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Let ABCDE be a convex pentagon such that
$\angle BAC= \angle CAD= \angle DAE$ and
$ \angle ABC= \angle ACD= \angle ADE$
If $BD$ and $CE$ intersect at $P$,then prove that $AP$ bisects $CD$.

It is easy to notice that $BC$ is a tangent to the circumcircle of \[\triangle ACD\].So is $ED$.Again note that $CD$ is a tangent the circumcircle of \[\triangle ADE\] and to that of \[\triangle ABC\].Since \[\angle BAC=\angle DAE\] and \[\angle ABC=\angle ADE\],we can say that \[\triangle ABC\sim \triangle ADE\].Again,\[\angle EAC=\angle EAD+\angle DAC=\angle DAC+\angle CAB=\angle DAB\].Both of this together implies that a spiral similarity takes \[\triangle AEC\] to \[\triangle ADB\] and both $A,P$ lies on the circumcircles of \[\triangle ADE,\triangle ABC\].Since $AP$ is the radical axis of these circles,it bisects their common tangent $CD$.

My Proof: (I am really very grateful to Tahmid Vaia for the last part.Go to this link for details:viewtopic.php?f=25&t=2075[Read his sketch of solution])

One can use similar quadrilaterals and ceva's theorem.