[ The ratio of the areas of triangles with a common base or a common height ] $]$  [ The area of a figure is equal to the sum of the areas of the figures into which it is divided ]

In parallelogram $A B C D$, the bisector of angle $A$ intersects side $B C$ at point $M$, and the bisector of angle $C$ intersects side $A D$ at point $N$. The area of the quadrilateral formed by the intersection of the bisectors $A M$ and $C N$ with segments $B N$ and $D M$ is $\frac{6}{5}$. Find the angles of parallelogram $A B C D$, if $A B=3, A D=5$.
Let $P$ be the intersection point of lines $C N$ and $D M$, and $Q$ be the intersection point of lines $A M$ and $B N$. Since

$$
\angle B M A = \angle D A M = \angle B A M
$$

triangle $A B M$ is isosceles,

$$
B M = A B = 3, \quad M C = B C - B M = 5 - 3 = 2
$$

Similarly, we can prove that $D N = 3$ and $A N = 2$. Therefore, $B M D N$ is a parallelogram. Hence, $M P \| N Q$, and since $A M \| C N$, $M P N Q$ is also a parallelogram. Thus,

$$
S_{\triangle \mathrm{MPN}} = \frac{1}{2} S_{\mathrm{MPNQ}} = \frac{3}{5}
$$

Moreover,

$$
\frac{M P}{P D} = \frac{C P}{P N} = \frac{M C}{N D} = \frac{2}{3}
$$

Therefore,

$$
\begin{gathered}
S_{\Delta \mathrm{MPC}} = \frac{C P}{P N} S_{\Delta \mathrm{MPN}} = \frac{2}{3} \cdot \frac{3}{5} = \frac{2}{5} \\
S_{\Delta \mathrm{CDP}} = \frac{D P}{P M} S_{\Delta \mathrm{MPC}} = \frac{3}{2} \cdot \frac{2}{5} = \frac{3}{5} \\
S_{\Delta \mathrm{DPN}} = \frac{D P}{P M} S_{\Delta \mathrm{MPN}} = \frac{3}{2} \cdot \frac{3}{5} = \frac{9}{10} \\
S_{\mathrm{ABCD}} = 2 S_{\mathrm{CDMN}} = 2 \left( S_{\Delta \mathrm{MPN}} + S_{\Delta \mathrm{MPC}} + S_{\triangle \mathrm{CDP}} + S_{\Delta \mathrm{DPN}} \right) = 2 \left( \frac{3}{5} + \frac{2}{5} + \frac{3}{5} + \frac{9}{10} \right) = 5
\end{gathered}
$$

On the other hand,

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

From the equation $15 \sin \angle B A D = 5$, we find that $\sin \angle B A D = \frac{1}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-32.jpg?height=349&width=924&top_left_y=1349&top_left_x=567)

## Answer

$\arcsin \frac{1}{3}, 180^{\circ} - \arcsin \frac{1}{3}$