48th Austrian Mathematical Olympiad Regional Competition (Qualifying Round)

Denote by $X$ the point of intersection of the diagonals $AC$ and $BD$, i.e. $AX$ is an altitude in the triangle $ABD$, see Figure 1.



Figure 1: Problem 2

$$\angle ABX=\frac{1}{2}\angle DOA.$$ Hence $$\angle XAB=90^{\circ }-\angle ABX=\frac{1}{2}\cdot (180^{\circ }-\angle DOA)=\angle OAD.$$ In the last step the angle sum in the equilateral triangle $DAU$ has been used. Since the lines $AB$ and $AD$ are symmetric with respect to the angle bisector $w_{\alpha }$, the same is true for $AX$ and $AU$. Hence the assertion follows. □