4. In triangle $ABC$, $AB = 3 \cdot BC$, point $M$ is the midpoint of side $AB$, and $BD$ is the angle bisector. Find the angle $MDB$.

---

Here is the translation of the given text, maintaining the original formatting and structure.
4. Answer. $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_73882b07206f9b28bd67g-2.jpg?height=234&width=532&top_left_y=310&top_left_x=206)

Let $X$ be the midpoint of segment $M B$, and draw segment $D X$. By the property of the angle bisector, we have

$$
A D: D C = A B: B C = 3: 1
$$

On the other hand, $A X: X B = 3: 1$, so $A D: D C = A X: X B$. Therefore, $X D \parallel B C$. Then $\angle X B D = \angle D B C = \angle X D B$, which means that $X D = X B = X M$. From this, we get that $\angle M D B = 90^{\circ}$.