Bakayev E.V.

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 $BKC$.

#
Since the picture is symmetrical with respect to the line $A C$, then $D K=B K=A C$. And since the diagonals in a square are equal, $A C=B D$. Thus, triangle $B K D$ is equilateral, and angle $B K D$ is $60^{\circ}$. Again, due to symmetry with respect to the line $A C$, $K$ is the bisector of this angle.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-28.jpg?height=409&width=360&top_left_y=1785&top_left_x=858)

Answer

$30^{\circ}$.