On the sides of angle $A B C$, points $M$ and $K$ are marked such that angles $B M C$ and $B K A$ are equal, $B M=B K, A B=15, B K=8, C M$ $=9$.

Find the perimeter of triangle $С O K$, where $O$ is the intersection point of lines $А K$ and $С M$.

#
Triangles $A B K$ and $C B M$ are equal (by side and adjacent angles, see figure). Therefore, $\angle B C M = \angle B A K$ and $C B = A B = 15$, so $C K = A M = 7$.

Considering also that $\angle C K O = \angle A M O$ (they supplement equal angles to straight angles), we get that triangles СОК and АОМ are equal. Therefore, $O K = O M$. Thus, $\mathrm{P}_{\text {СОК }} = \mathrm{CK} + \mathrm{CO} + \mathrm{OK} = \mathrm{CK} + C O + O M = C K + C M = 16$

![](https://cdn.mathpix.com/cropped/2024_05_06_34fd2032e75750f56a92g-09.jpg?height=452&width=397&top_left_y=1232&top_left_x=842)

Answer

16.