# Answer the whole 18th question."DON'T SEND STHE SIMILAR QUERIES.KINDLY PERFORM THE STEP BY STEP AND APPROPRIATE ANSWER".: 18. In a parallelogram ABCD, the bisector of $\angle$A meets DC in E and AB= 2AD. Prove that  (i) BE bisects $\angle$ B  (ii) $\angle$ AEB = a right angle.

Also AE is the bisector ∠A

∴∠1 = ∠2

Now, ∠2 = ∠5 (alternate angles)

∴∠1 = ∠5

Now AD = DE = x [∵ Sides opposite to equal angles are also equal]

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

∴ CD = 2x

⇒ DE + EC = 2x

⇒ x + PC = 2x

⇒ PC = x

Also, BC = x

In ΔBEC,

∠6 = ∠4 (Angles opposite to equal sides are equal)

Also, ∠6 = ∠3 (alternate angles)

∵ ∠6 = ∠4 and ∠6 = ∠3

⇒∠3 = ∠4

Hence, BE bisects ∠B.

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