如图：已知在等腰Rt △ABC中,∠CAB=90°,以AB为边向外作等边△ABD,AE⊥BD,CD、AE交于点M .求证：DM=二分之一BC

题目

如图：已知在等腰Rt △ABC中,∠CAB=90°,以AB为边向外作等边△ABD,AE⊥BD,CD、AE交于点M .求证：DM=二分之一BC

答案

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 =...