三角形ABC中,角B=60,E 为BC上的一点,若BE:EC=3:2,AB:EC=4:1,则AE:AC:BC=?

设EC=2x,则BE=3x、AB=8x、BC=5xAE^2=AB^2+BE^2-2AB*BEcos∠B=64x^2+9x^2-24x^2=49x^2AE=7xAC^2=AB^2+BC^2-2AB*BCcos∠B=64x^2+25x^2-40x^2=49x^2AC=7xAE:AC:BC=7x：7x：5x=7：7：5