\[\boxed{\mathbf{560.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано}:\ \]
\[\mathrm{\Delta}\text{ABC\ }и\ \mathrm{\Delta}A_{1}B_{1}C_{1}.\]
\[Доказать:\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1}.\]
\[Доказательство.\]
\[\textbf{а)}\ AB = 3\ см;BC = 5\ см;\]
\[CA = 7\ см;A_{1}B_{1} = 4,5\ см;\ \]
\[B_{1}C_{1} = 7,5\ см:\]
\[C_{1}A_{1} = 10,5\ см;\]
\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{3}{4,5} = \frac{2}{3};\]
\[\frac{\text{BC}}{B_{1}C_{1}} = \frac{5}{7,5} = \frac{2}{3};\]
\[\frac{\text{AC}}{A_{1}C_{1}} = \frac{7}{10,5} = \frac{2}{3};\]
\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{\text{BC}}{B_{1}C_{1}} = \frac{\text{AC}}{A_{1}C_{1}} = \frac{2}{3} \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1}\ \]
\[(по\ трем\ сторонам).\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{б)}\ AB = 1,7\ см;BC = 3\ см;\]
\[CA = 4,2\ см;A_{1}B_{1} = 34\ дм;\]
\[B_{1}C_{1} = 60\ дм;\]
\[C_{1}A_{1} = 84\ дм:\]
\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{1,7}{340} = \frac{1}{200};\]
\[\frac{\text{BC}}{B_{1}C_{1}} = \frac{3}{600} = \frac{1}{200};\]
\[\frac{\text{AC}}{A_{1}C_{1}} = \frac{4,2}{840} = \frac{1}{200};\]
\[\frac{\text{AB}}{A_{1}B_{1}} = \frac{\text{BC}}{B_{1}C_{1}} = \frac{\text{AC}}{A_{1}C_{1}} = \frac{1}{200} \Longrightarrow\]
\[\Longrightarrow \mathrm{\Delta}ABC\sim\mathrm{\Delta}A_{1}B_{1}C_{1}\ \]
\[(по\ трем\ сторонам)\mathbf{.}\]
\[\mathbf{Что\ и\ требовалось\ доказать}\mathbf{.}\]