Question 37
In triangle $ABC$, points $D$ and $E$ lie on side $BC$ and $AC$ respectively such that $AD \perp BC$ and $DE \perp AC$. The circumcircle of triangle $ABD$ meets segment $BE$ at point $F$ (other than $B$). Ray $AF$ meets segment $DE$ at point $P$. Prove that
$$
\frac{DP}{PE} = \frac{CD}{DB}.
$$
$$
\frac{DP}{PE} = \frac{CD}{DB}.
$$