TR-MO 2020 Problem 2

Let $P$ be an interior point of acute triangle $\Delta ABC$ , which is different from the orthocenter. Let $D$ and $E$ be the feet of altitudes from $A$ to $BP$ and $CP$ , and let $F$ and $G$ be the feet of the altitudes from $P$ to sides $AB$ and $AC$ . Denote by $X$ the midpoint of $[AP]$ , and let the second intersection of the circumcircles of triangles $\Delta DFX$ and $\Delta EGX$ lie on $BC$ . Prove that $AP$ is perpendicular to $BC$ or $\angle PBA = \angle PCA$ .