2010 China Girls Math Olympiad

In acute triangle $ABC$, $AB > AC$. Let $M$ be the midpoint of side $BC$. The exterior angle bisector of $\widehat{BAC}$ meet ray $BC$ at $P$. Point $K$ and $F$ lie on line $PA$ such that $MF \perp BC$ and $MK \perp PA$. Prove that $BC^2 = 4 PF \cdot AK$.
[asy] defaultpen(fontsize(10)); size(7cm); pair A = (4.6,4), B = (0,0), C = (5,0), M = midpoint(B--C), I = incenter(A,B,C), P = extension(A, A+dir(I--A)*dir(-90), B,C), K = foot(M,A,P), F = extension(M, (M.x, M.x+1), A,P); draw(K--M--F--P--B--A--C); pair point = I; pair[] p={A,B,C,M,P,F,K}; string s = "A,B,C,M,P,F,K"; int size = p.length; real[] d; real[] mult; for(int i = 0; i$"+k+"$",p,mult*dir(point--p)*dir(d)); }[/asy]

In triangle $ABC$, $AB = AC$. Point $D$ is the midpoint of side $BC$. Point $E$ lies outside the triangle $ABC$ such that $CE \perp AB$ and $BE = BD$. Let $M$ be the midpoint of segment $BE$. Point $F$ lies on the minor arc $\widehat{AD}$ of the circumcircle of triangle $ABD$ such that $MF \perp BE$. Prove that $ED \perp FD.$
[asy] defaultpen(fontsize(10)); size(6cm); pair A = (3,10), B = (0,0), C = (6,0), D = (3,0), E = intersectionpoints( Circle(B, 3), C--(C+100*dir(B--A)*dir(90)) )[1], M = midpoint(B--E), F = intersectionpoints(M--(M+50*dir(E--B)*dir(90)), circumcircle(A,B,D))[0]; dot(A^^B^^C^^D^^E^^M^^F); draw(B--C--A--B--E--D--F--M^^circumcircle(A,B,D)); pair point = extension(M,F,A,D); pair[] p={A,B,C,D,E,F,M}; string s = "A,B,C,D,E,F,M"; int size = p.length; real[] d; real[] mult; for(int i = 0; i$"+k+"$",p,mult*dir(point--p)*dir(d)); }[/asy]