Croatian Junior Mathematical Olympiad

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

Let us denote by $x$ the measure of angles along the bases in $ABC$ and $BCD$. Then $\angle ABC=\angle BCD=180^{\circ }-2x$ and $\angle ABD=\angle ACD=180^{\circ }-3x$, from which we get $\angle EBC=\angle ECB=90^{\circ }-\frac{x}{2}$ and $\angle BEC=x$. Hence, the point $E$ lies on the same circle as $A$, $B$, $C$ and $D$. We also have $\angle ADC=\angle ADB+\angle BDC=\angle ACB+\angle BDC=x+x=2x$ and $\angle ECD=\frac{1}{2}\angle ACD=90^{\circ }-\frac{3}{2}x$. Since $AE\parallel CD$, the cyclic quadrilateral $ACDE$ is an isosceles trapezium as well, so $\angle ADC=\angle ECD$ holds, and we get $7x=180^{\circ }$. Therefore, $\angle ABC=\frac{5}{7}\cdot 180^{\circ }$.