The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Comparing Sidelengths of Nested Triangles that Share the Same Base

Date: 09/07/2011 at 01:00:32
Subject: geometry

Prove that, for any point O inside a triangle ABC,

   AB + AC > OB + OC

I don't understand how to proceed. I know that the greater the sidelength,
the greater the angle opposite it.

Date: 09/07/2011 at 09:00:14
From: Doctor Floor
Subject: Re: geometry

Hi, Sandeep,

Thanks for your question.

We will make use of the triangle inequality. See, from the Dr. Math

If O is in the interior of ABC, we can construct BO to intersect AC at a
point Q.

      / \
     /  _Q
    / _O_ \
   /_-   -_\

From the triangle inequality, we know that

        OC < OQ + QC
Adding OB to both sides of the inequality,

   OB + OC < OB + OQ + QC
           = QB + QC

From the triangle inequality, we also know that

        BQ < AB + AQ

You finish.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum 
Associated Topics:
High School Euclidean/Plane Geometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.