CGMO 2010 Problem 2

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 $ED \perp FD.$

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<size; ++i) { d[i] = 0; mult[i] = 1;}
d[4] = -50;
string[] k= split(s,",");
for(int i = 0;i<p.length;++i) {
label(" $"+k[i]+"$ ",p[i],mult[i]*dir(point--p[i])*dir(d[i]));
}