Perpendicular bisector

In a plane, the perpendicular bisector of a line segment $AB$ is a line $l$ such that $AB$ and $l$ are perpendicular and $l$ passes through the midpoint of $AB$.

In 3-D space, for each plane containing $AB$ 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 $AB$.

In a triangle, the perpendicular bisectors of all three sides intersect at the circumcenter.

Locus

The perpendicular bisector of $AB$ is also the locus of points equidistant from $A$ and $B$.

To prove this, we must prove that every point on the perpendicular bisector is equidistant from $A$ and $B$, and also that every point equidistant from $A$ and $B$.

The first part we prove as follows: Let $P$ be a point on the perpendicular bisector of $AB$, and let $M$ be the midpoint of $AB$. Then we observe that the (possibly degenerate) triangles $APM$ and $BPM$ are congruent, by SAS congruence. Hence the segments $PA$ and $PB$ are congruent, meaning that $P$ is equidistant from $A$ and $B$.

To prove the second part, we let $P$ be any point equidistant from $A$ and $B$, and we let $M$ be the midpoint of the segment $AB$. If $P$ and $M$ are the same point, then we are done. If $P$ and $M$ are not the same point, then we observe that the triangles $PAM$ and $PBM$ are congruent by SSS congruence, so the angles $PAM$ and $PBM$ are congruent. Since these angles are supplementary angles, each of them must be a right angle. Hence $PM$ is the perpendicular bisector of $AB$, and we are done.

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