Question

In parallelogram ABCD, bisectors of

<c and <D meet at a point on side AB.

Prove that CD = 2AD ?
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Answer 1

Let AD = xAB = 2AD = 2x

Also AP is the bisector ∠A∴∠1 = ∠2

Now, ∠2 = ∠5 (alternate angles)

∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]

∵ AB = CD (opposite sides of parallelogram are equal)

∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = x

Also, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)

Also, ∠6 = ∠3 (alternate angles)

∵ ∠6 = ∠4 and ∠6 = ∠3⇒∠3 = ∠4

Hence, BP bisects ∠B.

(ii) To prove ∠APB = 90°∵ Opposite angles are supplementary..

Angle sum property,

Answer 2

Answer:

I hope it is help full for you