In acute triangle $ABC$, $AB>AC$. Let $I$ the incenter, $\Omega$ the circumcircle of triangle $ABC$, and $D$ the foot of perpendicular from $A$ to $BC$. $AI$ intersects $\Omega$ at point $M(\neq A)$, and the line which passes $M$ and perpendicular to $AM$ intersects $AD$ at point $E$. Now let $F$ the foot of perpendicular from $I$ to $AD$. Prove that $ID\cdot AM=IE\cdot AF$
What I thought: Notice that $\triangle IDE$ and $\triangle AFM$ are similar after some angle chasing.
COMMENT:
You correctly found that $\triangle IDE≈\triangle AFM$.We rewrite the required relation as:
$\frac{AF}{AM}=\frac{ID}{IE}$
That is triangles IAE and DIE must be similar. These two triangles have common angle $\angle IED$, therefore we must have:
$\angle IDE=\angle EIA$
This is not possible because:
$\angle IDE=\angle BDE (=90^o)+\angle BDI$
$\angle EIA=\angle EIF (≠90^o)+\angle FIA(=BDI)$