Question

如图,在△ABC中,已知∠ABC与∠ACB的平分线相交于点O,求证：

Answer 1

证明：【1】∴∠OBC=1/2∠ABC ∠OCB=1/2∠ACB∵∠ABC+∠ACB+∠A=180°∴1/2∠ABC+1/2∠ACB=90°-1/2∠A∴∠OBC+∠OCB=90°-1/2∠A∵∠OBC+∠OCB+∠BOC=180°∴∠BOC=180°-(∠OBC+∠OCB)=180°-(90°-1/2∠A)=90°+1...