In triangle with incircle prove that $\overline{CQ}$ is parallel to $\overline{AB}$

Question

We are given a triangle $ABC$ whose incircle touches side $AB$ at point $D$ and side $AC$ at point $E$.
Point $P$ lies on segment $AC$ such that segment $IP$ is parallel to the segment $DE$, where $I$ is center of incircle. Lines $DE$ and $BP$ intersect at point $Q$.

Prove that $CQ$ is parallel to $AB$.

I tried angle chasing, then some ideas with new special points, lines etc, but it doesn't give any solution. I would like any advice, any help. No solution.

Answer 1

Let us introduce a couple of additional points: $F$, as the intersection between $BC$ and the incircle; $R$, as the antipode of $D$ in the incircle. $PRI$ and $PEI$ are congruent and it is not difficult to check that $Q,R,F$ are collinear. Can you show $AB\parallel PR\color{red}{\parallel} QC$ by invoking Pascal's (or Brianchon's) theorem?