证明：作BC的垂直平分线,与AB交于点M,连接MC
则：MB = MC
∴∠MBC = ∠MCB
∴∠AMC = 2∠B = ∠A
∴AC =MC = MB
∵AB = 2AC,即：AB = 2BM
∴AM = BM = CM = AC
j即 △AMC为等边三角形
于是,∠A = 60°
∠B = 30°
故：∠ACB = 90°