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Help with a Proof

  1. Mar 3, 2006 #1
    Prove that the sum of the exterior angles at opposite vertices of any quadrilateral es equal to the sum of the interior angles at the other two vertices.

    THIS question is REAly really really really getting me frustrated...

    The way we're suppost to do it is like this....

    180n-360 is the interior angles and exterior angles is 180n-180(n-2)

    From there I don't know.....
  2. jcsd
  3. Mar 3, 2006 #2
    The four interior angles are labeled A, B, C, and D, and you know the sum of these. What happens if you subtract the exterior angles from opposite vertices?
  4. Mar 3, 2006 #3
    I don't know man I'm frigging confused...

    so you saying that A+B+C+D=360?
  5. Mar 3, 2006 #4
    If A and A' are two adjacent angles, then A + A' = 180deg.
    Similarly for C and C'. Now find A' + C' wrt A and C. Then find B + D wrt A and C.
  6. Mar 3, 2006 #5
    i'm lost :(
  7. Mar 4, 2006 #6
    You already know A+B+C+D=360, right?. Now take two opposite angles (assume A and C). What is the sum of the exterior angles of these 2?
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