
3. The sides $B C$ and $A D$ of a quadrilateral $A B C D$ are parallel and the diagonals intersect in $O$. For this quadrilateral $|C D|=|A O|$ and $|B C|=|O D|$ hold. Furthermore $C A$ is the angular bisector of angle $B C D$.

Determine the size of angle $A B C$.

Attention: the figure is not drawn to scale.

You have to write down your reasoning step by step in text and formulas. No points will be awarded for annotations in a picture

![](https://cdn.mathpix.com/cropped/2024_04_17_e07875fe43bd53054a56g-1.jpg?height=414&width=505&top_left_y=1483&top_left_x=1392)
alone.


3. First, we prove that some triangles in the figure are isosceles (the top angle coincides with the middle letter).

(1) Triangle $A D C$ is isosceles, because $\angle D A C=\angle A C B=\angle A C D$. The first equality holds because $A D$ and $B C$ are parallel and the second equality follows from the fact that $A C$ is the interior angle bisector of angle $B C D$.

(2) Triangle $D A O$ is isosceles, because $|A D|=|C D|=|A O|$. The first equality follows from (1) and the second equality is given in the problem statement.

(3) Triangle $B C O$ is isosceles, because it is similar to triangle $D A O$ (hourglass shape).

(4) Triangle $C O D$ is isosceles, because $|D O|=|B C|=|C O|$. Here the first equality is given in the problem statement and the second one follows from (3).

(5) Triangle $B D C$ is isosceles, because it is similar to triangle $B C O$ as two pairs of corresponding angles are equal: $\angle D B C=\angle O B C$ and $\angle B D C=\angle D C O=\angle O C B$. Here the last equality follows from (4).

(6) Triangle $A D B$ is isosceles, because $|A D|=|C D|=|B D|$ because of (1) and (5).

Denote $\angle A C B=\alpha$ and $\angle C B D=\beta$ (in degrees). From (5) it follows that $2 \alpha=\beta$. From (3) it follows that $180^{\circ}=2 \beta+\alpha=5 \alpha$, hence $\alpha=\frac{180^{\circ}}{5}=36^{\circ}$ and $\beta=72^{\circ}$. In the isosceles triangle $A D B$ the top angle is equal to $\angle A D B=\beta$, hence its equal base angles are $\frac{180^{\circ}-72^{\circ}}{2}=54^{\circ}$. The requested angle therefore equals $\angle A B C=\angle A B D+\angle D B C=54^{\circ}+72^{\circ}=126^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_ea22e3414b6ace40dc6dg-2.jpg?height=596&width=734&top_left_y=1301&top_left_x=367)
