Problem (2016CGMOp7): In acute triangle $ABC, AB<AC$, $I$ is its incenter, $D$ is the foot of perpendicular from $I$ to $BC$, altitude $AH$ meets $BI,CI$ at $P,Q$ respectively. Let $O$ be the circumcenter of $\triangle IPQ$, extend $AO$ to meet $BC$ at $L$. Circumcircle of $\triangle AIL$ meets $BC$ again at $N$. Prove that $\frac{BD}{CD}=\frac{BN}{CN}$.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; d = foot i b c; h = foot a b c; p = on_line a h, on_line b i; q = on_line a h, on_line c i; o = circumcenter i p q; l = on_line a o, on_line b c; n = on_line b c, on_circum a i l ? eqratio b d c d b n c n
HAGeo formalization: A B C = acute_triangle; I = incenter A B C; BC = line B C; D = foot I BC; H = foot A BC; AH = line A H; BI = line B I; P = intersection AH BI; CI = line C I; Q = intersection AH CI; O = circumcenter I P Q; AO = line A O; L = intersection AO BC; (AIL) = circle A I L; N = intersection (AIL) BC; Prove: equal_ratio B D C D B N C N
Difficulty score: 3.5