In the right-angled triangle $ABC$, the lengths of the legs $AC=3$ and $BC=4$. Point $A$ was moved to point $A_{1}$ parallel to $BC$, then point $B$ was moved to point $B_{1}$ parallel to $A_{1}C$, and finally, point $C$ was moved to point $C_{1}$ parallel to $A_{1}B_{1}$, such that the resulting triangle $A_{1}B_{1}C_{1}$ is right-angled at $B_{1}$, and the length of the leg $A_{1}B_{1}$ is 1 unit. What is the length of the leg $B_{1}C_{1}$?
Solution. The area of triangle $ABC$: $T_{ABC}=\frac{1}{2} \cdot 3 \cdot 4=6$. Since point $A$ is translated parallel to line $BC$, points $A_1$ and $A$ are at the same distance from line $BC$, so $T_{A_1BC}=T_{ABC}=6$. Then, point $B$ is translated parallel to line $A_1C$, so $B_1$ and $B$ are at the same distance from line $A_1C$, thus $T_{A_1B_1C}=T_{A_1BC}=6$. After translating point $C$ as well, we obtain triangle $A_1B_1C_1$, whose area, for similar reasons, is 6. Since we know that $\angle A_1B_1C_1=90^{\circ}$ and $B_1A_1=1$, we can write that

$$
T_{A_1B_1C_1}=\frac{B_1A_1 \cdot B_1C_1}{2}=6
$$

from which $B_1C_1=12$.