tag:blogger.com,1999:blog-6933544261975483399.post7928100423541849816..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 980. Equilateral Triangle, Vertices, Three Parallel, Equal Circles, ConstructionAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-42715376164286131442021-01-30T05:46:22.467-08:002021-01-30T05:46:22.467-08:00From simple angle chasing, < COD = 120 and <...From simple angle chasing, < COD = 120 and < DEC = 60 so CODE is a isosceles trapezoid with CO = DO. <br /><br />Similarly CF = OD (from isosceles trapezoid CODF)<br /><br />So Tr.s COF and ODE are congruent SAS, the included angle being 120<br /><br />Hence OE = OF and < DOE + < COF = < DOE + < DEO = 60<br /><br />So < FOE = 60 and the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42590769752317038152014-02-13T14:14:20.124-08:002014-02-13T14:14:20.124-08:00Because AC//DF and BD//CE, 60=<BOC=<BDF, 60=...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.Ivan Bazarovnoreply@blogger.com