Consider $M$.

$ABC$ is an isosceles right triangle. On the extension of the hypotenuse $AB$ beyond point $A$, a point $D$ is taken such that $AB = 2AD$. Points $M$ and $N$ on side $AC$ are such that $AM = NC$. On the extension of side $CB$ beyond point $B$, a point $K$ is taken such that $CN = BK$. Find the angle between the lines $NK$ and $DM$.
Let $L$ be the projection of $M$ onto $AB$. Note that $\frac{ML}{CN} = \frac{AL}{BK} = \frac{AD}{BC} = \frac{1}{\sqrt{2}}$; therefore, $\frac{LD}{CK} = \frac{AL + AD}{BK + BC} = \frac{1}{\sqrt{2}}$. Hence, the right triangles $MLD$ and $NCK$ are similar, and $\angle MLD = \angle NKC$ (see the figure). Therefore, the angle between the lines $NK$ and $MD$ is equal to the angle between the lines $KC$ and $LD$, which is $45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-23.jpg?height=674&width=824&top_left_y=995&top_left_x=617)

## Answer

$45^{\circ}$.