Saturday, August 20, 2016

Geometry Problem 1250: Cyclic or Inscribed Quadrilateral, Circle, Diameter, Congruence

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.

Geometry Problem 1250: Cyclic or Inscribed Quadrilateral, Circle, Diameter, Congruence

Posted by Antonio Gutierrez at 11:02 PM
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2 comments:

  1. APOSTOLIS MANOLOUDISAugust 21, 2016 at 12:57 AM

    Problem 1250
    Is 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

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  2. Sumith PeirisAugust 21, 2016 at 3:23 AM

    BCGE is a cyclic quadrilateral with diameter BC.
    So <GEC = <GBC = < DAC and so EG//FH

    Further < GHC = <GDC = <BAC = < BFE so < EFH = < GHF

    Hence EFHG is an isoceles trapezoid with = diagonals, EH = FG

    Sumith Peiris
    Moratuwa
    Sri Lanka

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