Perpendicular bisector

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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 $\displaystyle AB$ is also the locus of points equidistant from $\displaystyle A$ and $\displaystyle B$ .

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

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