Austria 2010

Let the point $C^{\prime }$ be chosen in such a way that triangles $△ABC$ and $△AC{C}^{\prime }$ are similar and have no common interior points. Furthermore, let $D^{\prime }$ be chosen on $C{C}^{\prime }$ such that triangles $△ABD$ and $△AC{D}^{\prime }$ are also similar. This means that $△AC{C}^{\prime }$ results from $△ABC$ by rotation and subsequent homothety with ratio $AC:AB$, both with center $A$.



Since $\angle D^{\prime }CD=\angle D^{\prime }CA+\angle ACD=\angle CBA+\angle ACB$ and $\angle D^{\prime }AD=\angle CAB$, we see that $\angle D^{\prime }CD+\angle D^{\prime }AD=180^{\circ }$. This means that the points $A$, $D$, $C$ and $D^{\prime }$ lie on a common circle. The circumcenter $V$ of $△CDA$ is therefore also the circumcenter of $△C{D}^{\prime }A$, and therefore results from the circumcenter $U$ of $△BDA$ by rotation and subsequent homothety with ratio $AC:AB$, both with center $A$.

We therefore see that $\angle UAV=\angle BAC$ and $AU:AV=AB:AC$. Triangles $ABC$ and $AUV$ are therefore similar, as claimed.