\[\boxed{\mathbf{535.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC};\]
\[\angle 1 = \angle 2.\]
\[\mathbf{Доказать:}\]
\[\frac{\text{BD}}{\text{AB}} = \frac{\text{CD}}{\text{AC}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ \text{AH} - высота,\ общая\ для\ \]
\[\mathrm{\Delta}\text{ABD}\ и\ \mathrm{\Delta}\text{ACD}:\]
\[\frac{S_{\text{ABD}}}{S_{\text{ACD}}} = \frac{\text{BD}}{\text{CD}}.\]
\[2)\ \angle 1 = \angle 2:\]
\[\frac{S_{\text{ABD}}}{S_{\text{ACD}}} = \frac{\text{AB} \bullet \text{AD}}{\text{AD} \bullet \text{AC}} = \frac{\text{AB}}{\text{AC}}.\]
\[3)\frac{\text{AB}}{\text{AC}} = \frac{\text{BD}}{\text{CD}}\]
\[\text{AC} \bullet \text{BD} = \text{AB} \bullet \text{CD}\]
\[\frac{\text{BD}}{\text{AB}} = \frac{\text{CD}}{\text{AC}}.\]
\[Что\ и\ требовалось\ доказать\]