是.证明如下：∵BD^2=AB^2+AD^2+2AB*ADcos∠ABD^2=CD^2+BC^2+2CD*BCcos∠C又AB=CD,∠A=∠C∴AD^2+2AB*ADcos∠A=BC^2+2AB*BCcos∠AAD*(AD+2ABcos∠A)=BC*(BC+2ABcos∠A)即AD=BC∵ΔBAD≌ΔDCB (边,角,边）∴∠ABD=∠B...