4. On the plane, there are four points $A, B, C, D$ such that $A B=B C=C D, B D=D A=A C$. Find the angles of the quadrilateral with vertices at these points.
Solution. Angles $\angle A B C$ and $\angle D C B$ are equal due to the congruence of triangles $A B C$ and $D C B$.

a) If vertices $A$ and $D$ are on opposite sides of line $B C$, then these angles are alternate interior angles. Therefore, segments $A B$ and $C D$ are parallel and equal, making $A B D C$ a parallelogram. However, each angle of the parallelogram would then be acute, as they are base angles of isosceles triangles ($\angle A$ in $\triangle A B C$, $\angle B$ in $\triangle A B D$). This is impossible.

b) Suppose $A$ and $D$ lie on the same side of $B C$. The distances from points $A$ and $D$ to $B C$ are equal, since $\triangle A B C = \triangle D C B$; thus, $A D \parallel B C$. This means we have a trapezoid with bases $A D$ and $B C$, and it is isosceles (the case $A D = B C$ is impossible, as then all six segments would be equal, which is not possible for such quadrilaterals). Inscribing it in a circle, let's assume, without loss of generality, that $B C < A D$. The arcs $A B, B C$, and $C D$ are subtended by equal chords (and do not intersect); let the degree measure of each be $\alpha$. Then the arc $A D$ is equal to the arc $A C$, which is $2 \alpha$. We get $5 \alpha = 360^{\circ}$, and the angles of the trapezoid are $2 \alpha, 2 \alpha, 3 \alpha, 3 \alpha$ as inscribed angles.
![](https://cdn.mathpix.com/cropped/2024_05_06_d4b15693c622237d278eg-2.jpg?height=644&width=326&top_left_y=1802&top_left_x=1570)

Answer: $72^{\circ}, 72^{\circ}, 108^{\circ}, 108^{\circ}$.

Criteria. 3 points - proof that it is not a parallelogram.

4 points - analysis of the trapezoid case and correctly found angles.