Note that the geometric locus of points $O$, from which the segment $A D$ is seen at an angle of $80^{\circ}$, and lying on the same side of $A D$ as point $B,$ is an arc of a circle passing through points $A$ and $D$, and the geometric locus of points $O$, from which the segment $B C$ is seen at an angle of $100^{\circ}$, and lying on the same side of $B C$ as point $A,$ is an arc of a circle passing through points $B$ and $C$. The desired point $O$ must lie at the intersection of these two arcs. Therefore, there can be no more than two such points. Let us indicate two points that satisfy the condition of the problem.

The first point $O 1$ lies on the diagonal $A C$, and $\angle B O 1 C=100^{\circ}$. Then

$$
\angle A O 1 B=180^{\circ}-\angle B O 1 C=80^{\circ}
$$

and since point $D$ is symmetric to point $B$ with respect to the line $A C$, then

$$
\angle A O 1 D=\angle A O 1 B=80^{\circ}
$$

which is required by the condition. Similarly, the second point $O 2$ lies on the diagonal $B D$, and $\angle B O 2 C=100^{\circ}$. In this case, $\angle A O 2 D=80^{\circ}$ and $\angle A O 2 B=100^{\circ}$. It remains to note that points $O 1$ and $O 2$ are distinct (they lie on different diagonals of the rhombus and are different from the point of intersection of the diagonals) and both lie inside the rhombus.

![](https://cdn.mathpix.com/cropped/2024_05_06_6c7d8d8bac098e64f10bg-11.jpg?height=883&width=1052&top_left_y=-1&top_left_x=523)

## Answer

$80^{\circ}$ or $100^{\circ}$.

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Volchkovich M.A.

Point $O$ lies inside rhombus $A B C D$. Angle $D A B$ is $110^{\circ}$. Angles $A O D$ and $B O C$ are $80^{\circ}$ and $100^{\circ}$, respectively. What can the angle $A O B$ be?