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 square $A B C D$, respectively. It satisfies: $\frac{A E}{E B} \cdot \frac{B F}{F C} \cdot \frac{C G}{G D} \cdot \frac{D H}{H A}=1$; and points $A, B, C, D$ are on the sides of square $E_{1} F_{1} G_{1} H_{1}$, with $E_{1} H_{1} / / H H_{1}$ and $H_{1} E_{1} / / H E$. Given $\frac{E_{1} A}{A H_{1}}=\lambda$, find the value of $\frac{F_{1} C}{C G_{1}}$. (Xiong Bin)
(1) If $E F / / A C$, then
$$
\frac{B E}{E A}=\frac{B F}{F C} .
$$

Substituting the known conditions, we can get $\frac{D H}{H} \frac{H}{A}=\frac{I(G}{C C}$, so $H\left(; / / A C\right.$. Therefore, $E_{1} F_{1} / / A C / / H_{1} G_{1}$, hence
$$
\frac{F_{1} C}{C C_{1}}-\frac{E_{1} A}{A H_{1}}=\lambda .
$$
(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
$$
\frac{C F}{F B} \cdot \frac{B E}{E A} \cdot \frac{A T}{T C}=1
$$

According to the proposition, we have
$$
\frac{C(G)}{(I I)} \cdot \frac{D H}{H A} \cdot \frac{A T}{T C}=1 .
$$

By the converse of Menelaus' theorem, points $T$, $H$, and $G$ are collinear. Let $TF$, $T(;$ intersect $E_{1} H_{1}$ at points $M$ and $N$ respectively. Since $E_{1} B / / E F$, we get $E_{1} A=\frac{B A}{E A} \cdot A M \cdot[$ similarly, $H_{1} A=\frac{A I)}{A H} \cdot A N$, so
$$
\frac{E_{1} A}{H_{1} A}=\frac{A M}{A N} \cdot \frac{A B}{A E} \cdot \frac{A H}{A I} .
$$
$\times \quad \frac{E Q}{Q H}=\frac{\triangle A E C}{\triangle A H C}=\frac{\triangle A B C}{\triangle A I C} \cdot \frac{A E}{A B} \cdot \frac{A D}{A H}$.
From (1) and (3), 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 get
$$
\frac{F_{1} C}{C_{C}}=\frac{\triangle A B C}{\triangle A D C} \text {. }
$$

Therefore,
$$
\frac{F_{1} C}{C C_{1}}=\frac{E_{1} A}{A H_{1}}=\lambda .
$$