Let $ABCD$ be a convex quadrilateral in which $AB = CD$ and $∠ABD + ∠ACD = 180^{\circ}$. Lines $AC$ and $BD$ intersect at $P$ and let $M$ be the midpoint of $AD$. Suppose that $MB$ and $MC$ intersect the circumcircle of $\triangle{BPC}$ again at $X$ and $Y$, respectively. Prove that lines $CX, BY , AD$ are concurrent.
When I first encountered the problem, I thought about some Ceva's Theorem. I drew the diagram, but couldn't make much sense of it. However, I couldn't find any more information about the lengths. Maybe something else could work. I haven't figured exactly what the first two conditions really can do, so can someone help?
Construct a point $B'$ such that $ABDB'$ is a parallelogram. Then, $$ \angle ACD+\angle AB'D=\angle ACD+\angle ABD=180^{\circ}, $$ so the quadrilateral $ACDB'$ is cyclic. We will prove that $X$ lies on the circumcircle of $ACDB'$. Indeed, note that (lines $BD$ and $AB'$ are parallel) $$ \angle B'XC=180^{\circ}-\angle BXC=180^{\circ}-\angle BPC=180^{\circ}-\angle APD=\angle B'AP=\angle B'AC, $$ so $X$ lies on the circumcircle of the quadrilateral $ACDB'$.
Similarly we can define the point $C'$ such that $ACDC'$ is parallelogram and analogously we can prove that points $A$, $B$, $D$, $C'$ and $Y$ are lying on the same circle. Thus, we have three circles: the circumcircle of $AXCDB'$, the circumcircle $ABDYC'$ and the circumcircle of $BXCPY$. Finally, note that corresponding radical axes are $AD$, $BY$ and $CX$. Hence, lines $AD$, $BX$ and $CY$ are concurrent, as desired.