# 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
- The **perpendicular bisector theorem** states that any point on the perpendicular bisector of a segment is equidistant from its two endpoints.
- The theorem is proved using **SAS congruence**. 
- Its **converse** is also true.
- The three perpendicular bisectors of a triangle meet at the **circumcentre**, equidistant from all three vertices.

## 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.