In trapezoid $A B C D$ angles $A$ and $D$ are right, $A B=1, C D=4, A D=5$. On side $A D$ a point $M$ is taken such that $\angle C M D=$ $2 \angle B M A$.

In what ratio does point $M$ divide side $A D$?
Let $\angle B M A=\alpha$. Then $\angle C M D=2 \alpha, A M=A B \operatorname{ctg} \alpha=\operatorname{ctg} \alpha, M D=C D \operatorname{ctg} 2 \alpha=4 \operatorname{ctg} 2 \alpha=2\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{ctg} \alpha$. Since $A M+M D=A D$, then $\operatorname{ctg} \alpha+2\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{ctg} \alpha=5$, or $2 \operatorname{tg}^{2} \alpha+5 \operatorname{tg} \alpha-3=0$, from which $\operatorname{tg} \alpha=1 / 2$. Therefore, $A M=2, M D=3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_04ea0aec75eb28164af1g-19.jpg?height=438&width=414&top_left_y=390&top_left_x=822)

## Answer

$2: 3$.