Comparing Sidelengths of Nested Triangles that Share the Same BaseDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/