Perpendicular Bisectors in Triangles

Learning Objectives

  • Construct the perpendicular bisector of a line segment.
  • Apply the Perpendicular Bisector Theorem to identify the point of concurrency of the perpendicular bisectors of the sides (the circumcenter).
  • Use the Perpendicular Bisector Theorem to solve problems involving the circumcenter of triangles.

Introduction

In our last lesson we examined midsegments of triangles. In this lesson we will examine another construction that can occur within triangles, called perpendicular bisectors.

The perpendicular bisector of a line segment is the line that:

  1. divides the line segment into two congruent sub-segments.
  2. intersects the line segment at a right angle.

Here is an example of a perpendicular bisector to line segment \overline{AB}.

Perpendicular Bisector Theorem and its Converse

We can prove the following pair of theorems about perpendicular bisectors.

Perpendicular Bisector Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

Proof. Consider \overline{AB} with perpendicular bisector l with points C and X on line l as follows:

We must show that \overline{AC} \cong \overline{BC}.

  1. Since l is the perpendicular bisector of \overline{AB}, it follows that \overline{AX}\cong \overline{XB} and angles \angle{CXA} and \angle{CXB} are congruent and are right angles.

By the SAS postulate, we have \triangle AXC \cong \triangle BXC.

So \overline{AC} \cong \overline{BC} by CPCTC (corresponding parts of congruent triangles are congruent). \blacklozenge

It turns out that we can also prove the converse of this theorem.

Converse of the Perpendicular Bisector Theorem: If a point is equidistant from the endpoints of a segment, then the point is on the perpendicular bisector of the segment.

Proof. Consider \triangle ABC as follows with \overline{AB} \cong\overline{AC}.

We will construct the midpoint X of \overline{BC} and show that \overline{AX} is the perpendicular bisector to \overline{BC}.

1. Construct the midpoint of \overline{BC} at point X. Construct \overline{XA}.

2. Consider \triangle ABX and \triangle ACX. These are congruent triangles by postulate SSS.

3. So by CPCTC, we have \angle{AXB} \cong \angle{AXC}.

4. Since \angle{AXB} and \angle{AXC} form a straight angle and are also congruent, then m\angle{AXB} = m\angle{AXC} = 90^\circ. Hence, X is on the perpendicular bisector to \overline{BC}. \blacklozenge

Notice that we just proved the Perpendicular Bisector Theorem and we also proved the Converse of the Perpendicular Bisector Theorem. When you prove a theorem and its converse you have proven a biconditional statement. We can state the Perpendicular Bisector Theorem and its converse in one step: A point is on the perpendicular bisector of a segment if and only if that point is equidistant from the endpoints of the segment.

We will now use these theorems to prove an interesting result about the perpendicular bisectors of the sides of a triangle.

Concurrency of Perpendicular Bisectors: The perpendicular bisectors of the sides of a triangle intersect in a point that is equidistant from the vertices.


Lesson Summary

In this lesson we:

  • Defined the perpendicular bisector of a line segment.
  • Stated and proved the Perpendicular Bisector Theorem.
  • Solved problems using the Perpendicular Bisector Theorem.


Points to Consider

If we think about three non-collinear points in a plane, we can imagine a triangle that has each point as a vertex. Locating the circumcenter, we can draw a circle that all three vertices will be on. What does this tell us about any three non-collinear points in a plane?

There is a unique circle for any three non-collinear points in the same plane.

Finding a circle through any three points will also work in coordinate geometry. You can use the circumcenter to find the equation of a circle through any three points. In calculus this method is used (together with some tools that you have probably not learned yet) to precisely describe the curvature of any curve.

Last modified: Monday, June 28, 2010, 3:49 PM