\[\boxed{\mathbf{536.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}\text{ABC};\]
\[\text{BD} - биссектриса\ \angle B;\]
\[\textbf{а)}\ \text{BC} = 9\ см;\]
\[\text{AD} = 7,5\ см;\]
\[\text{DC} = 4,5\ см;\]
\[\textbf{б)}\ \text{AB} = 30;\]
\[\text{AD} = 20;\]
\[\text{BC} = 16.\]
\[\mathbf{Найти:}\]
\[\textbf{а)}\ \text{AB} - ?\]
\[\textbf{б)}\ \text{DC} - ?\]
\[\mathbf{Решение.}\]
\[1)\ \text{BH} - высота,\ общая\ для\ \]
\[\mathrm{\Delta}\text{ABD}\ и\ \mathrm{\Delta}\text{BCD}:\ \]
\[\frac{S_{\text{ABD}}}{S_{\text{BCD}}} = \frac{\text{AD}}{\text{CD}}.\]
\[2)\ \angle\text{ABD} = \angle\text{DBC}:\]
\[\frac{S_{\text{ABD}}}{S_{\text{BCD}}} = \frac{\text{AB} \bullet \text{BD}}{\text{BD} \bullet \text{BC}} = \frac{\text{AB}}{\text{BC}}.\]
\[3)\ \frac{\text{AD}}{\text{CD}} = \frac{\text{AB}}{\text{BC}}\]
\[\text{AD} \bullet \text{BC} = \text{CD} \bullet \text{AB}\]
\[\frac{\text{AB}}{\text{AD}} = \frac{\text{BC}}{\text{CD}}.\]
\[\textbf{а)}\ \text{AB} = \frac{\text{AD} \bullet \text{BC}}{\text{CD}} = \frac{7,5 \bullet 9}{4,5} =\]
\[= 15\ см.\]
\[\textbf{б)}\ \text{DC} = \frac{\text{AD} \bullet \text{BC}}{\text{AB}} = \frac{20 \bullet 16}{30} =\]
\[= 10\frac{2}{3}.\]
\[\mathbf{Ответ:}\mathbf{а)}\ 15\ см;б)\ 10\frac{2}{3}.\]