ABC is an equilateral triangle and D is a point on BC. AE is the bisector of angle DAC (E on BC).
The circumcircle of triangle ABD cuts AE at F and DF extended
meets AC at G. Prove that (1) BF = AF; (2) FE = FG.
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
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