Question 3
(ROM) A right-angled triangle $ABC$ is given for which the hypotenuse $BC$ has length $a$ and is divided into $n$ equal segments, where $n$ is odd. Let $\alpha$ be the angle with which the point $A$ sees the segment containing the middle of the hypotenuse. Prove that
$$\tan\alpha=\frac{4nh}{(n^2-1)a},$$
where $h$ is the height of the triangle.
$$\tan\alpha=\frac{4nh}{(n^2-1)a},$$
where $h$ is the height of the triangle.