GeometryChallengeCircles & Lines ShareCircles & Lines · Challenge problemIn △ABC\triangle ABC△ABC, cevians ADADAD, BEBEBE, CFCFCF meet at one point, with BDDC=2\dfrac{BD}{DC} = 2DCBD=2 and CEEA=3\dfrac{CE}{EA} = 3EACE=3. What is AFFB\dfrac{AF}{FB}FBAF?A16\dfrac{1}{6}61B666C15\dfrac{1}{5}51D56\dfrac{5}{6}65E13\dfrac{1}{3}31Check Show answer Walk through it step by step Practice more like this Review this concept