Problem:
The isosceles triangle △ABC, with AB=AC, is inscribed in the circle ω. Let P be a variable point on the arc \overparen{B C} that does not contain A, and let IB​ and IC​ denote the incenters of triangles △ABP and △ACP, respectively.
Prove that as P varies, the circumcircle of triangle △PIB​IC​ passes through a fixed point.
Solution:
Let M be the midpoint of arc BC not containing A. We claim M is the desired fixed point.
Since ∠MPA=90∘ and ray PA bisects ∠IB​PIC​, it suffices to show that MIB​=MIC​. Let MB​,MC​ be the second intersections of PIB​ and PIC​ with circumcircle. Now MB​IB​=MB​B=MC​C=MC​IC​, and moreover MMB​=MMC​, and ∠IB​MB​M=21​PB=∠IC​MC​M, so triangles △IB​MB​M≅△IC​MC​M, done.
