A synthetic geometrical proof required on a result regarding Isosceles triangles.

I need a synthetic proof on this problem without the use of trigonometry.

Question:
Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ the foot of perpendicular drawn from $D$ to $AC$ and $F$ the midpoint of $DE$, prove that $AF\perp BE$.

The proof is very easy using coordinate geometry.

I'm stuck at: Since point $F$ is midpoint of $DE$, it is fixed and is not a trivial information. But I can't see a way to use this information.

Let $G$ be a midpoint for $CE$. Then $DG$ is a middle line in triangle $BCE$ so $DG||BE$. Also we see that $FG||DC$ so $FG\bot AD$. So $F$ is an orthocenter in triangle $ADG$. So ...
Let $K$ be a midpoint of $BD$.
Thus, since $$\Delta ABD\sim\Delta DCE,$$ we obtain: $$\frac{AB}{DC}=\frac{BD}{CE}$$ or $$\frac{2AB}{BC}=\frac{2BK}{CE}$$ or $$\frac{AB}{BC}=\frac{BK}{CE}$$ and since $\measuredangle ABK=\measuredangle BCE,$ we obtain $$\Delta ABK\sim\Delta BCE,$$ which gives $$\measuredangle BKA=\measuredangle BEC.$$ Now, let $BE\cap AK=\{L\}.$
Thus, $$\measuredangle BLA=\measuredangle LAE+\measuredangle BEA=\measuredangle KAD+180^{\circ}-\measuredangle BEC=$$ $$=\measuredangle KAD+180^{\circ}-\measuredangle BKA=180^{\circ}-90^{\circ}=90^{\circ}.$$