7. $2(2004 \mathrm{CMO})$ In the convex quadrilateral $E F G H$, the vertices $E, F, G, H$ are on the sides $A B, B C, C D, D A$ of the convex quadrilateral $A B C D$, respectively, and satisfy $\frac{A E}{E B} \cdot \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D H}{H A}=1$. The points $A, B, C, D$ are on the sides $H_{1} E_{1}, E_{1} F_{1}, F_{1} G_{1}, G_{1} H_{1}$ of the convex quadrilateral $E_{1} F_{1} G_{1} H_{1}$, respectively, and satisfy $E_{1} F_{1} \parallel E F, F_{1} G_{1} \parallel F G, G_{1} H_{1} \parallel G H, H_{1} E_{1} \parallel H E$. Given that $\frac{E_{1} A}{A H_{1}}=\lambda$, find the value of $\frac{F_{1} C}{C G_{1}}$.
7.2 (1) If $E F / / A C$, then $H G / / A C$. The value sought is still $\lambda$.
(2) If $E F$ is not parallel to $A C$, let $F E \cap C A=T$, apply Menelaus' theorem to $\triangle A B C$ and the transversal $T E F$. Combining with the given conditions, we have $\frac{O G}{G D} \cdot \frac{D H}{H A} \cdot \frac{A T}{T C}=1$, which implies that points $T$, $H$, and $G$ are collinear. Let $T F \cap E_{1} H_{1}=M, T G \cap E_{1} H_{1}=N$. From the parallel relationships given, we can deduce that $\frac{E_{1} A}{A H_{1}}=\frac{A M}{A N} \cdot \frac{A B}{A E} \cdot \frac{A H}{A D}$; also, $\frac{E Q}{Q H}=\frac{S_{A E C}}{S_{A H C}}=\frac{S_{A B C} \cdot A E \cdot A D}{S_{A C} \cdot A B \cdot A H}$. Therefore, we derive that $E_{1} A: A H_{1}=$ $S_{A B C}: S_{A D C}, F_{1} C: O G_{1}=S_{A B C}: S_{A D C}$. The ratio is $\lambda$.