Let $EFGH,ABCD$ and $E_1F_1G_1H_1$ be three convex quadrilaterals satisfying:

i) The points $E,F,G$ and $H$ lie on the sides $AB,BC,CD$ and $DA$ respectively, and $\frac{AE}{EB}\cdot\frac{BF}{FC}\cdot \frac{CG}{GD}\cdot \frac{DH}{HA}=1$;
ii) The points $A,B,C$ and $D$ lie on sides $H_1E_1,E_1F_1,F_1,G_1$ and $G_1H_1$ respectively, and $E_1F_1||EF,F_1G_1||FG,G_1H_1||GH,H_1E_1||HE$.

Suppose that $\frac{E_1A}{AH_1}=\lambda$. Find an expression for $\frac{F_1C}{CG_1}$ in terms of $\lambda$.

[i]Xiong Bin[/i]
1. **Given Conditions and Setup:**
   - We have three convex quadrilaterals \(EFGH\), \(ABCD\), and \(E_1F_1G_1H_1\).
   - Points \(E, F, G, H\) lie on the sides \(AB, BC, CD, DA\) respectively, and satisfy the condition:
     \[
     \frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CG}{GD} \cdot \frac{DH}{HA} = 1
     \]
   - Points \(A, B, C, D\) lie on the sides \(H_1E_1, E_1F_1, F_1G_1, G_1H_1\) respectively, with parallel conditions:
     \[
     E_1F_1 \parallel EF, \quad F_1G_1 \parallel FG, \quad G_1H_1 \parallel GH, \quad H_1E_1 \parallel HE
     \]
   - We need to find an expression for \(\frac{F_1C}{CG_1}\) in terms of \(\lambda\), where \(\frac{E_1A}{AH_1} = \lambda\).

2. **Using Menelaus' Theorem:**
   - Consider \(\triangle ABC\) with transversal \(EF\):
     \[
     \frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CP}{PA} = 1
     \]
   - Similarly, for \(\triangle ACD\) with transversal \(GH\):
     \[
     \frac{CG}{GD} \cdot \frac{DH}{HA} \cdot \frac{AP}{PC} = 1
     \]
   - Given \(\frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CG}{GD} \cdot \frac{DH}{HA} = 1\), it follows that \(P, H, G\) are collinear, implying \(AC, EF, GH\) concur at \(P\).

3. **Parallelism and Similarity:**
   - Since \(E_1F_1 \parallel EF\), \(F_1G_1 \parallel FG\), \(G_1H_1 \parallel GH\), and \(H_1E_1 \parallel HE\), the quadrilaterals \(E_1F_1G_1H_1\) and \(EFGH\) are homothetic.
   - Let \(O\) be the center of homothety. Then, the ratios of corresponding segments are equal.

4. **Homothety and Ratios:**
   - Given \(\frac{E_1A}{AH_1} = \lambda\), the homothety ratio is \(\lambda\).
   - By the properties of homothety, the ratio of corresponding segments in \(E_1F_1G_1H_1\) and \(EFGH\) is \(\lambda\).

5. **Finding \(\frac{F_1C}{CG_1}\):**
   - Since \(F_1G_1 \parallel FG\) and the homothety ratio is \(\lambda\), we have:
     \[
     \frac{F_1C}{CG_1} = \lambda
     \]

The final answer is \(\boxed{\lambda}\).