Task B-1.7. Let $A B C D$ be a quadrilateral with all angles less than $180^{\circ}$. Sides $A D, C D$ and diagonal $B D$ are of equal length, the measure of angle $\measuredangle A D B$ is equal to the measure of angle $\measuredangle D C A$ and the measure of angle $\measuredangle C B D$ is equal to the measure of angle $\measuredangle B A C$. Determine the measures of the angles of quadrilateral $A B C D$.
Solution. Let $\measuredangle A D B=\measuredangle D C A=\alpha, \measuredangle C B D=\measuredangle B A C=\beta$.

![](https://cdn.mathpix.com/cropped/2024_06_03_ade005c4283eff439485g-05.jpg?height=593&width=1005&top_left_y=625&top_left_x=533)

Triangle $D A C$ is isosceles because $|D A|=|D C|$. Therefore, $\measuredangle D A C=\measuredangle D C A=\alpha$.

(1 point)

Triangle $D A B$ is isosceles because $|D A|=|D B|$. Therefore, $\measuredangle D A B=\measuredangle D B A=\alpha+\beta$.

(1 point)

Triangle $B D C$ is isosceles because $|D C|=|D B|$. Therefore, $\measuredangle D B C=\measuredangle D C B=\beta$.

From this, $\measuredangle A C B=\beta-\alpha$.

In triangle $A B C$, we have

$$
\beta+\alpha+\beta+\beta+\beta-\alpha=180^{\circ}.
$$

Thus, $4 \beta=180^{\circ}, \beta=45^{\circ}$.

From triangle $A D B$, we get

$$
2(\alpha+\beta)+\alpha=180^{\circ}
$$

$3 \alpha=90^{\circ}, \alpha=30^{\circ}$.

The angles of the quadrilateral are

$$
\begin{aligned}
& \measuredangle D A B=\alpha+\beta=75^{\circ} \\
& \measuredangle D C B=\beta=45^{\circ} \\
& \measuredangle C B A=\alpha+2 \beta=120^{\circ} \\
& \measuredangle A D C=360^{\circ}-120^{\circ}-45^{\circ}-75^{\circ}=120^{\circ}
\end{aligned}
$$

(2 points)

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 2nd grade - high school - B variant

17. January 2013.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.