Problem 4. Given a square $ABCD$. On the extension of the diagonal $AC$ beyond point $C$, a point $K$ is marked such that $BK=AC$. Find the angle $BKC$. $[6$ points] (
Answer: $30^{\circ}$.

Solution: Since the picture is symmetric with respect to the line $A C$, we have $D K=B K$. By the condition, $B K=A C$.

And since the diagonals in a square are equal, $A C=B D$. Thus, in triangle $B K D$ all sides are equal, i.e., it is equilateral, and $\angle B K D=60^{\circ}$. Again, due to symmetry with respect to the line $A C$, we have $\angle B K C=\angle D K C$, and together these angles sum to $60^{\circ}$, i.e., each of them is $30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1bdd9b1df09d7b896302g-08.jpg?height=396&width=347&top_left_y=129&top_left_x=898)

Comment: The equalities $D K=B K$ and $\angle B K C=\angle D K C$ can also be proven without using the concept of symmetry.

Indeed, in triangles $B C K$ and $D C K$, side $C K$ is common, $B C=C D$ as sides of the square, $\angle B C K=\angle C K D$, since each is adjacent to a $45^{\circ}$ angle. Therefore, triangles $B C K$ and $D C K$ are congruent by two sides and the included angle. Consequently, $D K=B K$ and $\angle B K C=\angle D K C$ as corresponding elements of congruent triangles.