Let $ABC$ be an acute triangle. Denote by $D$ the foot of the perpendicular line drawn from the point $A$ to the side $BC$, by $M$ the midpoint of $BC$, and by $H$ the orthocenter of $ABC$. Let $E$ be the point of intersection of the circumcircle $\Gamma$ of the triangle $ ABC$ and the half line $MH$, and $F$ be the point of intersection (other than $E$) of the line $ED$ and the circle $\Gamma$ .

Prove that $\frac{BF}{CF} = \frac{AB}{AC}$ must hold.

[Here we denote by $XY$ the length of the line segment $XY$ ].