Stage 15 · Triangles

15.4  Reflection Symmetry

Fold and match — the axis of symmetry, the mirror image, and the perpendicular bisector hiding inside.

Ages 11–14 · Reasoning, one step at a time
Knowledge point page

Point 4 of 5 in this lesson: 15.4.4 The perpendicular bisector of a segment

15.4.4 The perpendicular bisector of a segment

Section 15.4.3 says points on the axis are equidistant from P and P′. Flip that statement around and it becomes a clean, distance-only definition:

Definition

The perpendicular bisector of a segment AB is the set of all points that are equidistant from A and B. A point lies on it if and only if its distance to A equals its distance to B.

This is the same line you build with compass and straightedge in 15.2: swing equal arcs from A and from B, and the two crossing points are each equidistant from A and B — join them and you have the perpendicular bisector. It runs straight through the midpoint of AB, square to it. Later, this set-of-equal-distances idea is exactly how you find the center of a circle through three points.

Try it Equal distances land on the bisector

Drag the test point M. When its distance to A equals its distance to B, it is sitting exactly on the perpendicular bisector — both arms light green.

Move M across
M's height
eastmath.com · 15.4 Reflection Symmetry · 15.4.4 The perpendicular bisector of a segment