[ Sum of angles in a triangle. Theorem about the exterior angle.]

$B K$ is the bisector of triangle $A B C$. It is known that $\angle A K B: \angle C K B=4: 5$. Find the difference between angles $A$ and $C$ of triangle $A B C$.

#
Find the angles $A K B$ and $C K B$ and use the theorem about the exterior angle of a triangle.

## Solution

Since $\angle A K B+\angle C K B=180^{\circ}$, then $\angle A K B=80^{\circ}, \angle C K B=100^{\circ}$.

Let $\angle A B K=\angle C B K=\beta$. By the theorem about the exterior angle of a triangle $80^{\circ}=\angle C+\beta, 100^{\circ}=\angle A+\beta, \angle A-\angle C$ $=20^{\circ}$.

## Answer

$20^{\circ}$.