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Slovenian Mathematical Olympiad, 2008

Geometry Difficulty 6.3 National Olympiad Prove it Slovenia

Let $ABC$ be a right triangle with the right angle at $C$ . On the segment $BC$ choose a point $D$ different from $B$ and $C$ . Denote the circumcircle of the triangle $ABD$ by $\mathcal{K}$ . Let $T$ be a point on the side $AB$ such that $DT$ is perpendicular to $AB$ . Let $E$ be the second intersection of the line $DT$ with $\mathcal{K}$ , denote the intersection of lines $CT$ and $EB$ by $F$ and let the line $DF$ meet $\mathcal{K}$ again at $G$ . Prove that triangles $CEF$ and $BEG$ are similar.

Solution

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.

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