Geometry Problem

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## Friday, May 27, 2011

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## Friday, May 27, 2011

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Problem 606: Isosceles Triangles, Perpendicular and Parallel Lines

Online Geometry theorems, problems, solutions, and related topics.

Labels:
isosceles,
parallel,
perpendicular,
triangle

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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