In 3-D space, for each plane containing there is a distinct perpendicular bisector in that plane. The set of lines which are perpendicular bisectors of form a plane which is the plane perpendicularly bisecting .
The perpendicular bisector of is also the locus of points equidistant from and .
To prove this, we must prove that every point on the perpendicular bisector is equidistant from and , and also that every point equidistant from and .
The first part we prove as follows: Let be a point on the perpendicular bisector of , and let be the midpoint of . Then we observe that the (possibly degenerate) triangles and are congruent, by SAS congruence. Hence the segments and are congruent, meaning that is equidistant from and .
To prove the second part, we let be any point equidistant from and , and we let be the midpoint of the segment . If and are the same point, then we are done. If and are not the same point, then we observe that the triangles and are congruent by SSS congruence, so the angles and are congruent. Since these angles are supplementary angles, each of them must be a right angle. Hence is the perpendicular bisector of , and we are done.