Let $ABC$ be a triangle. For every point $M$ belonging to segment $BC$ we denote by $B'$ and $C'$ the orthogonal projections of $M$ on the straight lines $AC$ and $BC$. Find points $M$ for which the length of segment $B'C'$ is a minimum.
1. **Identify the given elements and the goal:**
   - We have a triangle \(ABC\).
   - \(M\) is a point on segment \(BC\).
   - \(B'\) and \(C'\) are the orthogonal projections of \(M\) on lines \(AC\) and \(AB\), respectively.
   - We need to find the position of \(M\) on \(BC\) such that the length of segment \(B'C'\) is minimized.

2. **Use the cyclic quadrilateral property:**
   - Since \(AB'MC'\) is cyclic, we can use the properties of cyclic quadrilaterals.
   - In a cyclic quadrilateral, the opposite angles sum to \(180^\circ\).

3. **Apply the Law of Sines in the cyclic quadrilateral:**
   - By the Law of Sines in \(\triangle AB'M\) and \(\triangle AC'M\), we have:
     \[
     \frac{B'C'}{AM} = \frac{\sin \angle C'AB'}{\sin \angle AB'M}
     \]
   - Since \(\angle C'AB' = \angle BAC\) (both are angles subtended by the same arc \(B'C'\) in the cyclic quadrilateral), we get:
     \[
     \frac{B'C'}{AM} = \sin \angle BAC
     \]
   - This implies that \(B'C'\) is proportional to \(AM\) with a constant of proportionality \(\sin \angle BAC\).

4. **Minimize \(AM\):**
   - To minimize \(B'C'\), we need to minimize \(AM\) because \(\sin \angle BAC\) is a constant.
   - The minimum distance from a point to a line is the perpendicular distance. Therefore, \(AM\) is minimized when \(M\) is the foot of the perpendicular from \(A\) to \(BC\).

5. **Conclusion:**
   - The point \(M\) that minimizes \(B'C'\) is the foot of the perpendicular from \(A\) to \(BC\).

\(\blacksquare\)

The final answer is the point \( \boxed{ M } \) is the foot of the perpendicular from \(A\) to \(BC\).