已知,在△ABC中,∠ACB=90°,CD⊥AB,垂足为D,∠BCD=3∠ACD,CE是AB边上的中线.求证:∠DCE＝∠DEC

CE是Rt△ABC斜边AB上的中线,可得：BE = CE ,
所以,∠BCE = ∠B ,∠DEC = ∠BCE+∠B = 2∠B .
∠ACD = 90°-∠A = ∠B ,可得：∠BCD = 3∠ACD = 3∠B ,
所以,∠DCE = ∠BCD-∠BCE = 2∠B = ∠DEC .