The Math Forum

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

Exterior Angles in Triangles

Date: 09/11/99 at 20:08:47
From: Kimberly
Subject: Exterior angles

Dear Dr. Math,

I'm a sophomore taking 11th grade math and right now I'm trying to 
figure out how to prove that in any triangle, each exterior angle is 
equal to the sum of the two nonadjacent interior angles. Here's what 
the problem looks like:

             /d          I have to prove that d = b + c
          Q /a\
           /   \
          /     \
         /       \
        /         \
       /           \
      /b           c\

Thanks for your help,

Date: 09/18/1999 at 15:04:50
From: Doctor Lilla
Subject: Re: Exterior angles

Dear Kimberly,

 Thanks for writing to Dr. Math. You have included a very nice drawing 
here to explain your question, so I can see that you understood the 
problem. You understood the terms "exterior" and "interor" angles, 

Fortunately this problem is not as hard as it seems to be. I assume 
you know that the sum of the 3 interior angles in every triangle is 
180 degrees. So:

     a + b + c = 180

Next, we should realize that a and d are supplementary angles, so 
their sum is 180.

         a + d = 180

Try to compare these equations, and after some algebraic steps you 
should get a proof of your problem. 

I hope it helps you, and that you can finish from here. Please write 
back if you still have questions.

- Doctor Lilla, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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.