Nemotron-IMO-Bench 2026 G30

For a point $P$ inside the triangle $ABC$ the 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.