As shown in the figure: given ABO = 60, OC is the bisector of AOB, OD and OE bisector BOC and AOC respectively. When OC rotates around o point in AOB, ask if the degree of DOE is the same as before? Through this process, what conclusion do you come to?

As shown in the figure: given ABO = 60, OC is the bisector of AOB, OD and OE bisector BOC and AOC respectively. When OC rotates around o point in AOB, ask if the degree of DOE is the same as before? Through this process, what conclusion do you come to?

The same ∠ AOE = ∠ EOC ∠ cod = ∠ DOB ∠ AOE + ∠ EOC + ∠ doc + ∠ DOB = ∠ AOB
Therefore, EOC + doc = 1 / 2 AOB
No matter how OC changes, it is always = 1 / 2 ∠ AOB
Conclusion the bisector of the angle formed by the line segment (or ray) passing through a vertex in the triangle and the two adjacent triangle sides, and the angle of the bisector of the angle does not change (this is my summary, for reference only)