, 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
$$AC \cdot BD = R(AC + BD) \geq 2R\sqrt{AC \cdot BD},$$
using AM-GM inequality. This implies that $AC \cdot BD \geq 4R^2$. But $AC$ and $BD$ are the chords of $\Gamma$, so that $AC \leq 2R$ and $BD \leq 2R$. We obtain $AC \cdot BD \leq 4R^2$. 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 $\Gamma$.