∵AB=AC ∴∠B=∠C ∵AD=CD ∴∠1=∠C ∵AB=BD ∴∠2=∠3 = ∠1+∠C 设∠1=∠C=∠B= x 则 ∠2 = ∠3 = 2x △ABC内角和180° ∠C+∠B+∠1+∠2=180° x+x+x+2x=180° x = 36° ∴∠B=∠C=36° ∠BAC = 180° - 2x = 108°
