Geometry Problem
Click the figure below to see the complete problem 606.
Friday, May 27, 2011
Problem 606: Isosceles Triangles, Perpendicular and Parallel Lines
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Geometry Problem
Click the figure below to see the complete problem 606.
B, G, F, M, N, H are concyclic.
ReplyDeleteLet 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
MG= HN is not enough to show that GH//MN
ReplyDelete. please provide additional details to show that GH//MN
MGHN is a cyclic quadrilateral with the pair of
ReplyDeleteopposite 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
Due to cyclic quadrilaterals made from right angles, (<GHF=<GBF)+(<FNM=<FBM)=<GBM
ReplyDeleteBecause of isosceles triangles and cyclic condition of BHNF, <GBM=<NBH=<NFH
Left with <GHF+<FNM=<NFH from transitivity, if HF cuts MN at P then <FPN=<NFH-<FNM=<GHF and we are done