Question 95
Let $ABC$ be a non-equilateral triangle and let $M$ and $N$ be interior points of it such that $\measuredangle BAM = \measuredangle CAN$, $\measuredangle ABM = \measuredangle CBN$ and
$$
AM \cdot AN \cdot BC = BM \cdot BN \cdot CA = CM \cdot CN \cdot AB = k
$$
Prove that:
a) $3k = AB \cdot BC \cdot CA$;
b) the midpoint of the segment $MN$ is the centroid of $\triangle ABC$.
$$
AM \cdot AN \cdot BC = BM \cdot BN \cdot CA = CM \cdot CN \cdot AB = k
$$
Prove that:
a) $3k = AB \cdot BC \cdot CA$;
b) the midpoint of the segment $MN$ is the centroid of $\triangle ABC$.