â Â Â Â Â Â Â Therefore, by the converse of BasicProportionality Theorem we have EF || QR.

(iii) We have, PQ = 1.28 cm, PR = 2.56 cm

Â Â Â Â Â Â Â PF = 0.18 cm and, PF = 0.36 cm

Â Â Â Â Â Â Â Therefore EQ = PQ â€“ PE = (1.28 â€“ 0.18) cm.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.10 cm

Â Â Â Â Â Â Â Â and, ER = PR â€“ PF = (2.56 â€“ 0.36)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.20 cm

Â Â Â Â Â Â Â Â Now,

Â Â Â Â Â Â Â Â and,

Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Thus, EF divides sides PQ and PR of PQR in the same ratio.

Â Â Â Â Â Â Â Â Therefore, by the converse of Basic Proportionality Theorem, we have EF || QR

**Q.3Â Â Â Â Â Â Â In figure, if LM || CB and LN || CD, prove that .**

**Sol.**

In ABC , we have

LM || CB [Given]

Therefore by Proportionality Theorem, we have

In , we have

LN || CD [Given]

Therefore By Basic Proportionality Theorem, we have

.............. (2)

From (1) and (2), we obtain that

**Q.4Â Â Â Â Â In figure, DE || AC and DF || AE. Prove that **

*Sol.*

In BCA, we have

DE || AC [Given]

Therefore By Basic Proportionality Theorem, we have

............ (1)

In BEA, we have

DF || AE [Given]

Therefore by Basic Proportionality Theorem, we have

............. (2)

From (1) and (2), we obtain that

**Q.5 In figure, DE || OQ and DF || OR. Show that EF || QR.**Â **Sol.**

In PQO, we have

DE || OQ [Given]

Therefore By Basic Proportionality Theorem, we have

............ (1)

In POR, we have

DF || OR [Given]

Therefore By Basic Proportionality Theorem, we have

.............. (2)

From (1) and (2), we obtain that

[By the converse of BPT]

**Q.6Â Â Â Â Â 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.**

Sol.

Given : O is any point within , AB || PQ and AC || PR

To prove : BC || QR

Construction : Join BC

Proof : In , we have

AB || PQ [Given]

Therefore By Basic Proportionality Theorem, we have

......... (1)

In OPR, we have

AC || PR [Given]

Therefore by Basic Proportionality Theorem, we have

............. (2)

From (1) and (2), we obtain that

Thus, in OQR, B and C are points dividing the sides OQ and OR in the same ratio. Therefore, by

the converse of Basic Proportionality Theorem, we have,

BC || QR

**Q.7Â Â Â Â Â 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).**

**Sol. **

Given : A ABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E.Â To Prove : AE = EC

Proof : Since DE || BC

Therefore by Basic Proportionality Theorem, we have

............ (1)

But, AD = DB [Because D is the mid-point of AB]

= 1

From (1), = 1 AE = EC

Hence, E is the mid-point of the third side AC.

**Q.8Â Â Â Â Â 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).**

**Sol. **

Given : A ABC, in which D and E are the mid-points of sides AB and AC respectively.

To prove : DE || BC

Proof : Since, D and E are the mid-points of AB and AC respectively.

Therefore AD = DB and AE = EC

Since, AD = DB

Therefore = 1 and AE = EC

Therefore = 1

Therefore

i.e.

Thus, in ABC, D and E are points dividing the sides AB and AC in the same ratio, Therefore, by the converse

of Basic Proportionality Theorem, we have,

DE || BC.

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

**Sol. **

Given : A trapezium ABCD, in which AB || DC and its diagonals AC and BD intersect each other at O.

To prove :

Construction : Through O, draw OE || AB i.e. OE || DC

Proof : In ADC, we have OE || DC [Construction]

Therefore by Basic Proportionality Theorem, we have

................. (1)

Again, in ABD, we have

OE || AB [Construction]

Therefore by Basic Proportionality Theorem, we have

............... (2)

From (1) and (2), we obtain that

**Q.10Â Â Â The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.**

**Sol. **

Given : A quadrilateral ABCD in which its diagonals AC and BD intersect each other at O such that

i.e.

To prove : Quadrilateral ABCD is a trapezium.

Construction : Through O draw OE || AB meeting AD in E.

Proof : In ADB, we have

OE || AB [Construction]

Therefore by Basic Proportionality Theorem, we have

Thus, in ADC, E and O are points dividing the sides AD and AC in the same ratio. Therefore, by

the converse of Basic Proportionality Theorem, we have

EO || DC

But, EO || AB [Construction]

Hence, AB || DC

Therefore Quadrilateral ABCD is a trapezium.