∵AD=DC∴∠C=∴∠DAC又AB=BD∴∠BDA=BAD=2∠C又AB=AC∴∠C=∠B∴∠B+∠BDA+∠BAD=180°即5∠B=180°，∠B=36°∴∠BAC=∠BAD+∠DAC=3∠B=108°
