## Task A-1.4.

On the side $\overline{B C}$ of the square $A B C D$, an equilateral triangle $B E C$ is drawn such that the point $E$ is outside the square. Points $M$ and $N$ are the midpoints of segments $\overline{A E}$ and $\overline{C D}$, respectively.

Determine the measure of the angle $\varangle M N C$.
## Solution.

Let $P$ be the midpoint of the segment $\overline{B E}$.

Then $\overline{P M}$ is the midline of triangle $A B E$, so $|P M|=\frac{1}{2}|A B|$ and $P M \| A B$.

On the other hand, $|C N|=\frac{1}{2}|C D|=\frac{1}{2}|A B|$, which implies $|P M|=|C N|$.

Furthermore, $P M\|A B\| C N$, so we can conclude that $P C N M$ is a parallelogram.

![](https://cdn.mathpix.com/cropped/2024_05_30_14eaa2cf500c23a22c24g-03.jpg?height=508&width=871&top_left_y=2073&top_left_x=521)

Since triangle $B E C$ is equilateral, the median $\overline{C P}$ is also the altitude of this triangle, so $\varangle B C P=30^{\circ}$. It follows that $\varangle P C D=\varangle P C B+\varangle B C N=30^{\circ}+90^{\circ}=120^{\circ}$.

Since $P C N M$ is a parallelogram, it follows that

$$
\varangle M N C=180^{\circ}-\varangle M N D=180^{\circ}-\varangle P C D=60^{\circ}
$$