APMO Geometry

Let $AD$ be the distance from $A$ to $BC$, $H$ is the orthocenter of $\Delta ABC$
,$M$ is the middle point of $BC$,MH∩ABC=E ED∩○ABC=F

show that $AB:AC=BF:CF$

Here is a picture:

Let I be the intersection of EM and the circle，It is not hard to prove $HBIC$ is a parallelogram. By the law of sine, we have：

\[\angle{BAD}=\alpha\,\,\angle{DAC}=\beta%20\\\frac{BD}{\sin{\alpha}}=AB\,\,\,\,\frac{CD}{\sin{\beta}}=AC\]

\[\frac{\sin\beta}{HM}=\frac{\sin\angle{BHM}}{BM}\,\,\,\,\,\,\frac{\sin\alpha}{HM}=\frac{\sin\angle{HMC}}{MC}\iff\frac{\sin\beta}{\sin\alpha}=\frac{\sin{BHM}}{\sin{MHC}}\]

\[\frac{BF}{CF}=\frac{BD}{CD}\cdot\frac{\sin\beta}{\sin\alpha}=\frac{AB}{AC}\]