$\begin{aligned} & {\left[\begin{array}{l}\text { Sum of the angles of a triangle. Theorem about the exterior angle. } \\ \text { [ Criteria and properties of an isosceles triangle. }\end{array}\right]}\end{aligned}$

In triangle $ABC$, angle $B$ is $20^{\circ}$, angle $C$ is $40^{\circ}$. The bisector $AD$ is 2. Find the difference between sides $BC$ and $AB$.
On side $B C$, lay off segment $B M$ equal to $A B$.

## Solution

On side $B C$, we lay off segment $B M$ equal to $A B$. In the isosceles triangle $A B M$, the angles at the base $A M$ are each $80^{\circ}$, so

$\angle C A M = \angle A M D - \angle A C B = 40^{\circ} = \angle A C M$.

Moreover, $\angle A D M = \angle A B C + \angle B A D = 80^{\circ} = \angle A M D$.

Thus, triangles $A M C$ and $A M D$ are isosceles. Therefore, $B C - A B = C M = A M = A D = 2$.

## Answer

2.