Slovenija 2008

The sum of the two opposite angles $\angle ATD$ and $\angle ACD$ in the quadrilateral $ATDC$ is $\pi$, so this quadrilateral is cyclic. Set $\angle TCD=\alpha$. Then

$\angle TAD=\angle TCD=\alpha$. Points $A$, $B$, $E$, $D$ are concyclic, so $\angle BED=\angle BAD=\angle TAD=\alpha$. We have $\angle FED=\angle BED=\alpha =\angle TCD=\angle FCD$, so angles $FED$ and $FCD$ in the quadrilateral $FECD$ are equal and this quadrilateral is also cyclic.

Let $\angle ECF=\beta$. Since $FECD$ is cyclic, we have $\angle EDF=\angle ECF=\beta$. Points $E$, $D$, $B$ and $G$ all lie on $\mathcal{K}$, so $\angle EBG=\angle EDG=\angle EDF=\beta$.

Denote $\angle EFC=\gamma$. Since the quadrilateral $CDFE$ is cyclic, we have $\angle EDC=\angle EFC=\gamma$, so $\angle EDB=\pi -\angle EDC=\pi -\gamma$. Points $D$, $B$, $G$ and $E$ are concyclic, so $\angle EGB=\pi -\angle EDB=\pi -(\pi -\gamma )=\gamma$, and $\angle EFC=\angle EGB$. But we have already shown that $\angle ECF=\beta =\angle EBG$. Triangles $ECF$ and $EBG$ have two congruent angles, so they are similar.