Question

在三角形ABC中,∠ABC和∠ACB的平分线交与点I,求证：∠BIC=90度+1|2∠A

Answer 1

∠BIC=180°-∠IBC-∠ICB

=180°-∠ABC/2-∠ACB/2

=180°-∠ABC/2-∠ACB/2-∠BAC/2+∠BAC/2

=180°-(∠ABC/2+∠ACB/2+∠BAC/2)+∠BAC/2

=180°-(∠ABC+∠ACB+∠BAC)/2+∠BAC/2

=180°-180°/2+∠BAC/2

=90°+∠A/2