Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 606.
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 ∠NBHBut ∠MBG = ∠NBHSo MG = HN subtend equal angles at N, G respectivelyHence GH ∥ MN
MG= HN is not enough to show that GH//MN. please provide additional details to show that GH//MN
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 = ∠NHGBut ∠MGH and ∠MNH are supplementaryHence ∠NHG and ∠MNH are supplementary which implies GH ∥ MN
Due to cyclic quadrilaterals made from right angles, (<GHF=<GBF)+(<FNM=<FBM)=<GBMBecause of isosceles triangles and cyclic condition of BHNF, <GBM=<NBH=<NFHLeft with <GHF+<FNM=<NFH from transitivity, if HF cuts MN at P then <FPN=<NFH-<FNM=<GHF and we are done