\[\boxed{\mathbf{571.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AA_{1},\ BB_{1} - медианы;\]
\[AA_{1} \cap BB_{1} = O;\]
\[S_{\text{ABO}} = S.\]
\[\mathbf{Найти:}\]
\[S_{\text{ABC}} - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABC}} = \frac{1}{2}AB \bullet CH;\ \ \ \]
\[S_{\text{ABO}} = \frac{1}{2}AB \bullet OH_{1}.\]
\[2)\frac{\text{AO}}{OA_{1}} = \frac{\text{CO}}{OC_{1}} = \frac{\text{BO}}{OB_{1}} =\]
\[= \frac{2}{1}\ (по\ свойству\ медианы).\]
\[3)\frac{S_{\text{AOB}}}{S_{\text{OB}A_{1}}} = \frac{\text{AO}}{OA_{1}} = \frac{2}{1}\]
\[\frac{S}{S_{\text{OB}A_{1}}} = 2\]
\[S_{\text{OB}A_{1}} = \frac{S}{2}.\]
\[4)\frac{S_{\text{AOB}}}{S_{\text{AO}B_{1}}} = \frac{\text{BO}}{OB_{1}} = \frac{2}{1}\]
\[\frac{S}{S_{\text{AO}B_{1}}} = 2\]
\[S_{\text{AO}B_{1}} = \frac{S}{2}.\]
\[5)\ S_{\text{AB}A_{1}} = S_{AA_{1}C}\ \]
\[(как\ равновеликие).\]
\[6)\ S_{\text{AB}A_{1}} = S_{\text{AOB}} + S_{BA_{1}O} =\]
\[= S + \frac{S}{2} = \frac{3}{2}\text{S.}\]
\[7)\ S_{\text{ABC}} = S_{\text{AB}A_{1}} + S_{AA_{1}C} =\]
\[= \frac{3}{2}S + \frac{3}{2}S = 3S.\]
\[Ответ:S_{\text{ABC}} = 3S.\]
\[\boxed{\mathbf{571.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[m - прямая;\]
\[AB \parallel m;\]
\[D \in m;\]
\[C \in m.\]
\[\mathbf{Доказать:}\]
\[S_{\text{ABC}} = S_{\text{ABD}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ S_{\text{ABC}} = \frac{1}{2} \bullet AB \bullet CH;\]
\[S_{\text{ABD}} = \frac{1}{2} \bullet AB \bullet DE.\]
\[2)\ AB \parallel m;C \in m\ и\ D \in m;\]
\[CH\bot AE\ и\ DE\bot AE:\]
\[CH = DE.\]
\[3)\ CH = DE = h:\]
\[S_{\text{ABC}} = \frac{1}{2}AB \bullet CH = \frac{1}{2}AB \bullet DE =\]
\[= S_{\text{ABD}}.\]
\[4)\ S_{\text{ABC}} = S_{\text{ABD}}.\]
\[Что\ и\ требовалось\ доказать.\]