Slovenija 2008

Since triangles $ABD$ and $KCD$ are similar, we have $\angle ADB=\angle KDC$ and $\frac{|DA|}{|DB|}=\frac{|DK|}{|DC|}$. We see that $$\begin{gathered} \angle ADK=\angle ADB-\angle BDK= \\ =\angle KDC-\angle BDK=\angle BDC \end{gathered}$$ and since $\frac{|DA|}{|DK|}=\frac{|DB|}{|DC|}$, we conclude that triangles $ADK$ and $DBC$ are also similar (matching in one angle and the ratio of the two adjacent sides).