O is the outer center of the triangle ABC. If vector Ao multiplies vector AB = vector Bo multiplies vector BC, the shape of the triangle ABC can be helped by God

O is the outer center of the triangle ABC. If vector Ao multiplies vector AB = vector Bo multiplies vector BC, the shape of the triangle ABC can be helped by God

Because o is the outer center, then the distance to the three points is equal. Since Ao × AB = Bo × BC, Ao = Bo, then AB = BC, so the triangle ABC is isosceles triangle

If we know that the bisectors Bo and CO of angle ABC and angle ACB intersect point O in triangle ABC, what is the conclusion

∠BOC=90°+1/2∠A
Reason: in △ ABC, ∠ a + ABC + ACB = 180 ° and ∠ OBC = 1 / 2 ∠ ABC, ∠ OCB = 1 / 2 ∠ ACB
∠OBC+∠OCB=1/2∠A BC+1/2∠A CB=1/2（∠ABC +∠ACB）=1/2（180°-∠A）=90°-1/2∠A
So ∠ BOC = 180 ° - (∠ OBC + ∠ OCB) = 180 ° - (90 ° - 1 / 2 ∠ a) = 90 ° + 1 / 2 ∠ a

O is the outer center of the acute triangle ABC, the circumscribed circle of the straight line Bo intersects at another point D, and the circumscribed circle of the high extension intersection made by a intersects at point E, which proves that s △ becd = s △ ABC

Make ABC's perpendicular H
It is easy to prove that ahcd is a parallelogram, so CD = ah
EH is divided equally by BC, so if a's foot on BC is k
So EK + CD = AK
So the areas are equal

If △ ABC is an isosceles triangle, then the degree of ∠ A and ∠ B can be
A.∠A=60°,∠B=50°
B.∠A=70°,∠B=40°
C.∠A=80°,∠B=60°
D.∠A=90°,∠B=30°

B 70 + 70 + 40 =180