如图,在△ABC中,∠C=90°,∠A=30°,求证AB=2BC
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证明：过C点做 ∠ACD = 30° ,交AB于D,
因为 ∠A = ∠ACD = 30° ,
所以 AD = CD
又 ∠DCB = ∠ACB - ∠ACD = 90° - 30° = 60°
而 ∠B = ∠ACB - ∠A = 90° - 30° = 60°
所以 ∠DCB = ∠B = ∠CDB = 60°
所以 CD = BC = BD
所以 AD = CD
= BD = BC所以 AB = AD + BD = BC +BC =2BC