Let $ABCD$ be a square with side length $1$. 
Find the locus of all points $P$ with the property $AP\cdot CP + BP\cdot DP = 1$.
1. Let \(ABCD\) be a square with side length \(1\). We need to find the locus of all points \(P\) such that \(AP \cdot CP + BP \cdot DP = 1\).

2. Consider the point \(Q\) such that \(ABQP\) and \(CDPQ\) are parallelograms. By the properties of parallelograms, we know that \(BQ = AP\) and \(CQ = DP\).

3. Using Ptolemy's inequality for the quadrilateral \(BQCP\), we have:
   \[
   AP \cdot CP + BP \cdot DP = BQ \cdot CP + BP \cdot CQ \geq BC \cdot PQ
   \]
   Since \(BC = 1\), we get:
   \[
   AP \cdot CP + BP \cdot DP \geq PQ
   \]

4. Given that \(AP \cdot CP + BP \cdot DP = 1\), equality must hold in Ptolemy's inequality. This implies that \(B, Q, C, P\) are concyclic, meaning they lie on the same circle.

5. For \(B, Q, C, P\) to be concyclic, \(P\) must be inside the square \(ABCD\). Additionally, the equality condition in Ptolemy's theorem implies that:
   \[
   PX \cdot QX = BX \cdot CX
   \]
   where \(X = PQ \cap BC\).

6. Let \(Y\) be the foot of the perpendicular from \(P\) to \(AB\). Then, we have:
   \[
   BX \cdot CX = AY \cdot BY
   \]
   Since \(BX + CX = AY + BY = 1\), this implies that either \(AY = BX\) or \(BY = BX\).

7. The first equation \(AY = BX\) holds if and only if \(P\) lies on the diagonal \(AC\), and the second equation \(BY = BX\) holds if and only if \(P\) lies on the diagonal \(BD\).

8. Therefore, the locus of all points \(P\) satisfying the given condition is the union of the diagonals \(AC\) and \(BD\).

The final answer is the union of the diagonals \( \boxed{ AC } \) and \(BD\).