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As shown in the figure, OE bisects ∠ BOC, OD bisects ∠ AOC, ∠ BOE = 20 ° and ∠ AOD = 40 ° to calculate the degree of ∠ doe
∵ OE bisection ∠ BOC, ∵ BOE = ∠ COE = 20 °, ∵ od bisection ∠ AOC, ∵ AOD = ∠ cod = 40 degree. ∵ DOE = ∠ COE + ∠ cod = 20 ° + 40 degree = 60 degree. So the answer is 60 degree
As shown in the figure, the bisectors of the two external angles CBD and BCE of ABC intersect at O. if AB is parallel to CO, the ratio of cable angle ACB and angle BOC is tested
Let ∠ ACB = 1, ∠ BOC = 2, ∠ BCO = X
∵ ab ∥ OC, Bo bisection ∠ CBD
∴∠2=∠OBD=∠OBC
It can be obtained from △ BOC
2∠2+x=180°.①
∵ co bisection ∠ BCE
∴∠1+2x=180°.②
It can be obtained from (1) and (2)
4∠2-∠1=180°
That is, 4 ∠ BOC - ∠ ACB = 180 degree
As shown in the figure, a: the bisectors of ∠ CBD and ∠ BCE of the two outer angles of △ ABC intersect at O, then: ∠ BOC = 90 ° - half ∠ A. why? Why
Because Bo is divided into CBD and CO is divided equally
Therefore, CBO = CBD / 2 and BCO = BCE / 2
So ∠ CBO + ∠ BCO = ∠ CBD / 2 + ∠ BCE / 2 = (∠ CBD + ∠ BCE) / 2
So in △ BCO, from the theorem of the sum of internal angles of triangles, we get,
∠BOC=180-(∠CBO+∠BCO)
=180-(∠CBD+∠BCE)/2
=180-(180-∠ABC+180-∠ACB)/2
=180-(360-∠ABC-ACB)/2
=180-180+(∠ABC+∠ACB)/2
=(180-∠A)/2
=90-∠A/2
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