如图CE,CB分别为△ABC,△ADC的中线,AB=AC,∠ABC=∠ACB,求证CD=2CE

证明：延长CE到F,使EF=CE,连接FB．∵CE是△ABC的中线,∴AE=EB,又∵∠AEC=∠BEF,∴△AEC≌△BEF,（SAS）∴∠A=∠EBF,AC=FB．∵AB=AC,∴∠ABC=∠ACB,∴∠CBD=∠A+∠ACB=∠EBF+∠ABC=∠CBF；∵CB是ADC的中线,∴AB=BD,又...