IMO Hk TST

Suppose $DG, EF, AH$ are concurrent at $X$. Let $Q$ be the point on the line $CP$ such that $QH \parallel PG$. We only work on the configuration as shown since the other cases are similar.

Firstly, by the parallel lines $AH \parallel EG$ and $QH \parallel PG$, we have $\triangle CAH \sim \triangle CEG$ and $\triangle CPG \sim \triangle CQH$. Thus, we have
$$\frac{AE}{EC} = \frac{HG}{GC} = \frac{QP}{PC}.$$
By the angle bisector theorem, it suffices to prove $\frac{AP}{PC} = \frac{AE}{EC}$. Equivalently, we need to prove $PA = PQ$.

By similar triangles, we have $\frac{DF}{EG} = \frac{DX}{GX} = \frac{FH}{GH}$. Together with the included angles $\angle DFH = \angle EGH = 90^\circ$, we have $\triangle DFH \sim \triangle EGH$. This implies $\angle DHF = \angle EHG$, and hence $\angle AHD = \angle AHE$.

Note that $E, P, H, G$ are concyclic since $\angle EPH = \angle EGH = 90^\circ$. This yields
$$\angle PEH = \angle PGH = \angle QHB,$$
and so
$$\angle EHP = 90^\circ - \angle PEH = 90^\circ - \angle QHB = \angle AHQ.$$
It follows that $\angle DHQ = \angle EHA = \angle DHA$.

Next, note that $\angle PEG = \angle DHF = \angle GHE = \angle GPE$. This implies $PG = EG$. By similar triangles, we obtain
$$\frac{PG}{QH} = \frac{CG}{CH} = \frac{EG}{AH} = \frac{PG}{AH}.$$
Therefore, we have $QH = AH$. Thus, $HP$ is the perpendicular bisector of $AQ$, and hence $PA = PQ$ as desired.