Problem (2019KoMaLA742): Convex quadrilateral $ABCD$ is inscribed in circle $\Omega$. Its sides $AD$ and $BC$ intersect at point $E$. Let $M$ and $N$ be the midpoints of the circle arcs $AB$ and $CD$ not containing the other vertices, and let $I$, $J$, $K$, $L$ denote the incenters of triangles $ABD$, $ABC$, $BCD$, $CDA$, respectively. Suppose $\Omega$ intersects circles $IJM$ and $KLN$ for the second time at points $U \neq M$ and $V \neq N$. Show that the points $E$, $U$, and $V$ are collinear.
AlphaGeometry formalization: a b c = triangle; d = on_circum a b c; o = circumcenter a b c; e = on_line a d, on_line b c; m = on_bline a b, on_circle o a; n = on_bline c d, on_circle o a; i = incenter a b d; j = incenter a b c; k = incenter b c d; l = incenter c d a; o1 = circumcenter i j m; o2 = circumcenter k l n; u = on_circle o a, on_circle o1 i; v = on_circle o a, on_circle o2 k ? coll e u v
HAGeo formalization: A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; D = on_circle (O); AD = line A D; BC = line B C; E = intersection AD BC; M = midarc_no A B C; N = midarc_no C D A; I = incenter A B D; J = incenter A B C; K = incenter B C D; L = incenter C D A; O1 = circumcenter I J M; (O1) = circle_center_point O1 I; U = intersection (O1) (O); O2 = circumcenter K L N; (O2) = circle_center_point O2 K; V = intersection (O2) (O); Prove: collinear E U V
Difficulty score: 3.9