The quadrilateral $ABCD$ has $AB$ parallel to $CD$. $P$ is on the side $AB$ and $Q$ on the side $CD$ such that $\frac{AP}{PB}= \frac{DQ}{CQ}$. M is the intersection of $AQ$ and $DP$, and $N$ is the intersection of $PC$ and $QB$. Find $MN$ in terms of $AB$ and $CD$.
1. Given that $ABCD$ is a quadrilateral with $AB \parallel CD$, and points $P$ and $Q$ are on sides $AB$ and $CD$ respectively such that $\frac{AP}{PB} = \frac{DQ}{CQ}$.
2. Let $M$ be the intersection of $AQ$ and $DP$, and $N$ be the intersection of $PC$ and $QB$.
3. We need to find the length of $MN$ in terms of $AB$ and $CD$.

First, we note that $\triangle AMP \sim \triangle QMD$ and $\triangle BNP \sim \triangle QNC$ by the AA (Angle-Angle) similarity criterion. This is because:
- $\angle AMP = \angle QMD$ (corresponding angles since $AB \parallel CD$)
- $\angle MAP = \angle QDM$ (corresponding angles since $AB \parallel CD$)

Similarly, for $\triangle BNP \sim \triangle QNC$:
- $\angle BNP = \angle QNC$ (corresponding angles since $AB \parallel CD$)
- $\angle BPN = \angle QCN$ (corresponding angles since $AB \parallel CD$)

4. From the similarity of triangles, we have:
\[
\frac{PM}{MD} = \frac{AP}{DQ} = \frac{PB}{QC} = \frac{PN}{NC}
\]

5. This implies that $\overleftrightarrow{MN} \parallel \overleftrightarrow{CD}$ because the ratios of the corresponding segments are equal.

6. To find the length of $MN$, we use the fact that $MN$ is a segment parallel to $CD$ and divides the trapezoid $ABCD$ proportionally. Therefore:
\[
MN = \frac{PM}{PD} \cdot CD
\]

7. Since $\frac{PM}{MD} = \frac{AP}{DQ}$, we can write:
\[
\frac{PM}{PD} = \frac{AP}{AP + DQ}
\]

8. Given $\frac{AP}{PB} = \frac{DQ}{CQ}$, let $AP = k \cdot PB$ and $DQ = k \cdot CQ$ for some constant $k$. Then:
\[
AP = k \cdot PB \quad \text{and} \quad DQ = k \cdot CQ
\]

9. Since $AB = AP + PB$ and $CD = DQ + CQ$, we have:
\[
AP = \frac{k}{1+k} \cdot AB \quad \text{and} \quad DQ = \frac{k}{1+k} \cdot CD
\]

10. Substituting these into the expression for $MN$:
\[
MN = \frac{AP}{AP + DQ} \cdot CD = \frac{\frac{k}{1+k} \cdot AB}{\frac{k}{1+k} \cdot AB + \frac{k}{1+k} \cdot CD} \cdot CD
\]

11. Simplifying the expression:
\[
MN = \frac{\frac{k}{1+k} \cdot AB}{\frac{k}{1+k} \cdot (AB + CD)} \cdot CD = \frac{AB}{AB + CD} \cdot CD
\]

12. Therefore, the length of $MN$ is:
\[
MN = \frac{AB \cdot CD}{AB + CD}
\]

The final answer is $\boxed{\frac{AB \cdot CD}{AB + CD}}$