EASIEST OF ALL ...just use the fact that opposite angles of a cyclic quadrilateral add upto $180$ and that angle subtended at center is twice that at circumference

see attachment for my solution

Attachments:

easy angle chasing with cyclic quad. and fact that angle subtended by chord at centre is double the angle subtended by it at circumference proves the question.

correct your question it must be $AB=AE$

my solution = let centre of $Y$ be $O$. than we have $2\angle BEA=\angle BOA=180-\angle BCA=180-\angle EDB=180-\angle EAB=\angle BEA+\angle EBA$ which gives $\angle EBA=\angle BEA\leftrightarrow AB=AE$ so we are done. PS= that was the easiest geometry i have seen in RMO except RMO 2014 P1.

aditya21 wrote: my solution = let centre of $Y$ be $O$. than we have $2\angle BEA=\angle BOA=180-\angle BCA=180-\angle EDB=180-\angle EAB=\angle BEA+\angle EBA$ which gives $\angle EBA=\angle BEA\leftrightarrow AB=AE$ so we are done. PS= that was the easiest geometry i have seen in RMO except RMO 2014 P1. There is a mistake in your calculation.. $180^{\circ}-\angle{BCA}=360^{\circ}+\angle {EDB}$.. Rest what follows is correct..