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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1250.
Problem 1250Is BE perpendicular in AC and CG is perpendicular in BG.But <BEC=90=<BGC so BEGC is cyclic ie <GEC=<GBC=<DBC=<DAC so EG//AD therefore EGHF is trapezoid.Βut ABEF and CGHD are cyclic, then <EFH=<ABE=<90-<BAE=90-<BAC=90-<BDC=90-<GDC=<GCD=<GHF.So FEGH is isosceles trapezoid. Therefore EH=FG.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
BCGE is a cyclic quadrilateral with diameter BC.So <GEC = <GBC = < DAC and so EG//FHFurther < GHC = <GDC = <BAC = < BFE so < EFH = < GHFHence EFHG is an isoceles trapezoid with = diagonals, EH = FGSumith PeirisMoratuwaSri Lanka