Friday, May 27, 2011

Problem 606: Isosceles Triangles, Perpendicular and Parallel Lines

Geometry Problem
Click the figure below to see the complete problem 606.

Problem 606: Isosceles Triangles, Perpendicular and Parallel Lines.

Posted by Antonio Gutierrez at 7:05 PM
Labels: , , ,

3 comments:

  1. B, G, F, M, N, H are concyclic.
    Let R be the radius of the circumscribing circle.
    MG = 2R sin ∠MBG and HN = 2R sin ∠NBH
    But ∠MBG = ∠NBH
    So MG = HN subtend equal angles at N, G respectively
    Hence GH ∥ MN

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  2. MG= HN is not enough to show that GH//MN
    . please provide additional details to show that GH//MN

    ReplyDelete

  3. MGHN is a cyclic quadrilateral with the pair of
    opposite sides MG and NH of equal length.
    Triangles GHM and HGN are congruent by S.A.S property
    (since MG = HN, GH is common and ∠GHM = ∠GBM = ∠NBH = ∠NGH)
    So ∠MGH = ∠NHG
    But ∠MGH and ∠MNH are supplementary
    Hence ∠NHG and ∠MNH are supplementary which implies GH ∥ MN

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