Give a construction by straight-edge and compass of a point $C$ on a line $\ell$ parallel to a segment $AB$, such that the product $AC \cdot BC$ is minimum.
1. **Identify the Problem**: We need to find a point \( C \) on a line \( \ell \) such that \( \ell \) is parallel to a segment \( AB \) and the product \( AC \cdot BC \) is minimized.

2. **Geometric Insight**: The area of triangle \( \triangle ABC \) is constant for a given base \( AB \) and height from \( C \) to \( AB \). To minimize \( AC \cdot BC \), the angle \( \angle ACB \) must be a right angle. This is because the product of the segments of a chord through a point on a circle is minimized when the chord is a diameter.

3. **Construct the Circle**: Construct the circle with diameter \( AB \). This circle will have all points \( C \) such that \( \angle ACB = 90^\circ \).

4. **Construct Perpendicular Bisector**: Construct the perpendicular bisector of \( AB \). This can be done by:
   - Drawing two congruent circles with centers at \( A \) and \( B \) and radius greater than half of \( AB \).
   - The intersection points of these circles will be equidistant from \( A \) and \( B \). Connect these points to form the perpendicular bisector of \( AB \).

5. **Construct Parallel Line**: Construct a line parallel to \( AB \) through the intersection point of the circles. This line will be the line \( \ell \).

6. **Intersection Point**: The intersection of the perpendicular bisector of \( AB \) with the line \( \ell \) will give the midpoint of \( AB \). This point is equidistant from \( A \) and \( B \).

7. **Final Construction**: The point \( C \) on the line \( \ell \) such that \( AC \cdot BC \) is minimized will be the point where the perpendicular from the midpoint of \( AB \) intersects \( \ell \).

Conclusion:
\[
\boxed{\text{The point } C \text{ is the intersection of the perpendicular bisector of } AB \text{ with the line } \ell.}
\]