Question

如图,已知o是∠ABC,∠ACB的平分线BO、CO的交点,过O作EF//BC分别交AB于点E,交AC于F,

Answer 1

∵∠AEF=38°,∠AFE=80°

∴∠A=62°

∵CO平分 ∠ACB

∴∠OCB=1/2∠ACB

同理∠OBC=1/2∠ABC

∴∠BOC=180°-∠OCB-∠OBC

=180°-(1/2∠ACB+1/2∠ABC)

=180°-1/2(∠ACB+∠ABC)

=180°-1/2(180°-∠A)

=180°-1/2(180°-62°)=121°