Problem 1538

1. Identify Key Points and Properties:
- Let $ABCD$ be a quadrilateral.
- Let $P$ be the intersection of $AB$ and $CD$.
- Let $O$ be the intersection of the perpendicular bisectors of $AB$ and $CD$.
- $O$ does not lie on line $AB$ and $O$ does not lie on line $CD$.
- Let $B'$ and $D'$ be the reflections of $B$ and $D$ across $OP$.
- $AB'$ and $CD'$ meet on $OP$.

2. Define Intersection Point:
- Let $X = AB' \cap CD' \cap OP$.

3. **Properties of Point $O$:**
- Since $O$ lies on the perpendicular bisectors of $AB$ and $CD$, it is equidistant from $A$ and $B$, and from $C$ and $D$.
- $O$ is the midpoint of the arc $AB$ in the circumcircle of $\triangle APB'$.

4. Inversion and Symmetry:
- By inversion in $P$ with radius $\sqrt{PA \cdot PB'}$ and symmetry across $OP$, $X$ and $O$ are swapped.
- This implies that $PX \cdot PO = PA \cdot PB'$.

5. Equality of Products:
- Since $B'$ is the reflection of $B$ across $OP$, $PB' = PB$.
- Therefore, $PX \cdot PO = PA \cdot PB$.

6. Cyclic Quadrilateral Condition:
- To show that $ABCD$ is cyclic, we need to show that $PA \cdot PB = PC \cdot PD$.
- From the previous step, we have $PX \cdot PO = PA \cdot PB$.
- Similarly, by the same inversion and symmetry argument, $PX \cdot PO = PC \cdot PD$.

7. Conclusion:
- Since $PA \cdot PB = PC \cdot PD$, it follows that $ABCD$ is cyclic.

$\blacksquare$