# 5.2: Perpendicular Bisector

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Assume $$M$$ is the midpoint of the segment $$[AB]$$; that is, $$M \in (AB)$$ and $$AM = MB$$.

The line $$\ell$$ that passes thru $$M$$ and perpendicular to $$(AB)$$, is called the perpendicular bisector to the segment $$[AB]$$.

## Theorem $$\PageIndex{1}$$

Given distinct points $$A$$ and $$B$$, all points equidistant from $$A$$ and $$B$$ and no others lie on the perpendicular bisector to $$[AB]$$.

Proof

Let $$M$$ be the midpoint of $$[AB]$$.

Assume $$PA = PB$$ and $$P \ne M$$. According to SSS (Theorem 4.4.1), $$\triangle AMP \cong \triangle BMP$$. Hence

$$\measuredangle AMP = \pm \measuredangle BMP.$$

Since $$A \ne B$$, we have "-" in the above formula. Further,

$\begin{array} {rcl} {\pi} & = & {\measuredangle AMB \equiv} \\ {} & \equiv & {\measuredangle AMP + \measuredangle PMB \equiv} \\ {} & \equiv & {2 \cdot \measuredangle AMP.} \end{array}$

That is, $$\measuredangle AMP = \pm \dfrac{\pi}{2}$$. Therefore, $$P$$ lies on the perpendicular bisector.

To prove the converse, suppose $$P$$ is any point on the perpendicular bisector to $$[AB]$$ and $$P \ne M$$. Then $$\measuredangle AMP = \pm \dfrac{\pi}{2}$$, $$\measuredangle BMP = \pm \dfrac{\pi}{2}$$ and $$AM = BM$$. By SAS, $$\triangle AMP \cong \triangle BMP$$; in particular, $$AP = BP$$.

## Exercise $$\PageIndex{1}$$

Let $$\ell$$ be the perpendicular bisector to the segment $$[AB]$$ and $$X$$ be an arbitrary point on the plane.

Show that $$AX < BX$$ if and only if $$X$$ and $$A$$ lie on the same side from $$\ell$$.

Hint

Assume $$X$$ and $$A$$ lie on the same side of $$\ell$$.

Note that $$A$$ and $$B$$ lie on opposite side of $$\ell$$. Therefore, by Corollary 3.4.1, $$[AX]$$ does not intersect $$\ell$$ and $$[BX]$$ intersects $$\ell$$; suppose that $$Y$$ denotes the intersection point.

Note that $$BX = AY + YX \ge AX$$. Since $$X \not\in \ell$$, by Theorem $$\PageIndex{1}$$ we have $$BX \ne BA$$. Therefore $$BX > AX$$.

This way we proved the "if" part. To prove the "only if" part, you need to switch $$A$$ and $$B$$ and repeat the above argument.

## Exercise $$\PageIndex{2}$$

Let $$ABC$$ be a nondegenerate triangle. Show that

$$AC > BC \Leftrightarrow |\measuredangle ABC| > |\measuredangle CAB|.$$

Hint

Apply Exercise $$\PageIndex{1}$$, Theorem 4.2.1 and Exercise 3.1.2.

This page titled 5.2: Perpendicular Bisector is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform.