9. On the side $AC$ of an acute-angled triangle $ABC$, a point $M$ is taken. From point $M$, perpendiculars $MN$ and $MP$ are dropped to the sides $AB$ and $BC$. Where should point $M$ be located so that the length of the segment $NP$ is minimal?
Answer: Point $M$ should be the foot of the perpendicular dropped from point $B$.

Solution: Construct a circle with diameter $B M$. Obviously, points $N$ and $P$ will lie on it.

![](https://cdn.mathpix.com/cropped/2024_05_06_3fd104dfae7d68abe92bg-3.jpg?height=768&width=933&top_left_y=1815&top_left_x=573)

The chord $N P$ subtends a fixed angle $\angle A B C$, therefore, its length is minimized when the diameter of the circle is minimized. The diameter, equal to $M B$, is minimized when $B M$ is the height.