4. In rectangle $A B C D$, $A B: A D=1: 2$. Point $M$ is the midpoint of $A B$, and point $K$ lies on $A D$ and divides it in the ratio $3: 1$ starting from point $A$. Find the sum of $\angle C A D$ and $\angle A K M$.

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# Solution:

Complete the rectangle $A B C D$ to a square $A E F D$ with side $A D$.

$$
\begin{aligned}
& \text { Let } L \in E F, E L: L F=1: 3, \\
& \triangle M E L=\triangle A K M \Rightarrow \angle E M L=\angle A K M \\
& N \in F D, F N=N C, M R\|A D \Rightarrow M N\| A C \Rightarrow \\
& \Rightarrow \angle N M R=\angle C A D . \\
& \triangle E M L=\Delta L F N, M L=L N, \angle M L N=90^{\circ} \Rightarrow \\
& \Rightarrow \angle L M N=45^{\circ} \\
& \angle C A D+\angle A K M=\angle N M R+\angle E M L= \\
& =90^{\circ}-\angle L M N=90^{\circ}-45^{\circ}=45^{\circ} .
\end{aligned}
$$

Answer: 45.

![](https://cdn.mathpix.com/cropped/2024_05_06_04a9d2f4e9a143a4e5c1g-08.jpg?height=640&width=709&top_left_y=1025&top_left_x=1096)