Problem: Let $I$ be the incenter of $\triangle ABC$ and $P$ be a point such that $PI$ is perpendicular to $BC$ and $PA$ is parallel to $BC$. Let the line parallel to $BC$, which is tangent to the incircle of $\triangle ABC$, intersect $AB$ and $AC$ at points $Q$ and $R$ respectively. Prove that $\angle BPQ = \angle CPR$.
As I couldn't solve it, I looked at the synthetic solution given in AoPS.
The Solution: If $P' \in AP$ such that the perpendicular from $P'$ to $BC$ passes through the $A$-excircle's tangency point on $BC$, then $APQR \sim AP'BC$ by homothety and $BPP'C$ is an isosceles trapezoid, so $\angle{BPC} = \angle{BP'C} = \angle{QPR}$ which implies the desired result. $\square$(This solution is credited to v4913.) link
I understood the part $APQR \sim AP'BC$ by homothety. But I can't undertsand how it implies $BPP'C$ being isosceles trapezoid. if it is true, I understand the rest.
As I have seen in 3 AoPS solutions, One is purely synthetic. Another one is using complex numbers and I don't understand the last solution(it uses like this. $(JH;RB)=(PH;DG)=-1$). For this kind of problems, is it better to know advanced tools like complex numbers? Also, Some more solutions would be beneficial. Any suggestion is appreciated.
Source: BDMO 2024 national HS
