1 Convex quadrilateral $E F G H$ has vertices $E, F, G, H$ on the sides $A B, B C, C D, D A$ of 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 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 convex quadrilateral $E_{1} F_{1} G_{1} H_{1}$, respectively, such that $E_{1} F_{1} \parallel E F, F_{1} G_{1} \parallel F G, G_{1} H_{1} \parallel G H, H_{1} E_{1} \parallel H E$. Given $\frac{E_{1} A}{A H_{1}}=\lambda$, find the value of $\frac{F_{1} C}{C G_{1}}$.
(1) If $E F / / A C$, as shown in Figure 1, then
$$
\frac{B E}{E A}=\frac{B F}{F C}
$$

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

Combining the given conditions, by the converse of Menelaus' theorem, points $T, H, G$ 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

Also
$$
\frac{E_{1} A}{H_{1} A}=\frac{A M}{A N} \cdot \frac{A B}{A E} \cdot \frac{A H}{A D}
$$

From (1) and (2), we get $\frac{E_{1} A}{A H_{1}}=\frac{E Q}{Q H} \cdot \frac{A B}{A E} \cdot \frac{A H}{A D}=\frac{\triangle A B C}{\triangle A D C}$
Similarly, we can get
$$
\frac{F_{1} C}{C G_{1}}=\frac{\triangle A B C}{\triangle A D C}
$$

Therefore
$$
\frac{F_{1} C}{C G_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda
$$