In a parallelogram with sides $a$ and $b$ and angle $\alpha$, the bisectors of the four angles are drawn. Find the area of the quadrilateral bounded by these bisectors.
The quadrilateral bounded by the given angle bisectors is a rectangle.

## Solution

Let the bisector of angle $A$ of parallelogram $A B C D$ intersect side $B C$ at point $M, \angle B A D=\alpha\left(\alphaa\right)$. Then

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

Therefore, triangle $A B M$ is isosceles and $B M=A B=a$. Hence, $M C=b-a$.

The distance between the drawn bisector and the bisector of angle $B C D$ is

$$
M C \sin \frac{\alpha}{2}=(b-a) \sin \frac{\alpha}{2}
$$

Similarly, the distance between the bisectors of angles $B$ and $D$ is $(b-a) \cos \frac{\alpha}{2}$.

The quadrilateral bounded by the given bisectors is a rectangle with sides equal to

$$
(b-a) \sin \frac{\alpha}{2},(b-a) \cos \frac{\alpha}{2}
$$

$$
(b-a) \sin \frac{\alpha}{2} \cdot(b-a) \cos \frac{\alpha}{2}=\frac{1}{2}(a-b)^{2} \sin \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-23.jpg?height=308&width=778&top_left_y=192&top_left_x=640)

Answer

$\frac{1}{2}(a-b)^{2} \sin \alpha$.