Prove that $AF$ is perpendicular to $BE$ 
Let $ABC$ be a triangle with $AB=AC.$ If $D$ is the mid-point of $BC$, $E$ is the foot of perpendicular drawn from $D$ to $AC$ and $F$ is the midpoint of $DE, $then prove that $AF$ is perpendicular to $BE$.

This problem came in IITJEE exam. I know how to prove it algebraically: by taking $B$ as $(-a,0)$ and $C$ as $(a,0)$, $D$ as $(0,0)$, $A$ as $(0,h)$ etc. I want to know how to prove it geometrically. Can someone please help me? 
 A: Let $H$ be the foot of perpendicular drawn from $B$ to $AC$: by the intercept theorem we have $CE=EH$ so that $E$ is the midpoint of $CH$. The sides of triangle $BCH$ and $ADE$ are pairwise perpendicular by construction, it follows that their medians $BE$ and $AF$, relative to perpendicular sides $CH$ and $DE$, are perpendicular too.
A: The following lemma is needed for the required proof:-
Given that $⊿XYZ$ ~ $⊿X’Y’Z’$. If $YT$ and $Y’T’$ are the medians of the respective triangles, then $⊿XYT$ ~ $⊿X’Y’T’$ and $⊿YTZ$ ~ $⊿Y’T’Z’$.
This can be verified by applying “ratio of two sides and the included angle”.


After drawing the altitude $BH$ in $⊿ABC$, we have, by intercept theorem, $CE = EH$. This further means $BE$ is the median of $⊿BCH$. Note that from given, $AF$ is the median of $⊿ADE$.
By “A.A.A.”, $⊿BCH$ ~ $⊿ADE$.
By the above lemma, we have $⊿ADF$ ~ $⊿BCE$. This further means $x = y$.
Therefore, by converse of angles in the same segment, ABDT is cyclic (where T is the point of intersection of $AF$ and $BE$. Result follows.
