, 2010

Let $AB$ and $DC$ meet in $M$; let $FE$ meet $HG$ in $T$, $AB$ in $P$ and $DC$ in $Q$ respectively. We use the following fact: in a triangle $ABC$ with circumcircle $\gamma$, given a point $D$ outside $\gamma$, $AD$ is tangent to $\gamma$ if and only if

$$\frac{DB}{DC} = \frac{AB^2}{AC^2}.$$
Thus it suffices to prove that
$$\frac{TG}{TH} = \frac{GE^2}{EH^2}.$$
Let us put $AB = a$, $DC = b$, $GM = \lambda$, $HM = \mu$. Observe that $DB$, $CA$ and $FQ$ are concurrent cevians in the triangle $FDC$. Ceva's theorem gives
$$\frac{FA}{AD} \cdot \frac{DQ}{QC} \cdot \frac{CB}{BF} = 1.$$
Looking at the transversal $A-B-M$ in the triangle $FDC$, we can apply Menelaus' theorem to get
$$\frac{FA}{AD} \cdot \frac{DM}{MC} \cdot \frac{CB}{BF} = 1.$$
It follows that $\frac{DQ}{QC} = \frac{DM}{MC}$.
Similarly, we get $\frac{AP}{PB} = \frac{AM}{MB}$. Observe
$$\frac{DH+HQ}{CH-HQ} = \frac{DQ}{QC} = \frac{DM}{MC} = \frac{DH+HM}{HM-HC}.$$
Simplification gives (using $DH = CH$) $HC^2 = HQ \cdot HM$. In similar way, we get $GP \cdot GM = GB^2$. Thus
$$\begin{aligned} GP &= \frac{GB^2}{GM} = \frac{a^2}{4\lambda}, \\ PM &= \lambda - \frac{a^2}{4\lambda} = \frac{4\lambda^2 - a^2}{4\lambda}. \end{aligned}$$
Thus $\frac{PM}{GP} = \frac{4\lambda^2 - a^2}{a^2}$. Similarly, $\frac{QM}{HQ} = \frac{4\mu^2 - b^2}{b^2}$. Using the transversal $T-P-Q$ in the triangle $MGH$, we have $\frac{MP}{PG} \cdot \frac{GT}{TH} \cdot \frac{HQ}{QM} = 1$. Thus
$$\frac{GT}{TH} = \frac{PG}{MP} \cdot \frac{QM}{HQ} = \frac{a^2}{4\lambda^2 - a^2} \cdot \frac{4\mu^2 - b^2}{b^2}.$$
Suppose we show that $4\lambda^2 - a^2 = 4\mu^2 - b^2$. We get $GT/TH = a^2/b^2$. We observe that triangles $EBA$ and $ECD$ are similar; and they have respective medians $EG$ and $EH$. Hence $EG^2/EH^2 = AB^2/CD^2 = a^2/b^2$. It follows that $GT/TH = GE^2/EH^2$, giving what we required. We use coordinate geometry to prove $4\lambda^2 - a^2 = 4\mu^2 - b^2$.
Let us fix $M = (0,0)$. The circumcircle of $ABCD$ has equation $x^2 + y^2 + 2gx + 2fy + c = 0$. Let the equation of $MA$ be $y = mx$. Let $A = (x_1, y_1)$, $B = (x_2, y_2)$. Then $x_1, x_2$ are the solutions of $(1+m^2)x^2 + (2g+2fm)x + c = 0$, so that
$$x_1 + x_2 = -\frac{2g + 2fm}{1 + m^2}, \quad x_1x_2 = \frac{c}{1 + m^2}.$$
The coordinates of $G$ are $\left(\frac{x_1+x_2}{2}, m\left(\frac{x_1+x_2}{2}\right)\right)$. Thus
$$\lambda^2 = GM^2 = \left(\frac{x_1+x_2}{2}\right)^2 + m^2 \left(\frac{x_1+x_2}{2}\right)^2 = \frac{(g+fm)^2}{1+m^2}.$$
We also have
$$a^2 = AB^2 = (1+m^2)(x_1-x_2)^2 = (1+m^2)((x_1+x_2)^2 - 4x_1x_2) = 4\frac{(g+fm)^2}{1+m^2} - 4c.$$

Thus
$$4\lambda^2 - a^2 = 4c$$
Similarly, we get $4\mu^2 - b^2 = 4c$. It follows that $4\lambda^2 - a^2 = 4\mu^2 - b^2$.