Solution

Observe that $\measuredangle B M N=(180-60-60)^{\circ}=60^{\circ}$. In particular, $M B$ is the bisector of the angle $A M N$. If we draw the segment $\overline{B D}$, we see that this segment is the bisector of the right angle $A D C$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-33.jpg?height=505&width=539&top_left_y=514&top_left_x=838)

Thus, we conclude that point $B$ is the excenter of triangle $M N D$. This implies that $N B$ is necessarily the bisector of the angle $M N C$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-33.jpg?height=507&width=553&top_left_y=1203&top_left_x=823)

Since the sum of the measures of the interior angles of triangle $M D N$ must be $180^{\circ}$, we obtain that

$$
\measuredangle M N C=\measuredangle D M N+\measuredangle M D N=(60+90)^{\circ}
$$

and, therefore, $\measuredangle M N B=\measuredangle M N C / 2=75^{\circ}$. Finally, we use the fact that the sum of the interior angles of triangle $B M N$ is $180^{\circ}$ to conclude that $\measuredangle M B N=(180-$ $60-75)^{\circ}=45^{\circ}$.
The following figure shows a square $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-08.jpg?height=378&width=409&top_left_y=1685&top_left_x=697)

If $\measuredangle A M B=60^{\circ}$ and $\measuredangle D M N=60^{\circ}$, calculate $\measuredangle M B N$.

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