## [ Vectors help solve the task]

Auto: Zhenedarov $\mathbf{P . 5}$.

The lengths of the sides of triangle $A B C$ are $a, b$ and $c$ ( $A B=c, B C=a, C A=b$ and $a<b<c$ ). On the rays $B C$ and $A C$, points $B_{1}$ and $A_{1}$ are marked respectively such that $B B_{1}=A A_{1}=c$. On the rays $C A$ and $B A$, points $C_{2}$ and $B_{2}$ are marked respectively such that $C C_{2}=B B_{2}=a$. Find $A_{1} B_{1}: C_{2} B_{2}$.
Let $\alpha=c: a . \quad \overrightarrow{A_{1} B_{1}}=\overrightarrow{A_{1} A}+\overrightarrow{A B}+\overrightarrow{B B_{1}}=\alpha \overrightarrow{C C_{2}}+\alpha \overrightarrow{B_{2} B}+\alpha \overrightarrow{B C}=\alpha\left(\overrightarrow{B_{2} B}+\overrightarrow{B C}+\overrightarrow{C C_{2}}\right)=\alpha \overrightarrow{B_{2} C_{2}}$.

## Answer

$c: a$.