## Task B-2.4.

Let $A B C D$ be a parallelogram, and $M$ the midpoint of side $\overline{D C}$. If point $M$ lies on the angle bisector of $\angle D A B$, determine the measure of $\angle A M B$.
## Solution.

Let $\varangle B A D=\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_30_8e0ddf1d7bf4bfb115dcg-07.jpg?height=283&width=740&top_left_y=1349&top_left_x=584)

Since $D C \| A B$ and

$$
\varangle B A M=\varangle A M D=\varangle D A M=\frac{\alpha}{2},
$$

triangle $D A M$ is isosceles:

$$
|D M|=|D A| \quad \text{and} \quad \varangle D M A=\frac{\alpha}{2} .
$$

Since triangle $M C B$ is also isosceles (because $|M C|=|M D|=|D A|=|B C|$), it follows that

$$
\varangle B C M=\varangle B C D=\varangle D A B=\alpha
$$

and

$$
\varangle C M B=\varangle C B M=\frac{180^{\circ}-\varangle B C M}{2}=\frac{180^{\circ}-\alpha}{2}=90^{\circ}-\frac{\alpha}{2} .
$$

Now it is easy to calculate that $\varangle A M B=180^{\circ}-\varangle A M D-\varangle C M B=90^{\circ}$.