A point
is chosen on the side
of the triangle
in such a way that the radii of the circles inscribed in the triangles
and
are equal. Prove that
where X is the area of triangle
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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 $ABC.$
Školjka