Problem (2018KoMaLA726): In triangle $ABC$ with incenter $I$, line $AI$ intersects the circumcircle of $ABC$ at $S\ne A$. Let the reflection of $I$ with respect to $BC$ be $J$, and suppose that line $SJ$ intersects the circumcircle of $ABC$ for the second time at point $P\ne S$. Show that $AI=PI.$
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; o = circumcenter a b c; s = on_line a i, on_circle o a; j = reflect i b c; p = on_line s j, on_circle o a ? cong a i p i
HAGeo formalization: A B C = triangle; I = incenter A B C; AI = line A I; O = circumcenter A B C; (O) = circle_center_point O A; S = intersection AI (O); BC = line B C; J = reflect_point_wrt_line I BC; JS = line J S; P = intersection JS (O); Prove: cong A I I P
Difficulty score: 2.9