In parallelogram $A B C D$, where $\angle B A D$ equals $60^{\circ}, A B=2, A D=5$, the bisector of angle $B A D$ intersects with the bisector of angle $A B C$ at point $K$, with the bisector of angle $C D A$ at point $L$, and the bisector of angle $B C D$ intersects with the bisector of angle $C D A$ at point $M$, with the bisector of angle $A B C$ at point $N$. Find the ratio of the area of quadrilateral $K L M N$ to the area of parallelogram $A B C D$.
KLMN is a rectangle.

## Solution

Since the angle between the bisectors of the internal same-side angles at parallel lines is $90^{\circ}$, then $K L M N-$ is a rectangle.

Let $P$ be the point of intersection of the lines $A K$ and $B C$. Then

$$
\angle B P A=\angle D A P=\angle B A P .
$$

Therefore, triangle $A B P-$ is isosceles,

$$
B P=A B=2, P C=B C-B P=3 \text {. }
$$

If $F$ is the projection of point $P$ onto $C M$, then

$$
L M=P F=P C \sin \angle P C F=3 \sin 30^{\circ}=\frac{3}{2}
$$

Similarly, $M N=\frac{3 \sqrt{3}}{2}$. Therefore,

$$
S_{\mathrm{KLMN}}=L M \cdot M N=\frac{9 \sqrt{3}}{4}
$$

and since

$$
S_{\mathrm{ABCD}}=A B \cdot A D \sin \angle B A D=2 \cdot 5 \cdot \frac{\sqrt{3}}{2}=5 \sqrt{3}
$$

then

$$
\frac{S_{K L M N}}{S_{A B C D}}=\frac{\frac{9 \sqrt{3}}{4}}{5 \sqrt{3}}=\frac{9}{20}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-48.jpg?height=431&width=907&top_left_y=1831&top_left_x=587)

## Answer

$\frac{9}{20}$.