Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 980.
Thursday, February 13, 2014
Geometry Problem 980. Equilateral Triangle, Vertices, Three Parallel, Equal Circles, Construction
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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.
ReplyDeleteFrom simple angle chasing, < COD = 120 and < DEC = 60 so CODE is a isosceles trapezoid with CO = DO.
ReplyDeleteSimilarly CF = OD (from isosceles trapezoid CODF)
So Tr.s COF and ODE are congruent SAS, the included angle being 120
Hence OE = OF and < DOE + < COF = < DOE + < DEO = 60
So < FOE = 60 and the result follows
Sumith Peiris
Moratuwa
Sri Lanka