3. In quadrilateral $A B C D A D=B D=C D$. Angle $B D C$ is twice the angle $B D A$, and angles $B D A$ and $C A D$ are in the ratio $2: 3$. Find the angles of the quadrilateral.
Solution. Let angle $B D A$ be $x$ (see the figure), then by the condition angle $B D C$ is $2 x$, angle $C A D$ is $1.5 x$. Since $A D=B D=C D$, point $D$ is the center of the circle passing through points $A, B$, and $C$.

Triangle $\triangle A D B$ is isosceles $(A D=B D)$, so angle $D A B$ is $90^{\circ}-0.5 x$. Then $\angle C A B=\angle D A B-\angle D A C=90^{\circ}-0.5 x-1.5 x=90^{\circ}-2 x$. Angle $C D B$ is a central angle of the arc subtended by chord $B C$, and angle $C A B$ is an inscribed angle subtending the same arc. Therefore, $\angle C D B=2 \angle C A B$, which gives us the equation $2 x=2\left(90^{\circ}-2 x\right)$. Solving this equation, we find $x=30^{\circ}$. Now it is easy to find the angles of quadrilateral $A B C D$.

Answer: $75^{\circ}, 135^{\circ}, 60^{\circ}, 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6795fd51857d08c3c87eg-1.jpg?height=343&width=414&top_left_y=2330&top_left_x=770)