Two different points $X$ and $Y$ are chosen in the plane. Find all the points $Z$ in this plane for which the triangle $XYZ$ is isosceles.
To find all points \( Z \) in the plane such that the triangle \( XYZ \) is isosceles, we need to consider the geometric properties of isosceles triangles. An isosceles triangle has at least two sides of equal length. Therefore, \( Z \) must satisfy one of the following conditions:
1. \( |XZ| = |YZ| \)
2. \( |XY| = |XZ| \)
3. \( |XY| = |YZ| \)

However, since \( X \) and \( Y \) are fixed points, we focus on the condition \( |XZ| = |YZ| \), which implies that \( Z \) lies on the perpendicular bisector of the line segment \( \overline{XY} \).

1. **Perpendicular Bisector Condition:**
   - The perpendicular bisector of a line segment is the locus of all points equidistant from the endpoints of the segment. Therefore, any point \( Z \) on the perpendicular bisector of \( \overline{XY} \) will satisfy \( |XZ| = |YZ| \), making \( \triangle XYZ \) isosceles.

2. **Circle Condition:**
   - Additionally, \( Z \) can lie on the circles centered at \( X \) and \( Y \) with radius \( |XY| \). These circles are defined as:
     - Circle centered at \( X \) with radius \( |XY| \)
     - Circle centered at \( Y \) with radius \( |XY| \)
   - Any point \( Z \) on these circles will satisfy \( |XZ| = |XY| \) or \( |YZ| = |XY| \), respectively, making \( \triangle XYZ \) isosceles.

3. **Exclusion of Specific Points:**
   - The problem specifies that certain points must be excluded:
     1. The midpoint of \( \overline{XY} \): This point is equidistant from \( X \) and \( Y \) but does not form a valid triangle.
     2. The extreme endpoints of the line segments \( \overline{XY} \) and \( \overline{YX} \) extended by their own lengths: These points lie on the line through \( X \) and \( Y \) and do not form a valid triangle.
     3. The points \( X \) and \( Y \) themselves: These points do not form a valid triangle as they coincide with the vertices of the triangle.

Therefore, the points \( Z \) that make \( \triangle XYZ \) isosceles are all points on the perpendicular bisector of \( \overline{XY} \) and on the circles centered at \( X \) and \( Y \) with radius \( |XY| \), excluding the specified points.

The final answer is the set of points \( \boxed{ Z } \) on the perpendicular bisector of \( \overline{XY} \) and on the circles centered at \( X \) and \( Y \) with radius \( |XY| \), excluding the midpoint of \( \overline{XY} \), the extreme endpoints of the line segments \( \overline{XY} \) and \( \overline{YX} \) extended by their own lengths, and the points \( X \) and \( Y \).