Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, August 20, 2016

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

Labels:
circle,
congruence,
cyclic quadrilateral,
diameter

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Problem 1250

ReplyDeleteIs 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.

ReplyDeleteSo <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