结论是:∠B=2∠C
证明:
在AC上截取AE=AB
∵∠BAD=∠EAD,AD=AD
∴△ABD≌△AED
∴BD=DE,∠B=∠AED
∵AB=AC-BD
∴CE =BD =DE
∴∠C =∠EDC
∴∠AED=∠EDC+∠C=2∠C
∴∠B =2∠C
做DE=CD
∵AD平分∠BAC
∴∠DAC=∠DAB
∵AD=AD DE=CD
∴△ADC≌△ADE
∵AB=AC-BD
∴CD=BE=DE
∴∠B=∠EDC ∠C=∠DEA
∵∠B+∠C+∠A=180° ∠AED=∠B+EDB
∴∠C=∠B+∠A