Question

已知AB=AC,BD⊥AC,求证:∠DBC=1/2∠A

Answer 1

过点A作AE⊥BC,交BD于点F,交BC于点E
∵BD⊥AC
∴∠AEB=∠ADB=90º
∵∠BFE=∠AFD
∴90º-∠BFE=90º-∠AFD
即 ∠DBC=∠CAE
∵AB=AC,AE⊥BC
∴∠CAE=1/2∠BAC
∴∠DBC=1/2∠BAC

Answer 2

证明：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