Question 4
Let $ABC$ be an acute triangle. Let $D$ be the foot of the perpendicular from $A$ to $BC$, $M$ the midpoint of $BC$, and $H$ the orthocenter. Let $E$ be the intersection of the circumcircle $\Gamma$ of $ABC$ with ray $MH$, and $F\ne E$ the other intersection of line $ED$ with $\Gamma$. Prove
$$\frac{BF}{CF}=\frac{AB}{AC},$$
where $XY$ denotes the length of segment $XY$.
Newly drawn configuration:
$$\frac{BF}{CF}=\frac{AB}{AC},$$
where $XY$ denotes the length of segment $XY$.
Newly drawn configuration: