如图,在△ABC中,∠A=2∠B,AB=2AC,求证:∠ACB=90°
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如图,在△ABC中,∠A=2∠B,AB=2AC,求证:∠ACB=90°
感激不尽
感激不尽
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°
∴ ∠ACB = 90°
∴∠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°
∴ ∠ACB = 90°
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