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.

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## Monday, June 6, 2016

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

Labels:
angle bisector,
circumcircle,
congruence,
equilateral,
geometry problem,
triangle

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Problem 1225

ReplyDeleteIs <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

Draw BP P on circle between A and G => BF bisector => ang DAB = ABP or tr isosceles

ReplyDeleteExtend BF to Q on GC. FQ=FE (trAFQ congr tr BEF) . Ang QGF = arc PF+60,

ang GQF = arc FD + 60 => tr GFQ isosceles