Monday, May 19, 2008

Elearn Geometry Problem 73



See complete Problem 73
Three Intersecting Circles, Cyclic Quadrilateral, Angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. Thanks for the beautiful problem!
    It works also if EFGH are on a circle instead of a line.. but don't know why... :-(
    Ciao

    ReplyDelete
  2. join F to B, C to G ( name F1, C1 on the left...)

    A + F1 = 180, F2 + C1 = 180, C2 + H = 180

    A + H = 180 - F1 + 180 - C2
    A + H = 360 - F1 - C2
    A + H = 360 - ( 180 - F2 ) - ( 180 - C1 )
    A + H = F2 + C1
    A + H = 180 - C1 + C1

    A + H = 180

    ReplyDelete
  3. Using to circles 4 and 5 the statement proved in problem 72, we can say that BF and CH are paralel, so ang(EFB) = ang(GHD). Angles EAB and EFB are supplementary, so angles EAB and GHD are supplementary. That means that AEHC is cyclic.

    ReplyDelete
  4. Join BF & CF
    <EAB=180-<CHD
    Similarly <AEF=180-<HDC
    So AEHD is a cyclic quad

    ReplyDelete