5. Given trapezoid $A B C D$ with bases $A B$ and $C D$, angles $\angle C=30^{\circ}$ and $\angle D=80^{\circ}$. Find $\angle A C B$, if it is known that $D B$ is the bisector of angle $\angle D$.
Answer: $10^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_2ae48c21b42f492bf703g-09.jpg?height=691&width=740&top_left_y=1553&top_left_x=658)

Let $E$ be the intersection point of lines $A D$ and $B C$, and $D^{\prime}$ be the point symmetric to point $D$ with respect to line $B C$. Then $C D = C D^{\prime}$ and $\angle D C D^{\prime} = 2 \angle B C D = 60^{\circ}$. Therefore, triangle $D C D^{\prime}$ is equilateral and $D D^{\prime} = C D$. By symmetry, $D E = D^{\prime} E$ and triangle $D E D^{\prime}$ is isosceles. Thus, $\angle E D^{\prime} D = \angle E D D^{\prime} = \angle A D C - \angle C D D^{\prime} = 80^{\circ} - 60^{\circ} = 20^{\circ}$.

By Thales' theorem, $\frac{E A}{A D} = \frac{E B}{B C}$, and by the property of the angle bisector, $\frac{E B}{B C} = \frac{E D}{D C} = \frac{E D^{\prime}}{D D^{\prime}}$. Therefore, $\frac{E A}{A D} = \frac{E D^{\prime}}{D D^{\prime}}$ and $D^{\prime} A$ is the angle bisector of $\angle E D^{\prime} D$, so $\angle A D^{\prime} D = \angle A D^{\prime} E = 10^{\circ}$. Therefore, $\angle A D^{\prime} C = \angle A D^{\prime} D + \angle D D^{\prime} C = 10^{\circ} + 60^{\circ} = 70^{\circ} = \angle A E C$. Thus, quadrilateral $A E D^{\prime} C$ is cyclic and $\angle A C B = \angle A C E = \angle A D^{\prime} E = 10^{\circ}$.