Question

ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA=PD. Prove that AD is parallel to BC

Answer 1

(i) In triangle AED, we have

EA = ED   [given] 

⇒ ∠1 = ∠2  

[angles opposite to equal sides are equal]

Also, ∠3 = ∠2  [external angle of cyclic quad. = interior opposites angle]

 ∠1 = ∠2

and  ∠3 = ∠2

⇒  ∠1 = ∠3 

These are correspondence angles.

∴ AD || BC 

For diagram refer to the attachment !!

Hope it helps!!
Cheers!!

Answer 2

Step-by-step explanation:

answer is in the attachment