Austria 2010

The angle $\angle ABC$ is denoted by $\beta$. As $BC$ is normal to $AE$ and $DE$ is normal to Abbildung 1: Problem 4. $AB$, the angles $\angle ABC$ and $\angle AED$ are of equal measure. As we have $\angle ACB=\angle DCE=90^{\circ }$ and, by assumption, $\overline{AB}=\overline{DE}$, the triangles $ABC$ and $DEC$ are congruent.

This yields $\overline{BC}=\overline{CE}$ which implies that the triangle $BCE$ is an isosceles right-angled triangle with $\angle CEB=\angle CBE=45^{\circ }$.

Furthermore, we have $\beta =\angle CED=\angle DEB$, as $E$ lies on the perpendicular bisector of $AB$.

Thus we obtain $45^{\circ }=\angle CEB=2\beta$ and therefore $\beta =22.5^{\circ }$ and $\alpha =\angle CAB=67.5^{\circ }$.