Indija mo 2011

Let $R$ be the radius of the circle $\Gamma$. Observe that $\angle EDF=\frac{1}{2}\angle D$. Hence $EF=2R\sin \frac{D}{2}$. Similarly, $HG=2R\sin \frac{B}{2}$. But $\angle B=180^{\circ }-\angle D$. Thus $HG=2R\cos \frac{D}{2}$. We hence get $$EF\cdot GH=4R^2\sin \frac{D}{2}\cos \frac{D}{2}=2R^2\sin D=R\cdot AC.$$

Similarly, we obtain $EH\cdot FG=R\cdot BD$.

Therefore $$R(AC+BD)=EF\cdot GH+EH\cdot FG=EG\cdot FH,$$ by Ptolemy's theorem. By the given hypothesis, this gives $R(AC+BD)=AC\cdot BD$. Thus $$AC\cdot BD=R(AC+BD)\ge 2R\sqrt{AC\cdot BD},$$ using AM-GM inequality. This implies that $AC\cdot BD\ge 4R^2$. But $AC$ and $BD$ are the chords of $\Gamma$, so that $AC\le 2R$ and $BD\le 2R$. We obtain $AC\cdot BD\le 4R^2$. It follows that $AC\cdot BD=4R^2$, implying that $AC=BD=2R$. Thus $AC$ and $BD$ are two diameters of $\Gamma$. Using $EG\cdot FH=AC\cdot BD$, we conclude that $EG$ and $FH$ are also two diameters of $\Gamma$. Hence $AC$, $BD$, $EG$ and $FH$ all pass through the centre of $\Gamma$.