In the acute triangle $ABC$, the midpoint of side $BC$ is called $M$. Point $X$ lies on the angle bisector of $\angle AMB$ such that $\angle BXM = 90^o$. Point $Y$ lies on the angle bisector of $\angle AMC$ such that $\angle CYM = 90^o$. Line segments $AM$ and $XY$ intersect in point $Z$. Prove that $Z$ is the midpoint of $XY$ . [asy][asy] import geometry; unitsize (1.2 cm); pair A, B, C, M, X, Y, Z; A = (0,0); B = (4,1.5); C = (0.5,3); M = (B + C)/2; X = extension(M, incenter(A,B,M), B, B + rotate(90)*(incenter(A,B,M) - M)); Y = extension(M, incenter(A,C,M), C, C + rotate(90)*(incenter(A,C,M) - M)); Z = extension(A,M,X,Y); draw(A--B--C--cycle); draw(A--M); draw(M--interp(M,X,2)); draw(M--interp(M,Y,2)); draw(B--X, dotted); draw(C--Y, dotted); draw(X--Y); dot("$A$", A, SW); dot("$B$", B, E); dot("$C$", C, N); dot("$M$", M, NE); dot("$X$", X, NW); dot("$Y$", Y, NE); dot("$Z$", Z, S); [/asy][/asy]