Example 8. In a parallelogram with sides equal to $a$ and $b$, and the angle between these sides equal to $\alpha$, the bisectors of the four angles are drawn. Find the area of the quadrilateral bounded by the bisectors.

Given (Fig. 10): $ABCD$ is a parallelogram, $l_{A}, l_{B}, l_{C}, l_{D}$ are the bisectors of angles $A, B, C, D; l_{A} \cap l_{B}=P, l_{A} \cap l_{D}=M$, $l_{B} \cap l_{C}=K, l_{C} \cap l_{D}=E, |AB|=a, |BC|=b, \widehat{BAD}=\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-023.jpg?height=377&width=671&top_left_y=234&top_left_x=356)

Fig. 10

Find: $S$ - the area of the quadrilateral $PMEK$.
Solution. First, let's determine the type of quadrilateral $P M E K$. Since $(A D) \|(B C)$, $\widehat{M A D}=\frac{\alpha}{2}$ and $\widehat{B C K}=\frac{\alpha}{2}$, then $(A M) \|(C K)$. Similarly, $(B K) \|(D M)$, i.e., $P M E K$ is a parallelogram.

Furthermore, since $\widehat{B A M}=\frac{\alpha}{2}$ and $\widehat{A B K}=\frac{180^{\circ}-\alpha}{2}$, in $\triangle A B P$, $\widehat{A P B}=180^{\circ}-\frac{\alpha}{2}-\frac{180^{\circ}-\alpha}{2}=90^{\circ}$, and therefore, $\widehat{M P K}=90^{\circ}$. Thus, $P M E K$ is a rectangle. Therefore, $S=|P M| \cdot|P K|$.

Now, let's use direct calculation.

From the right triangles $A B P$ and $A M D$, we have respectively:

$$
|A P|=a \cos \frac{\alpha}{2},|A M|=b \cos \frac{\alpha}{2}
$$

Then

$$
|P M|=\|A M|-| A P\|=|a-b| \cos \frac{\alpha}{2}
$$

Similarly, from the right triangles $A B P$ and $B K C$:

$$
\begin{aligned}
& |B P|=a \sin \frac{\alpha}{2} \text { and }|B K|=b \sin \frac{\alpha}{2}, \text { i.e., } \\
& |P K|=\|B P|-| B K\|=|a-b| \sin \frac{\alpha}{2}
\end{aligned}
$$

Thus,

$$
S=|a-b| \cos \frac{\alpha}{2} \cdot|a-b| \sin \frac{\alpha}{2}=\frac{|a-b|^{2} \sin \alpha}{2}
$$

But $|a-b|^{2}=(a-b)^{2}$, therefore,

$$
S=\frac{(a-b)^{2} \sin \alpha}{2}
$$