Perpendicular Bisector Theorem: Statement, Proof, and Examples

Perpendicular Bisector Theorem: Statement, Proof, and Examples

TL;DR

The perpendicular bisector theorem states that any point on the perpendicular bisector of a segment is equidistant from the segment's two endpoints. This article covers the formal statement, the SAS-based proof, the converse, the link to the circumcentre, and worked examples.

What is a Perpendicular Bisector

A perpendicular bisector is a line that does two things to a segment at once: it crosses the segment at its midpoint (so it bisects, cutting the segment into two equal halves) and it meets the segment at a right angle (so it is perpendicular).

The Perpendicular Bisector Theorem

The perpendicular bisector theorem states:

Any point lying on the perpendicular bisector of a segment is equidistant from the two endpoints of that segment.

How Do You Know A Point is Equidistant Without Measuring?

You do not have to measure at all. If you can confirm a point lies on the perpendicular bisector, the theorem hands you the equal distances for free.

Proof of the Perpendicular Bisector Theorem

Given: Line ℓ is the perpendicular bisector of segment AB, meeting it at midpoint M. Point P is any point on ℓ.

To prove: PA = PB.

Step 1: AM = MB.

Step 2: ∠PMA = ∠PMB = 90°.

Step 3: PM = PM.

Step 4: By SAS, triangle PMA ≅ triangle PMB.

Step 5: Corresponding parts of congruent triangles are equal, so PA = PB.

The Converse of the Perpendicular Bisector Theorem

Converse: If a point is equidistant from the two endpoints of a segment, then that point lies on the perpendicular bisector of the segment.

Where the Theorem Leads: The Circumcentre

The theorem pays off most visibly inside a triangle. Each of a triangle's three sides has its own perpendicular bisector, and all three of those lines meet at a single point — the circumcentre.

Examples of the Perpendicular Bisector Theorem

Example 1

Point P lies on the perpendicular bisector of segment AB. If PA = 9 cm, find PB. So PB = PA = 9 cm.

Final answer: PB = 9 cm.

Example 2

Line ℓ is the perpendicular bisector of CD, meeting it at M. If CD = 16 cm, find CM. So CM = CD/2 = 16/2 = 8 cm.

Final answer: CM = 8 cm.

Example 3

A point Q is the same distance from both ends of segment EF. Is that correct? The correct reading: Q lies on the perpendicular bisector of EF, but it need not be the midpoint.

Final answer: No. Q lies on the perpendicular bisector of EF.

Example 4

Point P is on the perpendicular bisector of segment AB. PA = (3x + 5) and PB = (5x − 7). Find x. So set the two expressions equal: 3x + 5 = 5x - 7.

Final answer: x = 6.

Example 5

In a triangle, point O is the circumcentre. The distance from O to vertex A is 10 cm. What is the distance from O to vertex B? So OB = OA = 10 cm.

Final answer: OB = 10 cm.

Example 6

Point R lies on the perpendicular bisector of GH. RG = (2y + 4) and RH = (4y − 10). After finding y, give the common distance RG. So RG = RH.

Final answer: y = 7 and RG = 18.

Why the Perpendicular Bisector Theorem Matters

The theorem exists because "the set of points equally far from two fixed places" is one of the most useful objects in geometry — and it is always a straight line, the perpendicular bisector.

The Mistakes Students Make Most Often

Mistake 1: Dropping one of the two conditions

Don't do this: Treat a segment bisector alone, or a perpendicular line alone, as a perpendicular bisector.

Mistake 2: Confusing the equidistant point with the midpoint

Don't do this: Conclude "equidistant from A and B" means "the midpoint of AB."

Mistake 3: Setting the distance expressions equal incorrectly

Don't do this: Write PA + PB = 0 or solve only one expression on its own.

Conclusion

A Practical Next Step

  1. If P is on the perpendicular bisector of AB with PA=(4x−1) and PB=(2x+9), find x and the common distance.
  2. Explain why the circumcentre is equidistant from all three vertices of a triangle.