Perpendicular Bisector Theorem

Suppose $C M$ is the perpendicular bisector of $\overline{A B} .$ Then $M$ is the midpoint of $\overline{A B} .$

Consider a triangle with vertices $A,$ $M,$ and $C,$ and another triangle with vertices and $B,$ $M,$ and $C .$

Both $△ A C M$ and $△ B C M$ have a right angle and congruent legs $\overline{A M}$ and $\overline{B M} .$ Since all right angles are congruent, $∠ A M C ≅ ∠ B M C .$ Furthermore, by the Reflexive Property of Congruence, $\overline{C M}$ is congruent to itself.

By the Side-Angle-Side Congruence Theorem, the triangles are congruent. Therefore, since corresponding parts of congruent figures are congruent, their hypotenuses $\overline{A C}$ and $\overline{B C}$ are also congruent. By the definition of congruent segments, $\overline{A C}$ and $\overline{B C}$ have the same length. This means that $C$ is equidistant from $A$ and $B .$

Using this reasoning it can be proven that any point on a perpendicular bisector is equidistant from the endpoints of the segment.

Two-Column Proof

The proof can be summarized in the following two-column table.

Statements	Reasons
$\overline{A M} ≅ \overline{M B}$ $∠ A M C$ and $∠ B M C$ are right angles	Definition of a perpendicular bisector.
$∠ A M C ≅ ∠ B M C$	All right angles are congruent.
$\overline{C M} ≅ \overline{C M}$	Reflexive Property of Congruence.
$△ A C M ≅ △ B C M$	SAS Congruence Theorem.
$\overline{A C} ≅ \overline{B C}$	Corresponding parts of congruent figures are congruent.
$A C = B C$	Definition of congruent segments.

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Rule

Perpendicular Bisector Theorem

Proof

Two-Column Proof

Proof

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Statement	Reason
$\overline{A M} ≅ \overline{M B} \overline{A B} ⊥ C M$	Given
$△ A M C ≅ △ B M C$	SAS congruence theorem
$\overline{A C} ≅ \overline{B C}$	$△ A M C ≅ △ B M C$
$A C = B C$	Definition of congruent segments