Let $\triangle ABC$ has median $AD$.$P$ is a point on segment $AD$ such that $BD=PD$.$E$ is the foot of the perpendicular from $P$ to $BC$.The lines passing through $E$ perpendicular to $PB$ and $PC$ meet $AB$ and $AC$ at $F$ and $G$, respectively.Prove that the circumcircles of $\triangle AFG,\triangle BEF $ and $\triangle CEG $ meet at a single point on segment $PE$.
(It is a shortlisted problem and I've made some modifications of it.)
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