Consider a line segment $[AB]$ with midpoint $M$ and perpendicular bisector $m$. For each point$ X \ne M$ on m consider we are the intersection point $Y$ of the line $BX$ with the bisector from the angle $\angle BAX$. As $X$ approaches $M$, then approaches $Y$ to a point of $[AB]$. Which?
[img]https://cdn.artofproblemsolving.com/attachments/a/3/17d72a23011a9ec22deb20184717cc9c020a2b.png[/img]
[hide=original wording]Beschouw een lijnstuk [AB] met midden M en middelloodlijn m. Voor elk punt X 6= M op m beschouwenwe het snijpunt Y van de rechte BX met de bissectrice van de hoek < BAX . Als X tot M nadert, dan nadert
Y tot een punt van [AB]. Welk? [/hide]
1. **Identify the given elements and their properties:**
   - Line segment $[AB]$ with midpoint $M$.
   - Perpendicular bisector $m$ of $[AB]$.
   - Point $X \ne M$ on $m$.
   - Line $BX$ intersects the angle bisector of $\angle BAX$ at point $Y$.

2. **Use the Angle Bisector Theorem:**
   - The Angle Bisector Theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent sides.
   - For triangle $BAX$, the angle bisector of $\angle BAX$ intersects $BX$ at $Y$ such that:
     \[
     \frac{AY}{YB} = \frac{AX}{BX}
     \]

3. **Consider the limit as $X$ approaches $M$:**
   - As $X$ approaches $M$, $AX$ and $BX$ become nearly equal because $M$ is the midpoint of $[AB]$ and $X$ is on the perpendicular bisector $m$.
   - Therefore, $\frac{AX}{BX} \approx 1$ as $X$ approaches $M$.

4. **Apply the Angle Bisector Theorem in the limit:**
   - When $\frac{AX}{BX} = 1$, the angle bisector theorem implies that $Y$ divides $[AB]$ into two equal segments.
   - Hence, $Y$ must be the midpoint of $[AB]$.

5. **Conclusion:**
   - As $X$ approaches $M$, $Y$ approaches the midpoint of $[AB]$, which is $M$.

\[
\boxed{M}
\]