**NCERT Solutions Maths Class 10
Exercise 6.2**

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

**Solution: (i)** Since DE ∥
BC,

Therefore, AD/DB = AE/EC

⇒ 1.5/3 =
1/EC

⇒ EC =
3/1.5

⇒ EC = 2 cm

**(ii) **Since DE ∥ BC,

Therefore, AD/DB = AE/EC

⇒
AD/7.2 = 1.8/5.4

⇒ AD = (1.8 × 7.2)/5.4

⇒ EC = 2.4 cm

**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**

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

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

**Solution: (i) **The
figure for the above question is as follows:** **

Given: PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Now,
PE/EQ = 3.9/3 = 1.3 cm

And
PF/FR = 3.6/2.4 = 1.5 cm

Since, PE/EQ
≠ PF/FR

Therefore,
EF is not parallel to QR.

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

Now,
PE/EQ = 4/4.5 = 8/9 cm

And
PF/FR = 8/9 cm

Since, PE/EQ
= PF/FR

Therefore, EF
is parallel to QR.

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

⇒ EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And
FR = PR – PF = 2.56 – 0.36 = 2.20 cm

Now,
PE/EQ = 0.18/1.10 = 18/110 = 9/55 cm

And
PF/FR = 0.36/2.20 = 36/220 = 9/55 cm

Since, PE/EQ
= PF/FR

Therefore, EF
is parallel to QR.

**3. In figure, if LM **∥
**CB and LN **∥
**CD, prove that AM/AB = AN/AD****.**

**Solution: **In ∆ABC,
LM ∥ CB

Therefore, AM/AB = AL/AC
[From the Basic Proportionality Theorem] ………. (i)

And
in ∆ACD, LN ∥
CD

Therefore, AL/AC = AN/AD
[From the Basic
Proportionality Theorem] ………. (ii)

From
equations (i) and (ii), we have

AM/AB = AN/AD Hence proved.

**4. In figure, DE **∥
**AC and DF **∥
**AE. Prove that BF/FE = BE/EC****.**

**Solution: **In ∆BCA,
DE ∥ AC

Therefore, BE/EC = BD/DA
[From the Basic
Proportionality Theorem] ………. (i)

And
in ∆BEA, DF ∥
AE

Therefore, BF/FE = BD/DA
[From the Basic
Proportionality Theorem] ………. (ii)

From
equations (i) and (ii), we have

BF/FE
= BE/EC
Hence proved.

**5. In figure, DE **∥
**OQ and DF **∥
**OR. Show that EF **∥
**QR.**

**Solution: **In ∆PQO,
DE ∥ OQ

Therefore, PE/EQ = PD/DO
[From the Basic
Proportionality Theorem] ………. (i)

And
in ∆POR, DF ∥
OR

Therefore, PD/DO = PF/FR
[From the Basic
Proportionality Theorem] ………. (ii)

From
equations (i) and (ii), we have

PE/EQ
= PF/FR

Since in ∆PQR, PE/EQ = PF/FR

Therefore, EF
∥ QR
[By the converse of Basic Proportionality Theorem]

**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: Given**: O is any point in ∆PQR, in which AB ∥
PQ and AC ∥
PR.

**To prove**: BC ∥ QR

**Construction**: Join BC.

**Proof**: In ∆OPQ,
AB ∥ PQ

Therefore, OA/AP = OB/BQ
[From the Basic
Proportionality Theorem] ………. (i)

And
in ∆OPR, AC ∥
PR

Therefore, OA/AP = OC/CR
[From the Basic
Proportionality Theorem] ………. (ii)

From
equations (i) and (ii), we have

OB/BQ
= OC/CR

Since, in ∆OQR, B and C are points dividing the sides OQ
and OR in the same ratio.

Therefore, AB
∥ PQ
[By the converse of Basic Proportionality Theorem]

**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. (Recall that you have proved it in Class IX).**

**Solution: Given**: A ∆ABC,
in which D is the mid-point of side AB and the line DE is drawn parallel to BC,
meeting AC at E.

**To prove****:** AE = EC

**Proo****f:** Since DE ∥
BC

Therefore, AD/DB = AE/EC
[From the Basic
Proportionality Theorem] ………. (i)

But
AD = DB [Given]

⇒ AD/DB = 1

⇒ AE/EC = 1 [From equation (i)]

⇒ AE = EC

Hence,
E bisects the third side AC.

**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. (Recall that you have done it in Class IX).**

**Solution:
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

Now,
AD = DB

⇒ AD/DB = 1

And
AE = EC

⇒ AE/EC = 1

Therefore, AD/DB = AE/EC = 1

⇒ AD/DB = AE/EC

Thus,
in ∆ABC, D and E are points dividing the
sides AB and AC in the same ratio.

Therefore,
DE ∥ BC
[By the converse of Basic Proportionality Theorem]

**9. ABCD is a trapezium in which AB **∥ **DC and its diagonals intersect each other at
the point O. Show that AO/BO = CO/DO.**

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

**To Prove**: AO/BO
= CO/DO

**Construction****: **Through O, draw OE ∥ AB, and hence OE ∥ DC.

**Proof**: In ∆ADC,
we have OE ∥ DC

Therefore, AE/ED = AO/CO
[From the
Basic Proportionality Theorem] ………. (i)

Again,
in ∆ABD, we have OE ∥ AB [By
construction]

Therefore,
ED/AE = DO/BO [From the Basic Proportionality Theorem]
………. (ii)

⇒ AE/ED = BO/DO ………. (ii)

From
equations (i) and (ii), we get

AO/CO
= BO/DO

⇒ AO/BO = CO/DO

**10. The diagonals of a quadrilateral ABCD
intersect each other at the point O such that AO/BO = CO/DO. Show that
ABCD is a trapezium.**

**Solution:
Given**: A quadrilateral
ABCD, in which its diagonals AC and BD intersect each other at O such that AO/BO = CO/DO or AO/CO = BO/DO

**To Prove**: Quadrilateral ABCD is a trapezium.

**Construction****: **Through O, draw OE ∥ AB meeting AD at E.

**Proof**: In ∆ADB,
we have OE ∥ AB [By construction]

Therefore, DE/EA = DO/BO [From the Basic Proportionality Theorem]

⇒ EA/DE = BO/DO

⇒ EA/DE = BO/DO =
AO/CO [Since AO/CO = BO/DO]

⇒ EA/DE = AO/CO

Thus
in ∆ADC, E and O are points dividing
the sides AD and AC in the same ratio. Therefore, EO ∥ DC
[By the converse of Basic Proportionality Theorem]

But
EO ∥ AB
[By construction]

Therefore, AB
∥ DC

Therefore, the quadrilateral ABCD is a trapezium.