4. In triangle $ABC$, it holds that $\frac{h_{a} h_{b}}{h_{c}}=c$, where $h_{a}, h_{b}, h_{c}$ are the lengths of the altitudes, and $c$ is the length of one side of the triangle. Determine one of the angles of this triangle.
Solution. From the given equality, it follows that $h_{a} h_{b}=c h_{c}$, which means $\frac{h_{a} h_{b}}{2}=\frac{c h_{c}}{2}$. However, $\frac{c h_{c}}{2}=P$, where $P$ is the area of the triangle. Therefore, $\frac{h_{a} h_{b}}{2}=P$ and since $\frac{a h_{a}}{2}=P$ and $\frac{b h_{b}}{2}=P$, we get $\frac{h_{a} h_{b}}{2}=\frac{a h_{a}}{2}$ and $\frac{h_{a} h_{b}}{2}=\frac{b h_{b}}{2}$, so $a=h_{b}$ and $b=h_{a}$. This means that sides $a$ and $b$ are heights to each other, so the triangle is a right triangle with the right angle between sides $a$ and $b$.