7.4. In triangle $A B C$, the angles $A$ and $C$ at the base are $20^{\circ}$ and $40^{\circ}$, respectively. It is known that $A C - A B = 5$ (cm). Find the length of the angle bisector of angle $B$.
Answer: 5 cm. Solution. Let $B M$ be the bisector of angle $B$. Mark a point $N$ on the base $A C$ such that $A N=A B$. Then triangle $A B N$ is isosceles and $\angle A B N=\angle A N B=80^{\circ}$. Since $\angle A B M=\frac{180^{\circ}-20^{\circ}-40^{\circ}}{2}=60^{\circ}$, then $\angle B M N=\angle A+\angle A B M=20^{\circ}+60^{\circ}=80^{\circ}$. Thus, the angles at the base of $\triangle M B N$ are equal, which means $\triangle M B N$ is isosceles: $M B=B N$. In $\triangle B N C$, the angles $\angle N C B$ and $\angle C N B$ are equal to $40^{\circ}$ (since $\angle B N C=180^{\circ}-80^{\circ}=100^{\circ}$ and $\angle N B C=180^{\circ}-40^{\circ}-100^{\circ}=40^{\circ}$). Therefore, $\triangle B N C$ is isosceles: $B N=N C$. As a result, we have $B M=B N=N C=5$ cm (since $N C=A C-A N=A C-A B)$.

## 8th grade