Perpendicular Bisector and Equidistance

Trikona Team · v1.0

Every point on a perpendicular bisector is equidistant from its endpoints, and every point equidistant from two points lies on their perpendicular bisector.

Keywords: perpendicular bisector, equidistance, congruent triangles, SAS congruence, SSS congruence, midpoint

Prerequisites: congruent-triangles, midpoint · Difficulty: beginner

What is a Perpendicular Bisector?

A perpendicular bisector is a special line that:

1. Cuts a line segment exactly in half (at its midpoint)
2. Forms a 90° angle with the segment

This proof shows a beautiful two-way relationship:

• Forward: If you're on the perpendicular bisector, you're equally far from both endpoints
• Converse: If you're equally far from both endpoints, you must be on the perpendicular bisector

Try dragging point $P$ along line $L$ — notice how $PA$ and $PB$ always stay equal!

The Proof

Step 1: Part 1: Setting Up the Construction

We begin with segment $AB$ and construct its perpendicular bisector $L$ through midpoint $M$. Then we pick any point $P$ on $L$ and aim to prove $PA = PB$.

Step 2: Proving $PA = PB$ by SAS Congruence

We form triangles $\triangle PMA$ and $\triangle PMB$, then prove they are congruent using Side-Angle-Side (SAS):

• $AM = BM$ (M is the midpoint)
• $\angle PMA = \angle PMB = 90°$ (L is perpendicular)
• $PM = PM$ (common side)

Therefore $PA = PB$ by corresponding parts of congruent triangles (CPCTC).

Step 3: Part 2: The Converse

Now we prove the reverse: if a point $P$ satisfies $PA = PB$, then $P$ must lie on the perpendicular bisector of $AB$.

Step 4: Proving $P$ Lies on the Bisector by SSS

Let $M$ be the midpoint of $AB$. We prove $\triangle PMA \cong \triangle PMB$ using Side-Side-Side (SSS):

• $PA = PB$ (given)
• $AM = BM$ (M is the midpoint)
• $PM = PM$ (common side)

Since corresponding angles $\angle PMA = \angle PMB$ are equal and sum to $180°$, both must be $90°$. Thus $PM \perp AB$, and since $PM$ passes through midpoint $M$, point $P$ lies on the perpendicular bisector!

Conclusion

We've proven that the perpendicular bisector of a segment is exactly the set of all points equidistant from its endpoints. This is why perpendicular bisectors are so useful:

• Finding equidistant points: Any point on the perpendicular bisector works!
• Finding centers: The circumcenter of a triangle (center of its circumscribed circle) lies on all three perpendicular bisectors

This property is fundamental in geometry and has many applications in compass-and-straightedge constructions.