CGMO 2010 Problem 6

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 $BC^2 = 4 PF \cdot AK$ .

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<size; ++i) { d[i] = 0; mult[i] = 1;}
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]));
}