Question

Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$. Suppose $AC\cdot BD=EG\cdot FH$. Prove that $AC,BD,EG,FH$ are concurrent.

Answer 1

Let $R$ be the radius of the circle $\Gamma$
$\angle EDF=\frac{1}{2}\angle D$
$\therefore EF=2Rsin\frac{D}{2}$
Similarly, $HG=2Rsin\frac{B}{2}$
But $\angle B={180}^{\circ }-\angle D$
$\therefore HG=2Rcos\frac{D}{2}$
Hence we get
$EF\cdot GH=4R^{2}sin\frac{D}{2}cos\frac{D}{2}=2R^{2}sinD=R\cdot AC$.
Similarly, we have $EH\cdot FG=R\cdot BD$.
$\therefore R(AC+BD)=EF\cdot GH+EH\cdot FG=EG\cdot FH$......by Ptolemy's theorem
By the given hypothesis, $R(AC+BD)=AC\cdot BD$
$\therefore AC\cdot BD=R(AC+BD)\ge 2R\sqrt{AC\cdot BD}$........using $AM-GM$ inequality\
This implies that $AC\cdot BD\ge 4R^2$
But $ACandBD$ are the chords of $\Gamma$, so that $AC\le 2RandBD\le 2R$
We have $AC\cdot BD\le 4R^2$
It follows that $AC\cdot BD=4R^2$, implying that $AC=BD=2R$
$\therefore ACandBD$ are two diameters of $\Gamma$
Using $EG\cdot FH=AC\cdot BD$
Therefore we have $EGandFH$ are also two diameters of $\Gamma$
Hence $AC,BD,EGandFH$ all pass through the centre of $\Gamma$.