MHB Inscribed circle in the triangle

[Mathick]

In the triangle a point $$I$$ is a centre of inscribed circle. A line $$AI$$ meets a segment $$BC$$ in a point $$D$$. A bisector of $$AD$$ meets lines $$BI$$ and $$CI$$ respectively in a points $$P$$ and $$Q$$. Prove that heights of triangle $$PQD$$ meet in the point $$I$$.

I've tried to show that sides of triangle $$PQD$$ are parallel to sides of triangle $$ABC$$ but it didn't work out. That's why I ask you for help.

 

[Greg]

Can you provide a diagram?

 

[johng1]

I am convinced the OP has posted a true statement. I can't prove it, but hope someone here can. Here's a diagram.

Give a triangle ABC, let I be the incenter of the triangle (the intersection of the angle bisectors at A, B and C). Let D be the intersection of AI with BC and line L the perpendicular bisector of AD ($H_D$ is the midpoint of AD). Let P be the intersection of CI with L and Q the intersection of BI with L. Then the orthocenter of PDQ is I.

Clearly, if $H_Q$ is the intersection of BI with DP, it is sufficient to show angle $BH_QD$ is a right angle.