Inscribed circle in the triangle

[Mathick]

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

I've tried to show that sides of triangle \(\displaystyle PQD\) are parallel to sides of triangle \(\displaystyle ABC\) 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.