第2个回答 2010-11-30
延长DA到点F,则有:∠CAF = 180°-∠DAB-∠BAC = 180°-60°-90° = 30° 。
已知,AE是等边△ABD的高,可得:DE = EB = (1/2)BD = (1/2)AB 。
已知,AD = AB = AC ,可得:∠ADC = ∠ACD = (1/2)∠CAF = 15° 。
因为,∠DEM = 90° ,∠MDE = ∠ADB-∠ADC = 60°-15° =
可得:DM = (√2)DE = (√2)·(1/2)AB = (√2)·(1/2)·(√2/2)BC = (1/2)BC