The angle AOB = 80 degrees, OC is a ray outside the angle AOB, OD bisects the angle BOC, OE bisects the angle AOC, and the angle DOE needs geometric language There is another problem If the angle AOB = 80 degrees is changed to angle AOB = a degrees, what is the law obtained by finding ∠ doe?

The angle AOB = 80 degrees, OC is a ray outside the angle AOB, OD bisects the angle BOC, OE bisects the angle AOC, and the angle DOE needs geometric language There is another problem If the angle AOB = 80 degrees is changed to angle AOB = a degrees, what is the law obtained by finding ∠ doe?

∵ od bisection ∵ BOC, ∵ doc = (∵ BOC) / 2, ∵ OE bisection ∵ AOC, ∵ COE = (∵ AOC) / 2, and∵ DOE = ∵ doc - ∵ COE = (∵ BOC) / 2 - (? AOC) / 2 = (? BOC - ? AOC) / 2 = ? AOB / 2 = 80 ° / 2 = 40 °
Conclusion: if ∠ AOB = α, then ∠ DOE = α / 2, that is, the degree of ∠ DOE is always equal to half of ∠ AOB, which has nothing to do with the size of ∠ AOC. Similarly, it can be proved that if OC is a ray in ∠ AOB, the conclusion is the same