10.331. In a rectangle with sides $a$ and $b$, the angle bisectors of all angles are drawn until they intersect each other. Find the area of the quadrilateral formed by the angle bisectors.
Solution.

$K L M N$ is a quadrilateral formed by the mutual intersection of the angle bisectors of rectangle $A B C D$ (Fig. 10.118), $A B=a$, $A D=b$. Without loss of generality, assume that $b>a$. Since $A B C D$ is not a rhombus, it follows from the properties of the angle bisectors of a parallelogram that $K L M N$ is a rectangle. Points $P$ and $E$ are the midpoints of sides $A B$ and $C D$ of the rectangle, respectively. Since $\triangle A K B$ and $\triangle C M D$ are isosceles, $K P$ and $M E$ are the perpendicular bisectors of the opposite sides of the rectangle and, therefore, points $P, K, M, E$ lie on the same line. Then the line $K M$, which bisects $\angle A K B$, also bisects the vertical angle $\angle N K L$, and thus the rectangle $K L M N$ is a square. From $\triangle A K B: K P=\frac{1}{2} A B=\frac{a}{2} . M E=K P=\frac{a}{2}$; $P E=B C=b$.

Then $K M=b-a$ and the desired area $S=\frac{1}{2} K M^{2}=\frac{(b-a)^{2}}{2}$.

Answer: $\frac{(b-a)^{2}}{2}$.