Austria competition problems, 2010

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

Since $\angle D'CD = \angle D'CA + \angle ACD = \angle CBA + \angle ACB$ and $\angle D'AD = \angle CAB$, we see that $\angle D'CD + \angle D'AD = 180^\circ$. This means that the points $A$, $D$, $C$ and $D'$ lie on a common circle. The circumcenter $V$ of $\triangle CDA$ is therefore also the circumcenter of $\triangle CD'A$, and therefore results from the circumcenter $U$ of $\triangle BDA$ 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.