\[Схематический\ рисунок.\]
\[Дано:\]
\[\frac{AB_{1}}{B_{1}B_{2}} = \frac{AC_{1}}{C_{1}C_{2}}.\]
\[Доказать:\]
\[B_{1}C_{1} \parallel B_{2}C_{2}.\]
\[Доказательство.\]
\[1)\ Проведем\ прямую:\]
\[B_{2}X \parallel B_{1}C_{1};\]
\[B_{2}X \cap AC_{1} = X.\]
\[2)\ По\ теореме\ Фалеса:\]
\[\frac{AB_{1}}{AC_{1}} = \frac{B_{1}B_{2}}{C_{1}X}\]
\[C_{1}X = \frac{AC_{1} \bullet B_{1}B_{2}}{AB_{1}}.\]
\[3)\ Из\ условия:\]
\[C_{1}C_{2} = \frac{AC_{1} \bullet B_{1}B_{2}}{AB_{1}} = C_{1}\text{X.}\]
\[4)\ Точки\ X\ и\ C_{2}\ совпадают:\]
\[B_{2}C_{2} \parallel B_{1}C_{1}.\]
\[Что\ и\ требовалось\ доказать.\]