In triangle ABC, angle B is equal to 100. The bisector of angle C intersects AB at point E.

In triangle ABC, angle B is equal to 100. The bisector of angle C intersects AB at point E. Point D is taken on the AC side, so that angle CBD is 20. Find angle CED.

Let’s build a point P, which will be symmetric to point C with respect to line AB. The angle PBA is symmetrical to the angle CBA relative to the straight line AB, therefore the angle PBA = angle CBA = 100 °.
Angle PBD = angle PBA + angle DBA = 100 ° + 80 ° = 180 °, therefore points P, B and D lie on the same straight line PB.
The angle BPE is symmetrical to the angle ALL, therefore they are equal.
Points P, E, D, C lie on the same circle. RVS angle = 360 ° – РВА angle – СВА angle = 160 °. РВ = ВС, hence the RBC triangle is isosceles. It follows from this that the angle DPC = (180 ° – РВС angle) / 2 = (180 ° – 160 °) / 2 = 10 °.
Angle CED = angle DPC, since they are inscribed in one circle and rest on one arc, therefore the angle CED = 10 °.