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Home » Triangles » Similar Triangles » Perpendicular Bisector Theorem

Perpendicular Bisector Theorem

Last updated: Sep 30, 2019 by Ido Sarig · This website generates income via ads and uses cookies · Terms of use · Privacy policy

In today's geometry lesson, we will show a fairly easy way to prove the perpendicular bisector theorem.

A line that splits another line segment (or an angle) into two equal parts is called a "bisector." If the intersection between the two line segment is at a right angle, then the two lines are perpendicular, and the bisector is called a "perpendicular bisector".

The Perpendicular Bisector Theorem states that a point on the perpendicular bisector of a line segment is an equal distance from the two edges of the line segment.

Problem

Point C is on the perpendicular bisector of segment AB. Prove that |CA|=|CB|

perpendicular bisector

Strategy

We need to prove two line segments are equal, so let's draw them:

triangles for proving perpendicular bisector

And now it should be clear that the way to go here is to use congruent triangles. We have a common side (CD). We also have two segments given as equal (AD=DB) since AD is the bisector. And we have an equal angle, which is a right angle, in between them.

Proof

(1) CD=CD //Common side, reflexive property of equality
(2) AD=DB //Given, CD is a bisector
(3) m∠ADC= m∠BDC =90° //Given, CD is perpendicular to AB
(4) ∠ADC≅ ∠BDC //Side-Angle-Side postulate
(5) CA=CB //corresponding sides of congruent triangles (CPCTC)

And so we have proved the Perpendicular Bisector Theorem. Note that this is a converse theorem to one of the properties of an isosceles triangle - that the median to the base (which is a bisector) is perpendicular to the base. Here we prove the opposite - if we have a perpendicular median, the triangle is isosceles.

« Converse of the Angle Bisector Equidistant Theorem
Converse Angle Bisector Theorem for Isosceles Triangles »

About the Author

Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. His goal is to help you develop a better way to approach and solve geometry problems. You can contact him at [email protected]

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About

Welcome to Geometry Help! I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree in Management of Technology. I'm here to tell you that geometry doesn't have to be so hard! My goal with this website is to help you develop a better way to approach and solve geometry problems, even if spatial awareness is not your strongest quality. Read More…

Geometry Topics

  • Area of Geometric Shapes
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    • Isosceles Triangles
    • Pythagorean Theorem
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    • Similar Triangles
    • Triangle Inequalities

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