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

The perpendicular bisector of segment $AB$ has a beautiful two-way relationship with equidistance:

• Forward: every point on the bisector is equidistant from $A$ and $B$.
• Converse: every point equidistant from $A$ and $B$ lies on the bisector.

Strategy: the forward direction follows from SAS congruence of $△PMA$ and $△PMB$; the converse follows from SSS congruence, which forces the two angles at $M$ to be equal right angles.

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 $△PMA$ and $△PMB$, 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 $△PMA\cong △PMB$ 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!

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. ∎

Notes

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

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