In △ ABC, the bisectors of ∠ B and ∠ C intersect at point O. it is proved that ∠ BOC = a + ABO + ACO In △ ABC, the bisectors of ∠ B and ∠ C intersect at point O. it is proved that ∠ BOC = a + ABO + ACO All we need is the process, you know, complete@

In △ ABC, the bisectors of ∠ B and ∠ C intersect at point O. it is proved that ∠ BOC = a + ABO + ACO In △ ABC, the bisectors of ∠ B and ∠ C intersect at point O. it is proved that ∠ BOC = a + ABO + ACO All we need is the process, you know, complete@

Let ∠ B bisector intersect AC at point D and ∠ C bisector intersect AB at point E
∠BOC=∠BDC+∠ACO,∠BDC=∠A+∠ABO
Therefore, BOC = a + ABO + ACO

The triangle ABC is connected by any point O in the middle, which are Bo and Co. the angle ABO is angle 1 and the angle ACO is angle 2. The relation between the angle BOC and the angle 1, 2 and BAC is discussed

∠BOC=∠1+∠2+∠BAC
Aboc is a quadrilateral, so the sum of internal angles is 360 & ordm;
Except for the angle of ∠ 1 ∠ 2 ∠ BAC + BOC = 360 & ordm;
The angle of ∠ 1 + 2 + BAC + is 360 & ordm;
∴
∠BOC=∠1+∠2+∠BAC

O is the inner point of the triangle ABC. Can you name it

In △ ABC, if ∠ a = 50 ° and the bisectors of ∠ B and ∠ C intersect at point O, then the degree of ∠ BOC is ()
A. 65°B. 100°C. 115°D. 130°

The angle bisector be and CF intersect at O, with ∫ a = 50 °, OBC + OCB = 12 (∫ ABC + ACB) = 65 ° and BOC = 180 ° - 65 ° = 115 ° so C is selected