Question

In a parallelogram ABCD, the bisector of Angle A meets DC in E and AB = 2AD. Prove
that
(i) BE bisects ZB
(ii) angle AEB = a right angle



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Answer 1

$hello$

Question ⤵️

In a parallelogram ABCD the bisector of angle A meets DC in E and AB=2AD. Prove that BD bisects Angle B and angle AEB is a right angle

 

Answer ⤵️

(i)  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.

I hope my answer helps you.....✌

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Answer 2

Question ⤵️

In a parallelogram ABCD the bisector of angle A meets DC in E and AB=2AD. Prove that BD bisects Angle B and angle AEB is a right angle

Answer ⤵️

(i) 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.