In parallelogram $A B C D$, the bisector of angle $B A D$ intersects side $C D$ at point $M$, and $\frac{D M}{M C}=2$. It is known that angle $C A M$ is equal to $\alpha$. Find angle $BAD$.

#
Apply the Law of Sines to triangle $A C D$.

## Solution

Let $\angle B A D=\varphi, M C=x, D M=2 x$. Since $\angle A M D=\angle B A M=\angle M A D$, triangle $A M D$ is isosceles, $A D=D M=2 x$. In triangle $A D C$, it is known that

$$
\begin{gathered}
C D=3 x, A D=2 x, \angle C A D=\angle C A M+\angle M A D=\alpha+\frac{\varphi}{2} \\
\angle A C D=\angle A M D-\angle C A M=\frac{\varphi}{2}-\alpha
\end{gathered}
$$

By the Law of Sines

$$
\begin{gathered}
\frac{A D}{\sin \angle A C D}=\frac{C D}{\sin \angle C A D} \Rightarrow \frac{2 x}{\sin \left(\frac{\varphi}{2}-\alpha\right)}=\frac{3 x}{\sin \left(\frac{\varphi}{2}+\alpha\right)} \Rightarrow \\
\Rightarrow 2 \sin \left(\frac{\varphi}{2}+\alpha\right)=3 \sin \left(\frac{\varphi}{2}-\alpha\right) \Rightarrow \\
\Rightarrow 2 \sin \frac{\varphi}{2} \cos \alpha+2 \cos \frac{\varphi}{2} \sin \alpha=3 \sin \frac{\varphi}{2} \cos \alpha-3 \cos \frac{\varphi}{2} \sin \alpha
\end{gathered}
$$

Therefore,

$$
\sin \frac{\varphi}{2} \cos \alpha=5 \cos \frac{\varphi}{2} \sin \alpha
$$

Thus, $\operatorname{tg} \frac{\varphi}{2}=5 \operatorname{tg} \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-31.jpg?height=395&width=531&top_left_y=1246&top_left_x=772)

Answer

$2 \operatorname{arctg}(5 \operatorname{tg} \alpha)$.