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. 
 

 

(i)  Let AD = x

AB = 2AD = 2x

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.

$$\begin{gathered} \left(\mathrm{ii}\right). \\ \angle \mathrm{A} + \angle \mathrm{B} = 180° \left(\mathrm{adjacent} \mathrm{angles} \mathrm{of} {\parallel }^{\mathrm{gm}} \mathrm{are} \mathrm{equal}\right) \\ \Rightarrow \frac{1}{2}\left(\angle \mathrm{A} + \angle \mathrm{B}\right) = 90° \\ \Rightarrow \angle \mathrm{EAB} + \angle \mathrm{EBA} = 90° \\ \mathrm{In} ∆\mathrm{AEB}, \\ \angle \mathrm{AEB} + \angle \mathrm{EAB} + \angle \mathrm{EBA} = 180° \left(\mathrm{Angle} \mathrm{sum} \mathrm{property}\right) \\ \Rightarrow 90° + \angle \mathrm{AEB} = 180° \\ \Rightarrow \angle \mathrm{AEB} = 90° \end{gathered}$$