Problem

Source: SRMC 2002

Tags: geometry, circumcircle, angle bisector, geometry proposed



Let $ \triangle ABC$ be a triangle with incircle $ \omega(I,r)$and circumcircle $ \zeta(O,R)$.Let $ l_{a}$ be the angle bisector of $ \angle BAC$.Denote $ P=l_{a}\cap\zeta$.Let $ D$ be the point of tangency $ \omega$ with $ [BC]$.Denote $ Q=PD\cap\zeta$.Show that $ PI=QI$ if $ PD=r$.