因为AC=BC,所以 三角形ABC是等腰三角形,所以 ∠ACB=∠ABC 由余弦定理：a^2 = b^2+ c^2 -2*bc*cos(A); |BC|^2 = 2^2 + 1^2 -2*2*1*cos(120) = 7; |BC| = sqrt(7); |DC| = 2|BD|,|DC| = 2*sqrt(7)/3,|BD| = sqrt(7)/3; 考虑三角形ABD; cos(角ABC) = (|AB|^2+|BC|^2 - |AC|^2)|/(2|AB|*|BC|) = (4+7-1)/(2*2*sqrt(7)) = 5*sqrt(7)/14; |AD|^2 = |AB|^2 +|BD|^2 -2*|AB|*|BD|*cos(角ABD) = 13/9; cos(角ADC) = (|AD|^2 +|DC|^2 -|AC|^2)/(2*|AD|*|DC|) = 8*sqrt(91)/91; 由向量点乘法则：向量AD.向量BC = |AD|*|BC|*cos(向量AD与向量BC的夹角) = (sqrt(13)/3) * sqrt(7)*(-8*sqrt(91)/91) = -8/3.要注：sqrt 是开平方的意思