Let $B^{\prime}, C^{\prime}, E^{\prime}$ be the projections of $B, C$, and $E$ onto the line $A D$. The right triangles $\triangle D B B^{\prime}$ and $\triangle D C C^{\prime}$ are congruent, since $D$ is the midpoint of $B C$ and the acute angles at $D$ are equal. This implies that $B B^{\prime}=C C^{\prime}$.
In the triangle $\triangle A C^{\prime} C$, $E E^{\prime}$ is parallel to $C C^{\prime}$, and thus

$$\frac{A C}{A E}=\frac{C C^{\prime}}{E E^{\prime}}=\frac{B B^{\prime}}{E E^{\prime}}.$$

The right triangles $\triangle B B^{\prime} F$ and $\triangle E E^{\prime} F$ are similar, as the angles at $F$ are equal. This implies $\frac{B B^{\prime}}{E E^{\prime}}=\frac{B F}{F E}$, which, together with the previous relationship, leads to $\frac{A C}{A E}=\frac{B F}{E F}$.
Replacing $\frac{B F}{E F}$ in the initial relationship with $\frac{A C}{A E}$, we get:
$\frac{B C}{A B}+1=\frac{A C}{A E}=\frac{B C+A B}{A B}$, which simplifies to $\frac{A C-A E}{A E}=\frac{B C}{A B}$ and finally to $\frac{E C}{A E}=\frac{B C}{A B}$.
According to the converse of the Angle Bisector Theorem, $B E=w_{\beta}$.

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