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Mathematics LibreTexts

5.2: Perpendicular Bisector

( \newcommand{\kernel}{\mathrm{null}\,}\)

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

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

Theorem 5.2.1

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

Proof

截屏2021-02-03 下午4.38.06.png

Let M be the midpoint of [AB].

Assume PA=PB and PM. According to SSS (Theorem 4.4.1), AMPBMP. Hence

AMP=±BMP.

Since AB, we have "-" in the above formula. Further,

π=AMBAMP+PMB2AMP.

That is, AMP=±π2. Therefore, P lies on the perpendicular bisector.

To prove the converse, suppose P is any point on the perpendicular bisector to [AB] and PM. Then AMP=±π2, BMP=±π2 and AM=BM. By SAS, AMPBMP; in particular, AP=BP.

Exercise 5.2.1

Let 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 .

Hint

截屏2021-02-03 下午4.48.18.png

Assume X and A lie on the same side of .

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

Note that BX=AY+YXAX. Since X, by Theorem 5.2.1 we have BXBA. 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 5.2.2

Let ABC be a nondegenerate triangle. Show that

AC>BC|ABC|>|CAB|.

Hint

Apply Exercise 5.2.1, 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.

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