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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 980.
Because AC//DF and BD//CE, 60=<BOC=<BDF, 60=<AOQ=180-<ACF=<DFC, then <DFC=<BDF so OCDF is cyclic isosceles trapezoid. But 60=180-<QOD=<ECF so <DFC=<ECF which makes FEDC isosceles trapezoid as well. That means DEOCF is cyclic and OEF equilateral from subtended angles.