如图,在△ABC中,AB=AC,BD⊥AC,垂足为点D.求证,角DBC=二分之一∠A

方法一：作AF⊥BC于F∵AB=AC AF⊥BC∴∠CAF=∠BAF=1/2∠BAC∵AF⊥BC BD⊥AC∴∠CAF+∠C=∠DBC+∠C=90°∴∠DBC =∠CAF∴∠DBC=1/2*∠BAC方法二：∵AB=AC∴∠C=∠B =1/2(180°-∠A)=90°- (1/2*∠A)∵BD⊥AC∴∠DBC+...