8.3. $A B C D$ - a convex quadrilateral. It is known that $\angle C A D=\angle D B A=40^{\circ}, \angle C A B=60^{\circ}$, $\angle C B D=20^{\circ}$. Find the angle $C D B$.
Answer: $30^{\circ}$.

Solution. Since $\angle C A B=60^{\circ}, \angle A B C=\angle A B D+\angle D B C=60^{\circ}$, triangle $A B C$ is equilateral (see Fig. 8.3a). We can reason in different ways from here.

![](https://cdn.mathpix.com/cropped/2024_05_06_83dbb8f64176429cc0ceg-2.jpg?height=368&width=374&top_left_y=384&top_left_x=404)

Fig. 8.3a

![](https://cdn.mathpix.com/cropped/2024_05_06_83dbb8f64176429cc0ceg-2.jpg?height=392&width=434&top_left_y=358&top_left_x=1219)

Fig. 8.3b

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 $\angle B D A=180^{\circ}-(40^{\circ}+100^{\circ})=40^{\circ}$. Therefore, triangle $A B D$ is isosceles (see Fig. 8.3a). Thus, $A B=B C=C A=A D$, so triangle $C A D$ is also isosceles. Then $\angle A D C=\angle A C D=(180^{\circ}-\angle C A D): 2=70^{\circ}, \angle C D B=\angle C D A-\angle B D A=70^{\circ}-40^{\circ}=30^{\circ}$.

Second method. Draw a circle with center $A$ and radius $A B=A C$ and prove that it contains point $D$. Suppose this is not the case, then the circle intersects ray $A D$ at some point $E$, different from $D$ (see Fig. 8.3b). By the inscribed angle theorem, $\angle C B E=\frac{1}{2} \angle C A E=20^{\circ}$, so rays $B E$ and $B D$ coincide. Therefore, points $E$ and $D$ coincide, which is a contradiction. Since the circle passes through point $D$, $\angle C D B=\frac{1}{2} \angle C A B=30^{\circ}$.

It is also possible to use a similar solution where the isosceles nature of triangle $C A D$ is first proven, and then the circle with center $A$ and radius $A C=A D$ is used. Note that attempting to use the auxiliary circle immediately (without proving the equality of any two segments) is unlikely to be successful.

## Grading Criteria:

+ A complete and well-reasoned solution is provided

$\pm$ A generally correct reasoning is provided, with minor inaccuracies or gaps

干 An auxiliary circle with center $A$ containing points $B, C$, and $D$ is drawn, but it is not proven that it exists, and the correct answer is obtained

- Only the answer is provided
- The problem is not solved or is solved incorrectly