8,9 |

On the sides $C B$ and $C D$ of the square $A B C D$, points $M$ and $K$ are taken such that the perimeter of triangle $C M K$ is equal to twice the side of the square.

Find the measure of angle $M A K$.

#
Let's rotate triangle $ABM$ around point $A$ by $90^{\circ}$ so that vertex $B$ moves to $D$. Let $M'$ be the image of point $M$ under this rotation. Since by the condition

$MK + MC + CK = (BM + MC) + (CK + KD)$, then $MK = BM + KD = KM'$. Moreover, $AM = AM'$, therefore triangles $AMK$ and $AM'K$ are congruent, which means,

$\angle MAK = \angle M'AK = 1/2 \angle MAM' = 45^{\circ}$.