Prove that A(FKP)=A(EDP).

Question

ABCD is circumcircled in W. It is trapezoid. Mid points of parallel sides, BC and AD are E and F respectively. AB and DC intersect at P, W(circle) and BF intersect at K. Then prove A(FKP)=A(EDP) Where A(FKP) is angle FKP. My try: It is clear that ABE and DEC are equal triangles. Thus A(BAE)=A(EDP). Now we need to prove A(BAE)=A(FKP). If we can prove W(circle), AE and KP intersect at the same point then we will have desired result.

Answer 1

Let $PK\cap W=\{H,K\}$ and $AH\cap BC=\{E_1\}$.

Now, for $ADCBKH$ we can use the Pascal's theorem:

$$AD\cap BK=\{F\},$$ $$DC\cap HK=\{P\}$$ and $$BC\cap AH=\{E_1\}.$$ Thus, $F$, $P$ and $E_1$ are placed on the same line, which says $$E\equiv E_1,$$ $$\measuredangle FKP=\measuredangle BAH=\measuredangle EDP$$ and we are done!