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Comparing Sidelengths of Nested Triangles that Share the Same Base

Date: 09/07/2011 at 01:00:32
From: SANDEEP
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
archives,

    http://mathforum.org/library/drmath/view/55276.html 

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

       A
      / \
     /  _Q
    / _O_ \
   /_-   -_\
  B---------C

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
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Euclidean/Plane Geometry

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