Find the locus of the points $ M $ that are situated on the plane where a rhombus $ ABCD $ lies, and satisfy:
$$ MA\cdot MC+MB\cdot MD=AB^2 $$


[i]Ovidiu Pop[/i]
1. **Define the coordinates of the vertices of the rhombus and the point \( M \):**
   Let \( A(0, b) \), \( B(a, 0) \), \( C(0, -b) \), and \( D(-a, 0) \) be the coordinates of the vertices of the rhombus \( ABCD \) with \( a, b > 0 \). Let \( M(x, y) \) be a point on the plane of the rhombus.

2. **Express the given condition in terms of coordinates:**
   The given condition is:
   \[
   MA \cdot MC + MB \cdot MD = AB^2
   \]
   Using the distance formula, we have:
   \[
   MA = \sqrt{x^2 + (y - b)^2}, \quad MB = \sqrt{(x - a)^2 + y^2}
   \]
   \[
   MC = \sqrt{x^2 + (y + b)^2}, \quad MD = \sqrt{(x + a)^2 + y^2}
   \]
   The side length of the rhombus \( AB \) is given by:
   \[
   AB = \sqrt{a^2 + b^2}
   \]
   Therefore, \( AB^2 = a^2 + b^2 \).

3. **Rewrite the given condition using the distance expressions:**
   \[
   \sqrt{x^2 + (y - b)^2} \cdot \sqrt{x^2 + (y + b)^2} + \sqrt{(x - a)^2 + y^2} \cdot \sqrt{(x + a)^2 + y^2} = a^2 + b^2
   \]

4. **Simplify the first product:**
   \[
   \sqrt{x^2 + (y - b)^2} \cdot \sqrt{x^2 + (y + b)^2} = \sqrt{(x^2 + y^2 - 2by + b^2)(x^2 + y^2 + 2by + b^2)}
   \]
   \[
   = \sqrt{(x^2 + y^2 + b^2)^2 - (2by)^2} = \sqrt{(x^2 + y^2 + b^2)^2 - 4b^2y^2}
   \]
   \[
   = \sqrt{(x^2 + y^2 + b^2)^2 - 4b^2y^2} = x^2 + y^2 + b^2
   \]

5. **Simplify the second product:**
   \[
   \sqrt{(x - a)^2 + y^2} \cdot \sqrt{(x + a)^2 + y^2} = \sqrt{((x - a)^2 + y^2)((x + a)^2 + y^2)}
   \]
   \[
   = \sqrt{(x^2 - 2ax + a^2 + y^2)(x^2 + 2ax + a^2 + y^2)}
   \]
   \[
   = \sqrt{(x^2 + y^2 + a^2)^2 - (2ax)^2} = \sqrt{(x^2 + y^2 + a^2)^2 - 4a^2x^2}
   \]
   \[
   = x^2 + y^2 + a^2
   \]

6. **Combine the simplified expressions:**
   \[
   (x^2 + y^2 + b^2) + (x^2 + y^2 + a^2) = a^2 + b^2
   \]
   \[
   2(x^2 + y^2) + a^2 + b^2 = a^2 + b^2
   \]
   \[
   2(x^2 + y^2) = 0
   \]
   \[
   x^2 + y^2 = 0
   \]

7. **Solve for \( x \) and \( y \):**
   Since \( x^2 + y^2 = 0 \), we have \( x = 0 \) and \( y = 0 \).

8. **Conclusion:**
   The only point that satisfies the given condition is the origin, which is the intersection of the diagonals of the rhombus.

The final answer is \( \boxed{ (0, 0) } \).