Bariz - Problem

let $ K$ be the intesection of $ BB_2$ and $ A_1C_1$ we prove that $ k$ lies on $ (IB_2)$. since $ AKB$ and $ BKC$ has the same area and $ area(AKB)=KC_1.AB.sin(<BC_1K)$ and $ area(AKB)=KA_1.CB.sin(<BA_1K)$ and : $ sin(<BA_1K)=sin(<BC_1K)$, we have :$ KC_1.AB=KA_1.CB$. now consider the line parallel to $ IA_1$ through $ C_1$, intersect $ IK$ on $ I_1$, so $ \frac{C_1I_1}{IA_1}=\frac{C_1K}{KA_1}=\frac{BC}{AB}$ and also $ <I_1C_1I=<ABC$ so $ ABC$ and $ C_1II_1$ are similar so $ <C_1IK=<C_1II_1=<BAC =180-<C_1IB_1$.