(1)
取A为原点, D在+x轴上, B, C分别在第一和第四象限, ∠BAD = ∠CAD = 30°
令AC的三等分点为E, 且AC = 3AE
在射线AB上取点F, 使得AF = λAB
按题意, E, F关于x轴对称, 即AFsin∠BAD = AEsin∠CAD, λ|AB|= (1/3)|AC|
|AC| = 3λ*3 = 9λ
AD = (1/3)|AC|cos∠BAD + λ|AB|cos∠BAD = (1/3)9λ*(√3/2) + λ*3*√3/2 = 3√3λ
(2)
令OA+OB=OC', C'在外接圆上, 且OAC'B为平行四边形, OA = OB = AC' = BC' = 外接圆的半径, 即该四边形为菱形
按题意, OA+OB+ CO =0, OA+OB =-CO = OC
即OC = OC'， 即C, C'重合. 在菱形OACB中, 对角线OC是外接圆的半径, 与菱形的边长相等, 于是△OAC和△OBC均为正三角形, ∠C=120°， △ABC为等腰三角形, ∠A = (180° - ∠C)/2 = 30°
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