Question 154
Points $M$ and $N$ are considered in the interior of triangle $ABC$ such that $\widehat{MAB} = \widehat{NAC}$ and $\widehat{MBA} = \widehat{NBC}$. Prove that
$$
\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = 1.
$$
$$
\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{BA \cdot BC} + \frac{CM \cdot CN}{CA \cdot CB} = 1.
$$