Problem 2. Given a right triangle $ABC$, with a right angle at vertex $C$ and the angle at vertex $B$ equal to $20^{\circ}$. The angle bisector of $\measuredangle BAC$ intersects the leg $BC$ at point $D$, and the angle bisector of $\measuredangle ABC$ intersects the leg $AC$ at point $F$. From points $D$ and $F$, perpendiculars are drawn to the hypotenuse and intersect the hypotenuse at points $M$ and $N$. Calculate $\measuredangle MCN$.
Solution. Triangle $F C N$ is isosceles because from the property of the perpendicular bisector $B F$ we have $\overline{C F}=\overline{F N}$. Triangle $C D M$ is isosceles, be-
cause from the property of the perpendicular bisector $A D$ we have $\overline{C D}=\overline{D M}$. Therefore, we have

![](https://cdn.mathpix.com/cropped/2024_06_05_e235c78af50361de0ec1g-10.jpg?height=404&width=971&top_left_y=375&top_left_x=435)

$\measuredangle F C N=\measuredangle C N F$ and $\measuredangle D C M=\measuredangle D M C$.

We have, $\measuredangle A=90^{\circ}-20^{\circ}=70^{\circ}$, so $\measuredangle A F N=20^{\circ}$. Similarly, we find $\measuredangle B D M=70^{\circ}$. Therefore, $\measuredangle F C N=\measuredangle C N F=\frac{\measuredangle A F N}{2}=\frac{20^{\circ}}{2}=10^{\circ}$. Similarly, we find $\measuredangle D C M=\measuredangle D M C=\frac{70^{\circ}}{2}=35^{\circ}$. Finally, for the desired angle we find

$$
\measuredangle M C N=90^{\circ}-\left(10^{\circ}+35^{\circ}\right)=90^{\circ}-45^{\circ}=45^{\circ}
$$