$D$ is any point on side $BC$ of $\triangle{ABC}$. Points $E, I, F$ are respectively the incentres of $\triangle{ADB},\triangle{ABC},\triangle{ADC}$. $P$ is the foot of perpendicular from $I$ on $BC$. Prove that $\angle{EPF}$ is right angle.
I have two approaches to solve this question. first, since $\angle{EDF}$ is right angle, if we prove that quadrilateral $EFDP$ is cyclic then we are done. secondly, if $\angle{EPI}=\angle{FPC}$ then the question is solved. I tried both the approaches but I could not get the solution.
Fact#1) By considering tangent lengths, we have $a’ = s – a$ where s = semi-perimeter of $\triangle ABC$.
Fact#2) By applying Fact-1 and by considering tangent lengths again, we have $I’J’ = DK’$ with $I’D$ as the common portion.
Let O be the midpoint of EF. Then, O is the center of the circle EDF with OE = OD = OF.
Through O, draw $OM \bot E’F’$ cutting $E’F’$ at M. By intercept theorem, M is the midpoint of $E’F’$. Then, by Fact#2, M is also the midpoint of the common portion DP. These make OD = OP. Therefore, P is also a point on the circle EDF.