Here is my question:
Let $\triangle ABC$ and let $K,L$ be midpoints of $AC$ and $AB$ respectively. Let $D$ be some point on $AC$ (between $K$ and $C$) such that $KD=AL$.
Show that the perpendicular line from $D$ to the angle bisector of $A$ halves $BC$.
First of all, I've drawn the following drawing:
I tried to connect $KE$ and $KL$ and to prove that $KE$ is parallel to $AB$, but to no avail.
Please give a hint, I find this question very hard.