证明：在AC上截取AE=AB
∵AD平分∠BAC,∴∠EAD=∠BAD.
AD=AD
∴△AED≌△ABD（SAS）∴∠AED=∠ABD
而∠ABD=2∠C,∴∠AED=2∠C
又∵∠AED=∠C+∠EDC,∴∠C=∠EDC,∴DE=CE
∴AC=AE+CE=AE+DE=AB+BD,∴AC=AB+BD