International Mathematical Olympiad 1960 Problem 3

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