[Sum of angles in a triangle. Theorem about the exterior angle. ]  [ Lengths of sides, heights, medians, and angle bisectors ]

In triangle $A B C \angle A=40^{\circ}, \angle B=20^{\circ}$, and $A B-B C=4$. Find the length of the angle bisector of angle $C$.
Let's set aside segment $BD$ on side $AB$, equal to $BC$. Then triangle $BCD$ is isosceles with the angle at the vertex $20^{\circ}$, so the angles at the base are $80^{\circ}$ (see the figure). Let $CE$ be the bisector of angle $C$. Then $\angle BCE = 60^{\circ}$, so $\angle AEC = 20^{\circ} + 60^{\circ} = 80^{\circ}$. Thus, in triangle $DEC$, two angles are equal, so it is isosceles. The angle at its vertex $C$ is $20^{\circ}$, so $\angle ACD = 40^{\circ}$. Therefore, triangle $ACD$ is also isosceles, hence $CE = CD = AD = AB - BC = 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-37.jpg?height=426&width=1334&top_left_y=2011&top_left_x=362)

## Answer

4