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 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 $\frac{F_{1} C}{C G_{1}}$. (Xiong Bin)
(1) As shown in Figure 1, if
$E F / / A C$, then
$$
\frac{B E}{E A}=\frac{B F}{F C} .
$$

Substituting the known conditions, we get $\frac{D H}{H A}=\frac{D G}{G C}$, so,
$$
H G / / A C \text {. }
$$

Thus, $E_{1} F_{1} / / A C / / H_{1} G_{1}$. Therefore, $\frac{F_{1} C}{C G_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda$.
(2) As shown in Figure 2,

if $E F$ is not
parallel to $A C$. Let the
extension of $F E$ and the
extension of $C A$ intersect at point
$T$. By Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{C F}{F B} \cdot \frac{B E}{E A} \cdot \\
\frac{A T}{T C}=1 \text {. } \\
\end{array}
$$

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

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

Also, $\frac{E Q}{Q H}=\frac{S_{\triangle E C}}{S_{\triangle A H C}}=\frac{S_{\triangle A B C} \cdot A E \cdot A D}{S_{\triangle A D C} \cdot A B \cdot A H}$, hence
$$
\frac{E_{1} A}{A H_{1}}=\frac{E Q}{Q H} \cdot \frac{A B}{A E} \cdot \frac{A H}{A D}=\frac{S_{\triangle A B C}}{S_{\triangle A D C}} \text {. }
$$

Similarly, $\frac{F_{1} C}{C G_{1}}=\frac{S_{\triangle A B C}}{S_{\triangle C C}}$.
Therefore, $\frac{F_{1} C}{C G_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda$.