**Problem 1 :**

P and Q are points on sides AB and AC respectively, of triangle ABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.

**Solution :**

AP/PB = AQ/AC

3/9 = 5/15

1/3 = 1/3

In triangles APQ, and ABC we get

(AP/AB) = (AQ/AC)

∠ A = ∠A

By using SAS criterion ∆ APQ ~ ∆ ABC

(AP/AB) = (Q/AC) = (PQ/BC)

(AP/PB) = (PQ/BC)

PQ/BC = 3/9

PQ/BC = 1/3

3 PQ = BC

BC = 3 PQ

**Problem 2 :**

In triangle ABC, AB = AC and BC = 6 cm. D is a point on the side AC such that AD = 5 cm and CD = 4 cm. Show that triangle BCD and triangle ACB are congruent and hence find BD.

**Solution :**

In triangle ABC, AB = AC

(BC/AC) = (CD/CB)

(6/9) = (4/6)

From the triangles ∆ BCD, ∆ ACB we get

(BC/AC) = (CD/CB)

∠ C = ∠ C (Common angle)

By using SAS criterion, we get

∆ BCD ~ ∆ ACB

(BD/AB) = (BC/AC)

Here, AB = AC

BD/AC = 6/9

BD/9 = 6/9

BD = 6 cm

**Problem 3 :**

The points D and E are on the sides AB and AC of triangle ABC respectively, such that DE and BC are parallel. If AB = 3 AD and the area triangle ABC is 72 cm², then find the area of the quadrilateral DBCE.

**Solution :**

From the diagram,

the sides DE and BC are parallel and AB = 3 AD

(AD/AB) = 1/3

By taking the two triangles, ∆ ADE and ∆ ABC

∠ ADE = ∠ ABC

∠A = ∠A (Common angle)

By using AA similarity criterion ∆ADE ~ ∆ABC

Area of ∆ ADE/Area of ∆ ABC = AD²/AB²

Area of ∆ ADE/72 = 1/9

Area of ∆ ADE = (1/9) ⋅ 72 = 8 cm

Area of quadrilateral = area of ∆ ABC – Area of ∆ DE

= 72 – 8

= 64 cm^{2}

Therefore the required area of quadrilateral is 64 cm^{2}.

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