[ Inscribed quadrilaterals ] [Median proportionals in a right triangle ]

A circle with center $O$, inscribed in quadrilateral $A B C D$, touches sides $A B, B C, C D$ and $A D$ at points $K$, $L, M$ and $N$ respectively. Segment $K N$ bisects $O A$, segment $K L$ bisects $O B$, and segment $M N$ divides $O D$ in the ratio $1: 3$, counting from point $O$. Find the angles of quadrilateral $A B C D$.
Let segments $O D$ and $M N$ intersect at point $P$. Denote $O P=x$, then $D P=3 x$. Since $O D \perp M N$, segment $M P$ is the altitude of the right triangle $O M D$ drawn from the vertex of the right angle, therefore

$$
O M=\sqrt{O P \cdot O D}=\sqrt{x \cdot 4 x}=2 x
$$

The leg $O P$ of the right triangle $O P M$ is half the hypotenuse $O M$, hence $\angle O M P=30^{\circ}$. Then $\angle M D O=30^{\circ}$, and since $D O$ is the bisector of angle $A D C$, $\angle A D C=60^{\circ}$. The diagonals of quadrilateral $A K O N$ are bisected by their point of intersection, so this quadrilateral is a parallelogram, and since $A K=A N$, it is a rhombus with a right angle $A N O$, i.e., a square. Therefore, $\angle B A D=90^{\circ}$. Similarly, $\angle A B C=90^{\circ}$. Lines $A D$ and $B C$ are perpendicular to the same line $A B$, so $B C \| A D$. Therefore,

$$
\angle B C D=180^{\circ}-\angle A D C=180^{\circ}-60^{\circ}=120^{\circ} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1f91526468b1b4d32730g-37.jpg?height=711&width=938&top_left_y=-1&top_left_x=603)

## Answer

$60^{\circ}, 120^{\circ}, 90^{\circ}, 90^{\circ}$