Problem 2. Given a square $A B C D$. On the extension of the diagonal $A C$ is chosen a point $E$ such that $\overline{D E}=\overline{A C}$. Determine the size of the angle $D E C$.
Solution. Let point $O$ be the intersection of the diagonals. Then we have that $\overline{B O}=\overline{D O}$. From the condition of the problem, we have that $\overline{D E}=\overline{A C}$. From the fact that $A B C D$ is a square, we have that $\overline{B D}=\overline{A C}$. Therefore,

$$
\overline{B D}=\overline{D E}
$$

On the other hand, we have that segment $B E$ is the image of segment $D E$ under axial symmetry with respect to the line $A C$. Thus, we obtain that

$$
\overline{B E}=\overline{D E}
$$

From (1) and (2), we conclude that triangle BED is equilateral. Therefore, $\angle B E D=60^{\circ}$. From the congruence of triangles $B O E$ and $D O E$, we have that $\angle D E O=\angle B E O=30^{\circ}$. Therefore, $\angle D E C=\angle D E O=30^{\circ}$.