Auswahlwettbewerb Zur IMO

Solution:

Let $B'$, $C'$, $E'$ be the projections of $B$, $C$ and $E$ onto the line $A D$. The right triangles $\triangle D B B'$ and $\triangle D C C'$ are congruent, since $D$ is the midpoint of $B C$ and the acute angles at $D$ are equal in size. From this it follows that $B B' = C C'$.
In triangle $\triangle A C' C$, $E E'$ is parallel to $C C'$ and consequently
$$\frac{A C}{A E} = \frac{C C'}{E E'} = \frac{B B'}{E E'}.$$
The right triangles $\triangle B B' F$ and $\triangle E E' F$ are similar, since the angles at $F$ are equal. From this it follows that $\frac{B B'}{E E'} = \frac{B F}{F E}$, which together with the second-to-last relation leads to $\frac{A C}{A E} = \frac{B F}{F E}$.

If we now replace $\frac{B F}{F E}$ in the initial relation by $\frac{A C}{A E}$, we obtain:
$\frac{B C}{A B} + 1 = \frac{A C}{A E} = \frac{B C + A B}{A B}$, which leads to $\frac{A C - A E}{A E} = \frac{B C}{A B}$ and finally to $\frac{E C}{A E} = \frac{B C}{A B}$.
By the converse of the angle bisector theorem, $B E = w_{\beta}$.