证明：在AB上截取AE=AC,并连接DE
∵∠CAD=∠BAD
AD=AD
∴△CAD全等于△EAD
∴∠C=∠AED
CD=ED
∵∠AED=∠B+∠BDE
∠C=2∠B
∴∠AED=2∠B=∠B+∠BDE
∴∠B=∠BDE
∴BE=DE=CD
∵AB=BE+AE
∴AB=AC+CD