如图所示,在△ABC的外侧作Rt△ABD和Rt△ACE,∠ABD=∠ACE=90°,且∠BAD=∠CAE,M是DE的中点

取AD中点P,连接BP、MP.
则有：BP是Rt△ABD斜边上的中线,MP是△ADE的中位线,
可得：BP = AP = (1/2)AD = MQ ,∠BAD = ∠ABP ,MP‖AE .
取AE中点Q,连接CQ、MQ.
则有：CQ是Rt△ACE斜边上的中线,MQ是△ADE的中位线,
可得：CQ = AQ = (1/2)AE = MP ,∠CAE = ∠ACQ ,MQ‖AD .
在平行四边形APMQ中,有：∠APM = ∠AQM ；
而且,∠BPD = ∠BAD+∠ABP = 2∠BAD = 2∠CAE = ∠CAE+∠ACQ = ∠CQE ；
可得：∠BPM = 180°-∠APM-∠BPD = 180°-∠AQM-∠CQE = ∠MQC .
在△BPM和△MQC中,BP = MQ ,∠BPM = ∠MQC ,MP = CQ ,
所以,△BPM ≌ △MQC ,
可得：BM = CM .