证明：延长BD到F，使BF=BA，连接AF，CF，∵∠ABD=60度，∴△ABF为等边三角形，∴AF=AB=AC=BF，∠AFB=60°，∴∠ACF=∠AFC，又∵∠ACD=60°，∴∠AFB=∠ACD=60°∴∠DFC=∠DCF，∴DC=DF．∴BD+DC=BD+DF=BF=AB，即BD+DC=AB．
