![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-19.jpg?height=186&width=2015&top_left_y=1598&top_left_x=17)

$ABCD$ - a convex quadrilateral. It is known that $\angle CAD = \angle DBA = 40^{\circ}, \angle CAB = 60^{\circ}, \angle CBD = 20^{\circ}$. Find the angle $\angle CDB$.
Since $\angle C A B=60^{\circ}, \angle A B C=\angle A B D+\angle D B C=60^{\circ}$, the triangle $A B C$ is equilateral (see the left figure). We can reason in different ways.

The first method. In triangle $A B D$, $\angle A B D=40^{\circ}, \angle B A D=\angle B A C+\angle C A D=100^{\circ}$, so $\angle B D A=180^{\circ}-$ $\left(40^{\circ}+100^{\circ}\right)=40^{\circ}$. Therefore, this triangle is isosceles (see the left figure). Thus, $A B=B C=$ $C A=A D$, so triangle $C A D$ is also isosceles. Hence,

$\angle A D C=\angle A C D=1 / 2\left(180^{\circ}-\angle C A D\right)=70^{\circ}, \angle C D B=\angle C D A-\angle B D A=70^{\circ}-40^{\circ}=30^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-19.jpg?height=560&width=1272&top_left_y=2310&top_left_x=402)

The second method. Draw a circle with center $A$ and radius $A B=A C$. Let it intersect the ray $A D$ at some point $E$ (see the right figure). By the inscribed angle theorem, $\angle C B E=1 / 2 \angle C A E=20^{\circ}$, so the rays $B E$ and $B D$ coincide. Therefore, points $E$ and $D$ coincide. Since the circle passes through point $D$, $\angle C D B=1 / 2$ $C A B=30^{\circ}$.

## Answer

$30^{\circ}$.