IMO 1960 Problem 3

In a given right triangle $ABC$ , the hypotenuse $BC$ , of length $a$ , is divided into $n$ equal parts ( $n$ and odd integer). Let $\alpha$ be the acute angel subtending, from $A$ , that segment which contains the mdipoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse fo the triangle. Prove that: $$\tan{\alpha}=\dfrac{4nh}{(n^2-1)a}.$$