## Task 3.

On a circle, points $A, B, C$, and $D$ are given such that $|A B|=|B C|=|C D|$. Let the angle bisectors of $\varangle A C D$ and $\varangle A B D$ intersect at point $E$.

If lines $A E$ and $C D$ are parallel, determine $\varangle A B C$.
## Solution.

Due to symmetry, isosceles triangles $ABC$ and $BCD$ are congruent, and the cyclic quadrilateral $ABCD$ is an isosceles trapezoid.

![](https://cdn.mathpix.com/cropped/2024_05_30_1bf59d0666071a572bf4g-2.jpg?height=512&width=528&top_left_y=1126&top_left_x=770)

Let $x$ be the measure of the angles at the bases of triangles $ABC$ and $BCD$. Then $\angle ABC = \angle BCD = 180^\circ - 2x$ and $\angle ABD = \angle ACD = 180^\circ - 3x$, from which $\angle EBC = \angle ECB = 90^\circ - \frac{x}{2}$ and $\angle BEC = x$. Therefore, point $E$ lies on the same circle as $A, B, C$, and $D$.

Additionally, $\angle ADC = \angle ADB + \angle BDC = \angle ACB + \angle BDC = x + x = 2x$ and $\angle ECD = \frac{1}{2} \angle ACD = 90^\circ - \frac{3}{2}x$.

Since $AE \parallel CD$, the cyclic quadrilateral $ACDE$ is also an isosceles trapezoid, and we have $\angle ADC = \angle ECD$, from which $7x = 180^\circ$. Finally, $\angle ABC = \frac{5}{7} \cdot 180^\circ$.