It is known that, as shown in the figure, in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o

It is known that, as shown in the figure, in △ ABC, the bisector of ∠ ABC and ∠ ACB intersects at point o

It is proved that: the bisector of ∵ ABC and ∵ ACB intersects at point O, ∵ OBC = 12 ∵ ABC, ∵ OCB = 12 ∵ ACB, ∵ OBC + ∵ OCB = 12 (∵ ABC + ∵ ACB). In △ OBC, ∵ BOC = 180 ° - (∵ OBC + ? OCB) = 180 ° - 12 (? ABC + ? ACB) = 180 ° - 12 (180 ° - a) = 90 ° + 12 ? a

Given the intersection o of bisector be and CF of angle B and angle C in triangle ABC, we prove that the angle BOC = 90 ° and half angle A

In the triangle ABC, there are: 180 degree angle a = angle ABC + angle ACB. In the triangle BOC, there are: 180 degree angle BOC = angle EBC + angle FCA. Because be and CF bisect angle ABC and angle ACB respectively, angle EBC + angle FCB = 1 / 2 (angle ABC + angle ACB)

It is known that in the triangle ABC, if angle a = 60 degrees, the bisector of angle ABC and angle ACB intersects at point 0, then the degree of angle BOC is

Angle ABC + angle ACB = 120 degrees
So angle ABO + angle ACO = 1 / 2 * 120 = 60 degrees
So the angle BOC = 120 degrees

In the triangle ABC, the bisector of angle ABC and angle ACB intersects at point O. try to explain that the angle BOC = 90 ° + & frac12; angle A
It's a question on page 47 of the basic training in Volume 2 of Grade 7 - I can't think about it for a long time.

The bisector of angle ABC and angle ACB intersects at point o
∴∠OBC=∠ABC/2,∠OCB=∠ACB/2,
∴∠OBC+OCB=∠ABC/2+∠ACB/2=(1/2)(∠ABC+∠ACB)=(1/2)(180-∠A)=90-∠A/2
That is, the angle BOC = 90 ° + & frac12; the angle A