Points $E$, $F$, $G$, and $H$ lie on sides $AB$, $BC$, $CD$, and $DA$ of a convex quadrilateral $ABCD$ such that
$\frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CG}{GD} \cdot \frac{DH}{HA}=1$. 

Points $A$, $B$, $C$, and $D$ lie on sides $H_1E_1$, $E_1F_1$, $F_1G_1$, and $G_1H_1$ of a convex quadrilateral $E_1F_1G_1H_1$ such that $E_1F_1 \parallel EF$, $F_1G_1 \parallel FG$, $G_1H_1 \parallel GH$, and $H_1E_1 \parallel HE$. Given that $\frac{E_1A}{AH_1}=a$, express $\frac{F_1C}{CG_1}$ in terms of $a$. 
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To solve the problem, we need to express $\frac{F_1C}{CG_1}$ in terms of $a$, given that $\frac{E_1A}{AH_1} = a$. We will use the given conditions and properties of similar triangles and parallel lines.

1. **Given Condition:**
   \[
   \frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CG}{GD} \cdot \frac{DH}{HA} = 1
   \]
   This condition implies that the points $E, F, G, H$ divide the sides of quadrilateral $ABCD$ in such a way that the product of the ratios of the segments they create is 1.

2. **Parallel Lines and Similar Triangles:**
   Since $E_1F_1 \parallel EF$, $F_1G_1 \parallel FG$, $G_1H_1 \parallel GH$, and $H_1E_1 \parallel HE$, the quadrilateral $E_1F_1G_1H_1$ is similar to quadrilateral $EFGH$.

3. **Using the Given Ratio:**
   Given $\frac{E_1A}{AH_1} = a$, we need to find $\frac{F_1C}{CG_1}$ in terms of $a$.

4. **Concurrent Lines:**
   Since $\frac{AE}{EB} \cdot \frac{BF}{FC} = \frac{AH}{HD} \cdot \frac{DG}{GC}$, the lines $FE$, $CA$, and $GH$ are concurrent at a point $Y$.

5. **Intersection and Parallel Lines:**
   Let $AC$ intersect $BD$ at $X$. Draw the line through $E_1$ parallel to $DH_1$ which intersects $DA$ at $Z$. 

6. **Using Sine Rule in Triangles:**
   By the sine rule in triangles, we have:
   \[
   \frac{\sin \angle EYA}{\sin \angle AYH} \cdot \frac{\sin \angle AYH}{\sin \angle AHY} \cdot \frac{\sin \angle AHE}{\sin \angle AEH} = 1
   \]
   This simplifies to:
   \[
   \frac{\sin \angle EYA}{\sin \angle AEH} = 1
   \]

7. **Angles and Parallel Lines:**
   Note that $\angle E_1BA = \angle YEA$, $\angle E_1AB = \angle AEH$, $\angle E_1AZ = \angle AHE$, and $\angle E_1ZA = \angle AHY$. Thus, $\angle E_1BZ = \angle AYE$ and $\angle E_1ZB = \angle AYH$, so $BZ \parallel AC$.

8. **Using Ratios:**
   Therefore, we have:
   \[
   \frac{E_1A}{AH_1} = \frac{ZA}{AD} = \frac{BX}{XD}
   \]
   Similarly, for the other side:
   \[
   \frac{F_1C}{CG_1} = \frac{BX}{XD}
   \]

9. **Conclusion:**
   Since $\frac{E_1A}{AH_1} = a$, it follows that:
   \[
   \frac{F_1C}{CG_1} = a
   \]

The final answer is $\boxed{a}$