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Slovenia — National Math Olympiad, 2012

Geometry Difficulty 6.4 National Olympiad Prove it Slovenia

Let $ABCD$ be a cyclic quadrilateral with longest side $AB$ . Let the bisectors of the angles $\angle DCB$ and $\angle ADC$ additionally intersect the circumscribed circle of the quadrilateral $ABCD$ in points $E$ and $F$ . Call $G$ the intersection point of the lines $CE$ and $DF$ , and $H$ the intersection point of the lines $AE$ and $BF$ .
Prove that the lines $EF$ and $GH$ intersect at right angles.

Solution

Because the points $C$ , $D$ , $E$ and $F$ are concyclic, the angles $\angle FDC$ and $\angle FEC$ are equal. Because the points $A$ , $E$ , $F$ and $D$ are concyclic, the angles $\angle ADF$ and $\angle HEF$ are equal. Hence the angles $\angle HEF$ and $\angle FEG$ are equal. We can show similarly that the angles $\angle GFE$ and $\angle EFH$ are equal. The triangles $EFG$ and $EFH$ coincide in one side and the adjacent angles, so they are congruent. We have $|EG| = |EH|$ , hence the triangle $HGE$ is isosceles with the top angle at $E$ . From this we conclude that the lines $EF$ and $GH$ intersect at right angles.

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