Example 10 (2004 China Mathematical Olympiad) In a 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 a convex quadrilateral $A B C D$, respectively, satisfying: $\frac{A E}{E B} \cdot \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D H}{H A}=1$; and 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 a convex quadrilateral $E_{1} F_{1} G_{1} H_{1}$, respectively, satisfying: $E_{1} F_{1} \parallel E F, F_{1} G_{1} \parallel F G, G_{1} H_{1} \parallel G H, H_{1} E_{1} \parallel H E$. Given that $\frac{E_{1} A}{A H_{1}}=\lambda$, find the value of $\frac{F_{1} C}{C G_{1}}$.
(1) If $E F / / A C$, then
$$
\frac{B E}{E A}=\frac{B F}{F C}
$$

Substituting the known conditions, we can get $\frac{D H}{H A}=\frac{D G}{G C}$.
Therefore, $H G / / A C$, so $E_{1} F_{1} / / A C / / G_{1} H_{1}$,
hence $\frac{F_{1} C}{C G_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda$
(2) If $E F$ is not parallel to $A C$, let the extension of $F E$ intersect the extension of $C A$ at point $T$, then by Menelaus' theorem we have
$$
\frac{C F}{F B} \cdot \frac{B E}{E A} \cdot \frac{A T}{T C}=1
$$

Combining with the given conditions, we have
$$
\frac{O G}{G D} \cdot \frac{D H}{H A} \cdot \frac{A T}{T C}=1
$$

By the converse of Menelaus' theorem: $T, H, G$ are collinear. Let $T F, T G$ intersect $E_{1} H_{1}$ at points $M, N$ respectively.
By $E_{1} B / / E F$, we get $E_{1} A=\frac{B A}{E A} \cdot A M$.
Similarly, $H_{1} A=\frac{A D}{A H} \cdot A N$, so
$$
\frac{E_{1} A}{H_{1} A}=\frac{A M}{A N} \cdot \frac{A B}{A E} \cdot \frac{A H}{A D}
$$

Also,
$$
\frac{E Q}{Q H}=\frac{S_{\triangle A E C}}{S_{\triangle A H C}}=\frac{S_{\triangle A B C} \cdot A E \cdot A D}{S_{\triangle A D C} \cdot A B \cdot A H}
$$

By $M N / / E H$, we get $\frac{A M}{A N}=\frac{E Q}{Q H}$, combining (1) and (2) we get
$$
\frac{E_{1} A}{H_{1} A}=\frac{E Q}{Q H} \cdot \frac{A B}{A E} \cdot \frac{A H}{A D}=\frac{S_{\triangle A B C}}{S_{\triangle A D C}}
$$

Similarly, we can get $\quad \frac{F_{1} C}{C G_{1}}=\frac{S_{\triangle A B C}}{S_{\triangle A D C}}$
Therefore, $\frac{F_{1} C}{C G_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda$
Figure 21-9