In a quadrilateral $ABCD$ the intersection of the diagonals is called $P$. Point $X$ is the orthocentre of triangle $PAB$. (The orthocentre of a triangle is the point where the three altitudes of the triangle intersect.) Point $Y$ is the orthocentre of triangle $PCD$. Suppose that $X$ lies inside triangle $PAB$ and $Y$ lies inside triangle $PCD$. Moreover, suppose that $P$ is the midpoint of line segment $XY$ . Prove that $ABCD$ is a parallelogram. [asy][asy] import geometry; unitsize (1.5 cm); pair A, B, C, D, P, X, Y; A = (0,0); B = (2,-0.5); C = (3.5,2.2); D = A + C - B; P = (A + C)/2; X = orthocentercenter(A,B,P); Y = orthocentercenter(C,D,P); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(A--extension(A,X,B,P), dotted); draw(B--extension(B,X,A,P), dotted); draw(P--extension(P,X,A,B), dotted); draw(C--extension(C,Y,D,P), dotted); draw(D--extension(D,Y,C,P), dotted); draw(P--extension(P,Y,C,D), dotted); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N); dot("$P$", P, E); dot("$X$", X, NW); dot("$Y$", Y, SE); [/asy][/asy]