## Chapter 6 Triangles Exercise 6.1

**Question 1: In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).**

**Answer:**

For this question we will use Basic Proportionality theorem.

i) According to Basic Proportionality Theorem,

AD/DB = AE/EC

= 1.5/3 = 1/EC

= 1.5(EC) = 1(3)

= 1.5 EC = 3

= EC = 3/1.5

= EC = 2

Therefore, EC = 2 cm

ii) According to Basic Proportionality Theorem,

AD/DB = AE/EC

= AD/7.2 = 1.8/5.4

= AD(5.4) = 1.8(7.2)

= 5.4 AD = 12.96

= AD = 12.96/5.4

= AD = 2.4

Therefore, 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
(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
Answer:**

For this question we will use converse of Basic Proportionality Theorem,

i)

= 3.9/3 = 3.6/2

Since 3.9/3 ≠ 3.6/2.4. Therefore by converse of Basic Proportionality Theorem, EF is not parallel to QR.

ii)To prove that EF || QR in case (ii), we have to show that PE/EQ = PF/FR

= 4/4.5 = 8/9

= 4/4.5 = 4/4.5

Since 4/4.5 = 8/9. Therefore by converse of Basic Proportionality Theorem, EF is parallel to QR.

iii)

= PE/PQ = PF/PR (by corollary of Basic Proportionality Theorem)

= 0.18/1.28 = 0.36/2.56

= 18/128 = 36/256 (multiplying with 100)

Simplifying both sides,

= 9/64 = 18/128

= 9/64 = 9/64

Since 0.18/1.28 = 0.36/2.56. Therefore by converse of Basic Proportionality Theorem, EF is parallel to QR.

**Question 3: In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.**

**Answer:**

Considering ∆ABC,

Since ML || BC,

∴ By Basic Proportionality
Theorem,

AM/AB = AL/AC …(1)

Considering ∆ADC,

Since LN || CD

∴ By Basic Proportionality
Theorem,

AN/AD = AL/AC …(2)

From (1) and (2)

AM/AB = AN/AD

Hence proved.

**Question 4: In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.**

**Answer:**

Considering ∆ABE
with DF || AE.

Since DF || AE

∴ By Basic Proportionality
Theorem,

BD/DA = BF/FE …(1)

Considering ∆ABC
with DE || AC.

Since DE || AC

∴ By Basic Proportionality
Theorem,

BD/DA = BE/EC …(2)

From (1) and (2)

BF/FE = BE/EC

Hence proved.

**Question 5: In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.**

**Answer:**

Considering ΔPQO,

Since ED || QO,

Therefore, by Basic Proportionality Theorem,

PE/QE = PD/DO ...(1)

Considering ΔPOR,

Since DF || OR,

Therefore, by Basic Proportionality Theorem,

PD/DO = PF/FR ...(2)

From (1) and (2),

PE/QE = PF/FR

Since, PE/QE = PF/FR, therefore, by converse of Basic Proportionality Theorem,
EF || QR.

Hence proved.

**Question 6: In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively
such that AB || PQ and AC || PR. Show that BC || QR.**

**Answer:**

Considering ΔOPQ,

Since AB || PQ,

Therefore, by Basic Proportionality Theorem,

OA/AP = OB/BQ ...(1)

Considering ΔOPR,

Since AC || PR,

Therefore, by Basic Proportionality Theorem,

OA/AP = OC/CR ...(2)

From (1) and (2)

OB/BQ = OC/CR

Since OB/BQ = OC/CR, therefore by converse of Basic Proportionality theorem, BC
|| QR.

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

DE || BC

D is midpoint of AB

To prove:

E intersects at AC

Since DE || BC,

Therefore, by Basic Proportionality Theorem,

AD/DB = AE/EC

⇒ 1 = AE/EC (since D is midpoint of AB)

⇒ EC = AE

Since EC = AE, therefore E intersects AC and is the midpoint of AC.

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

Answer:Given:

Answer:

AD/DB = AE/EC

D is midpoint of AB and E is midpoint of AC

To prove:

DE || BC

Since it is given AD/DB = AE/EC

⇒ 1 = 1 (as D and E are midpoints of AB and AC respectively)

Now,

Since AD/DB = AE/EC, therefore, by converse of Basic Proportionality Theorem, DE || BC

Hence proved.

**Question 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.**

Answer:

Answer:

Trapezium ABCD where AB || CD and AB & CD intersect each other at O.

To prove:

AO/BO = CO/DO

Construction:

Draw a line EO through O such that EO || DC || AB

Proof:

In ΔADC,

OE || DC (by construction)

Therefore, by Basic Proportionality Theorem,

AE/ED = AO/CO ...(1)

In ΔABD,

OE || AB (by construction)

Therefore, by Basic Proportionality Theorem,

DE/EA = DO/BO or EA/DE = BO/DO ...(2)

From (1) and (2)

AO/CO = BO/DO

= (AO)(DO) = (BO)(CO)

= AO/BO = CO/DO

Hence proved.

**Question 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.**

Answer:Given:

Answer:

Quadrilateral ABCD where AC & BD intersect each other at O, such that AO/BO = CO/DO.

To prove:

ABCD is a trapezium

Construction:

Draw a line EO through O such that EO || AB, which meets AD at E.

Proof:

In Δ DAB,

EO || AB (by construction)

Therefore, by Basic Proportionality Theorem,

DE/EA = DO/OB ...(1)

Also it is given,

AO/BO = CO/DO

= (AO)(DO) = (CO)(BO)

= AO/CO = BO/DO or CO/AO = DO/BO ...(2)

From (1) and (2)

DE/EA = CO/AO

Therefore by converse of Basic Proportionality Theorem, EO is parallel to DC also EO is parallel to AB

⇒ DC || AB

Since two pair of sides (AB and CD) are parallel to each other, therefore, ABCD is a trapezium.

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