ILL 1988 Problem 84

A point $M$ is chosen on the side $AC$ of the triangle $ABC$ in such a way that the radii of the circles inscribed in the triangles $ABM$ and $BMC$ are equal. Prove that $$BM^{2} = X \cot \left( \frac {B}{2}\right)$$

where X is the area of triangle $ABC.$