A point $P$ in the interior of a cyclic quadrilateral $ABCD$ satisfies $\angle BPC = \angle BAP + \angle PDC$. Denote by $E, F$ and $G$ the feet of the perpendiculars from $P$ to the lines $AB, AD$ and $DC$, respectively. Show that the triangles $FEG$ and $PBC$ are similar.

See here. Best regards, sunken rock