Rioplatense Mathematical Olympiad

Let $\angle BCA=\alpha$ and $X$ be a point on ray $CA$ such that $BX=BC$. Since $BX=BC$, we have that $\angle BXC=\alpha$, and using that $\angle BAC=\angle AXB+\angle ABX$, we obtain that $\angle ABX=\alpha$ and hence $AX=AB$. We can observe that $\angle XAD=\angle DAB$, as $\angle XAD=180^{\circ }-\angle DAC=90^{\circ }+\alpha$, $\angle DAB=90^{\circ }-\alpha +2\alpha =90^{\circ }+\alpha$. Therefore, using SAS congruence we have $△XAD\cong △BAD$, and $XD=DB$. This shows that $△XDB$ is equilateral, and $\angle XDA=\angle ADB=30^{\circ }$.