On the side $ AB$ of a cyclic quadrilateral $ ABCD$ there is a point $ X$ such that diagonal $ BD$ bisects $ CX$ and diagonal $ AC$ bisects $ DX$. What is the minimum possible value of $ AB\over CD$?

[i]Proposed by S. Berlov[/i]
1. **Define the problem and setup the geometry:**
   - We are given a cyclic quadrilateral \(ABCD\) with a point \(X\) on side \(AB\).
   - Diagonal \(BD\) bisects \(CX\) and diagonal \(AC\) bisects \(DX\).
   - We need to find the minimum possible value of \(\frac{AB}{CD}\).

2. **Introduce parallel lines and intersections:**
   - Draw a line through \(X\) parallel to \(CD\) and let it intersect \(AC\) at \(C_0\) and \(BD\) at \(D_0\).
   - Let \(X_0\) be the intersection of \(CD_0\) and \(DC_0\).

3. **Analyze the triangles and medians:**
   - Since \(CA\) and \(DB\) are medians of \(\triangle XCD\), it follows that \(\triangle X_0C_0D_0\) is anticomplementary to \(\triangle XCD\).
   - This implies that \(X\) is the midpoint of \(D_0C_0\) and \(D_0X = XC_0 = CD\).

4. **Establish angle relationships:**
   - \(\angle AC_0X = \angle CC_0D_0 = \angle DCC_0 = \angle DCA = \angle DBA = \angle D_0BX \pmod{\pi}\)
   - \(\angle BD_0X = \angle BD_0C_0 = \angle CDD_0 = \angle CDB = \angle CAB = \angle C_0AX \pmod{\pi}\)
   - \(\angle C_0XA = \angle D_0XB\)

5. **Prove similarity and cyclic nature:**
   - \(\triangle XAC_0 \sim \triangle XD_0B\) are oppositely similar, implying that \(AC_0BD_0\) is cyclic.
   - Therefore, \(AX \cdot XB = D_0X \cdot XC_0 = CD^2\).

6. **Apply the AM-GM inequality:**
   - By the Arithmetic Mean-Geometric Mean Inequality (AM-GM), we have:
     \[
     AB = AX + XB \geq 2 \sqrt{AX \cdot XB} = 2 \sqrt{CD^2} = 2CD
     \]
   - Thus, \(\frac{AB}{CD} \geq 2\).

7. **Conclusion:**
   - The minimum possible value of \(\frac{AB}{CD}\) is \(2\).

The final answer is \(\boxed{2}\).