# Perpendicular Bisectors

## What are perpendicular bisectors?

Consider two fixed points \(A\) and \(B\) in the plane. Suppose that \(C\) is a **moving** point which moves in such a way so that it is always equidistant (at the same distance) from \(A\) and \(B\).

**Think:** What will be the path traced out by \(C\)? In technical language, what will be the **locus** of \(C\), given this constraint? **Locus** is a latin word, and you can think of it as the path traced out by a point under a given set of conditions.

The answer to this question is: The path traced out by \(C\) will be the **perpendicular bisector** of the segment joining \(A\) and \(B\). In other words, if \(C\) lies anywhere on the** perpendicular bisector** of \(AB\), it will necessarily be equidistant from \(A\) and \(B\).

Let us see how. Consider the following figure, in which \(C\) is an arbitrary point on the perpendicular bisector of \(AB\) (which intersects \(AB\) at \(D\)):

Compare \(\Delta ACD\) and \(\Delta BCD\). We have:

- \(AD = BD\)
- \(CD = CD\) (common)
- \(\angle ADC = \angle BDC = 90^\circ \)

We see that \(\Delta ACD \cong \Delta BCD\) by the SAS congreunce criterion. Thus, \(CA = CB\), which means that \(C\) is equidistant from \(A\) and \(B\).

**✍****Note:** Refer SAS congreunce criterion to understand why \(\Delta ACD\) and \(\Delta BCD\) are congruent.

To repeat this extremely important point once again: If a point \(C\) is equidistant from two fixed points \(A\) and \(B\), then it **must** lie on the perpendicular bisector of \(A\) and \(B\). Every point on the perpendicular bisector of \(AB\) will be equidistant from \(A\) and \(B\).

## Definition of Perpendicular Bisector:

A **perpendicular bisector** is a special kind of segment, ray, or line that:

- intersects a given segment at a \(90^\circ \)
**angle**, and, - passes through the given
**segment's midpoint**.

**✍****Note:** In the above figure, segment \(CD\) is the **perpendicular bisector** to segment \(AB\).