Markelov S.v.

In parallelogram $A B C D$, angle $A C D$ is equal to $30^{\circ}$. It is known that the centers of the circles circumscribed around triangles $A B D$ and $B C D$ are located on diagonal $A C$. Find angle $A B D$.
Consider two cases: the centers of the circles are different points; the centers of the circles coincide.

## Solution

If the centers of the given circles do not coincide, then the line of centers (i.e., the diagonal $AC$ of the parallelogram) is perpendicular to the common chord (diagonal $BD$). Therefore, the given parallelogram is a rhombus and $\angle ABD = 60^{\circ}$. If the centers of the circles coincide, then the circles coincide as well, since triangles $ABD$ and $BCD$ are equal. In this case, the parallelogram $ABCD$ is a rectangle and

$$
\angle ABD = \angle ACD = 30^{\circ}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-20.jpg?height=450&width=1334&top_left_y=1458&top_left_x=366)

Answer

$60^{\circ}$ or $30^{\circ}$.