Monday, June 6, 2016

Geometry Problem 1225: Equilateral Triangle, Circumcircle, Angle Bisector, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1225.

Geometry Problem 1225: Equilateral Triangle, Circumcircle, Angle Bisector, Congruence

Posted by Antonio Gutierrez at 10:09 AM
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2 comments:

  1. APOSTOLIS MANOLOUDISJune 6, 2016 at 11:27 AM

    Problem 1225
    Is <CAF=<FAD=<FBD=x, so <FAB=<FBA then BF=AF.
    he straight BF intersects AC in M <BMA=60+x, <MGF=<GAF+<FAD+<ADG=60+x,so FG=FM.
    But tri ABM=tri ABE then BM=AE or FE=FM=FG.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  2. c.t.e.oMay 9, 2018 at 9:04 AM

    Draw BP P on circle between A and G => BF bisector => ang DAB = ABP or tr isosceles
    Extend BF to Q on GC. FQ=FE (trAFQ congr tr BEF) . Ang QGF = arc PF+60,
    ang GQF = arc FD + 60 => tr GFQ isosceles

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