4. Right triangles $\triangle M D C$ and $\triangle A D K$ have a common right angle $\angle D$. Point $K$ lies on $C D$ and divides it in the ratio $2: 3$ from point $C$. Point $M$ is the midpoint of side $A D$. Find the sum of $\angle A K D$ and $\angle M C D$, if $A D: C D=2: 5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0d43a086e0070a45fd08g-08.jpg?height=220&width=574&top_left_y=978&top_left_x=798)
# Solution

Complete $\triangle A D C$ to a square $L J C D$. 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 $C B: B J=2: 3$. Then $\triangle A H N$ is a right isosceles triangle with $\angle A=45^{0}$, $\angle L A H=\angle A K D$, $\angle N A B=\angle M C D$. $\angle L A B=\angle L A H+\angle H A N+\angle N A B=90^{\circ}$.

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

![](https://cdn.mathpix.com/cropped/2024_05_06_0d43a086e0070a45fd08g-08.jpg?height=514&width=605&top_left_y=1622&top_left_x=774)

Answer: 45.