The inscribed circle of triangle $ABC$, with centre $I$, touches sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. Let $P$ be a point, on the same side of $FE$ as $A$, for which $\angle PFE = \angle BCA$ and $\angle PEF = \angle ABC$. Prove that $P$, $I$ and $D$ lie on a straight line.
Problem
Source: SAMO 2016 Q3
Tags: geometry
mathprince2000
24.09.2016 14:53
It's an easy problem! It is obvious that $\Delta PEF\sim\Delta ABC$, then $PFIE$ is a cyclic quadrilateral. Also, $AFIE$ is a cyclic quadrilateral, so $A$, $E$, $I$, $F$, $P$ are concyclic. Thus, $AP\perp IP$. In other hand, $\widehat{PAE}=\widehat{PFE}=\widehat{BCA}$, then $AP//BC$. Hence, $IP\perp BC$, but $ID\perp BC$, so $P$, $I$ and $D$ lie on a straight line.
Attachments:
fighter
24.09.2016 15:03
same as me
finn123
25.09.2016 06:17
This problem can be solve by using spiral similarity too!
fighter
25.09.2016 09:14
here is another solution from me. <PEF = <ABC and <PFE = <ACB so, <EPF = <BAC = <EAF thus A,P,E,F cyclic. but, we have A,E,I,F cyclic then P,E,I,F cyclic. <PIF = <PEF = <ABC = <DBF = 180 - <DIF so, P,I,D is collinear