Question

已知三角形ABC是边长为4的等边三角形,点O在边AB上,圆O过点B且分别与边AB BC相交于点DEEF垂直AC,垂足为F

Answer 1

1、连接OE,OB = OE∵△ABC为等边三角形∴∠OBE = 60°∴∠OEB = ∠C = 60°∴OE//AC∵EF⊥AC∴OE⊥ EF∴ EF是圆O的切线2 、∵FD为圆O的切线 ∴∠ADF = 90°∴∠AFD = 90° - ∠A = 30°∵EF⊥AC∴∠DFE = 60°连接OF则：∠DFO = ∠EFO = ∠DFE/2 = 30°∴∠FOD = ∠A = ∠AFO = 60°∴△AFO为等边三角形于是,AD = DO = OB = AB/3 = 4/3即：圆半径为4/3