\[\boxed{\mathbf{473.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC};\]
\[m - прямая;\]
\[\text{AB} \parallel m;\]
\[D \in m;\]
\[C \in m.\]
\[\mathbf{Доказать:}\]
\[S_{\text{ABC}} = S_{\text{ABD}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ S_{\text{ABC}} = \frac{1}{2} \bullet \text{AB} \bullet \text{CH};\]
\[S_{\text{ABD}} = \frac{1}{2} \bullet \text{AB} \bullet \text{DE}.\]
\[2)\ \text{AB} \parallel m;C \in m\ и\ D \in m;\]
\[\text{CH}\bot\text{AE}\ и\ \text{DE}\bot\text{AE}:\]
\[\text{CH} = \text{DE}.\]
\[3)\ \text{CH} = \text{DE} = h:\]
\[S_{\text{ABC}} = \frac{1}{2}\text{AB} \bullet \text{CH} = \frac{1}{2}\text{AB} \bullet \text{DE} =\]
\[= S_{\text{ABD}}.\]
\[4)\ S_{\text{ABC}} = S_{\text{ABD}}.\]
\[Что\ и\ требовалось\ доказать.\]