Question 161
Consider a triangle $ABC$ and a point $P$ in its interior. Lines $PA$, $PB$, $PC$ intersect $BC$, $CA$, $AB$ at $A'$, $B'$, $C'$, respectively. Prove that
$$
\frac{BA'}{BC} + \frac{CB'}{CA} + \frac{AC'}{AB} = \frac{3}{2}
$$
if and only if at least two of the triangles $PAB$, $PBC$, $PCA$ have the same area.
$$
\frac{BA'}{BC} + \frac{CB'}{CA} + \frac{AC'}{AB} = \frac{3}{2}
$$
if and only if at least two of the triangles $PAB$, $PBC$, $PCA$ have the same area.