8. Right triangles $M D C$ and $A D K$ have a common right angle $D$. Point $K$ lies on $C D$ and divides it in the ratio $2: 3$, counting from point $C$. Point $M$ is the midpoint of side $A D$. Find the sum of the degree measures of angles $A K D$ and $M C D$, if $A D: C D=2: 5$.
Solution. Extend triangle $A D C$ to form a square $L J C D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b77a4ddf94595f81d054g-09.jpg?height=677&width=694&top_left_y=535&top_left_x=721)

Choose point $H$ on side $L J$ such that $L H: H J=2: 3$, point $N$ on side $C J$ such that $C N: N J=3: 2$, and point $B$ on side $C J$ such that $C B: B J=2: 3$. We obtain equal triangles: $\triangle A K D=\triangle A L H=\triangle H J N$.

Next, $\quad A H=H N, \angle A H N=90^{\circ}$, therefore, $\quad \angle H A N=45^{\circ} . \quad$ From the equality of triangles $M D C$ and $N A B$, we get $\angle N A B=\angle M C D$.

Next, $\angle L A B=\angle L A H+\angle H A N+\angle N A B=90^{\circ}$.

From this, finally, $\angle L A H+\angle N A B=\angle A K D+\angle M C D=45^{\circ}$.

Answer: 45.