55rd Ukrainian National Mathematical Olympiad - Third Round

Let $K$ be the intersection point for the bisectors of the angles $ABD$ and $AED$ (Fig. 3). Then this point lies on the segment $AD$ and is equidistant from rays $BA$ and $BC$, as well as from $EA$ and $ED$. Hence, it's equidistant from rays $DE$ and $DC$. Then $DA$ is the bisector of $\angle CED$ (in other words, $K$ is an excenter of the triangle $EBD$). With the initial conditions, this implies $\angle ADC=60^{\circ }$. Since $\angle DCA=2\angle DAC$, we have that $\angle DCA=80^{\circ }$. Finally, we can write $\angle BAC=\angle BCA=80^{\circ }$, $\angle ABC=20^{\circ }$.



Fig. 3