In a convex quadrilateral circumscribed around a circle, the products of opposite sides are equal. The angle between a side and one of the diagonals is $20^{\circ}$. Find the angle between this side and the other diagonal.
Let $A B C D$ be the given quadrilateral, $\angle B A C=20^{\circ}$. Denote $A B=a, B C=b, C D=c, A D=d$. From the condition of the problem, it follows that $a c=b d$ and $a+c=b+d$ (property of a cyclic quadrilateral). Then, by the theorem converse to Vieta's theorem, either $a=b$ and $c=d$, or $a=d$ and $c=b$. Suppose that $a=b$ and $c=d$, i.e., $A B=B C$ and $C D=A D$. Then points $B$ and $D$ are equidistant from the ends of segment $A C$. Therefore, $B D$ is the perpendicular bisector of diagonal $A C$. Hence,

$$
\angle A B D=90^{\circ}-\angle B A C=90^{\circ}-20^{\circ}=70^{\circ} .
$$

Similarly for the case $a=d$ and $c=b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_808033c11c6fe49d1738g-22.jpg?height=618&width=623&top_left_y=254&top_left_x=749)

## Answer

$70^{\circ}$.