\[\boxed{\mathbf{537.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC};\]
\[\angle\text{BAD} = \angle\text{DAC};\]
\[\text{AB} = 14\ см;\]
\[\text{BC} = 20\ см;\]
\[\text{AC} = 21\ см.\]
\[\mathbf{Найти:}\]
\[\text{BD} - ?\]
\[\text{DC} - ?\]
\[\mathbf{Решение.}\]
\[1)\ \text{AH} - высота,\ общая\ для\ \]
\[\mathrm{\Delta}\text{ABD}\ и\ \mathrm{\Delta}\text{ACD}:\]
\[\frac{S_{\text{ABD}}}{S_{\text{ACD}}} = \frac{\text{BD}}{\text{CD}}.\]
\[2)\ \angle\text{BAD} = \angle\text{DAC}:\]
\[\frac{S_{\text{ABD}}}{S_{\text{ACD}}} = \frac{\text{AB} \bullet \text{AD}}{\text{AD} \bullet \text{AC}} = \frac{\text{AB}}{\text{AC}}.\]
\[3)\frac{\text{AB}}{\text{AC}} = \frac{\text{BD}}{\text{CD}}\]
\[\text{AC} \bullet \text{BD} = \text{AB} \bullet \text{CD}\]
\[\frac{\text{BD}}{\text{DC}} = \frac{\text{AB}}{\text{AC}}.\]
\[4)\ Пусть\ \text{BD} = x;\]
\[\text{DC} = \text{BC} - x = 20 - x.\]
\[5)\frac{x}{20 - x} = \frac{14}{21}\]
\[\frac{x}{20 - x} = \frac{2}{3}\]
\[3x = 2(20 - x) = 40 - 2x\]
\[5x = 40\]
\[x = 8\ см.\]
\[3)\ \text{DC} = 20 - 8 = 12\ см.\]
\[\mathbf{Ответ:}\text{BD} = 8\mathbf{\ см};\text{DC} = 12\ см.\]