\[\boxed{\mathbf{643.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[BD - биссектриса\ \angle B;\]
\[\textbf{а)}\ BC = 9\ см;\]
\[AD = 7,5\ см;\]
\[DC = 4,5\ см;\]
\[\textbf{б)}\ AB = 30;\]
\[AD = 20;\]
\[BC = 16.\]
\[\mathbf{Найти:}\]
\[\textbf{а)}\ AB - ?\]
\[\textbf{б)}\ DC - ?\]
\[\mathbf{Решение.}\]
\[1)\ BH - высота,\ общая\ для\ \]
\[\mathrm{\Delta}\text{ABD\ }и\ \mathrm{\Delta}BCD:\ \]
\[\frac{S_{\text{ABD}}}{S_{\text{BCD}}} = \frac{\text{AD}}{\text{CD}}.\]
\[2)\ \angle ABD = \angle DBC:\]
\[\frac{S_{\text{ABD}}}{S_{\text{BCD}}} = \frac{AB \bullet BD}{BD \bullet BC} = \frac{\text{AB}}{\text{BC}}.\]
\[3)\ \frac{\text{AD}}{\text{CD}} = \frac{\text{AB}}{\text{BC}}\]
\[AD \bullet BC = CD \bullet AB\]
\[\frac{\text{AB}}{\text{AD}} = \frac{\text{BC}}{\text{CD}}.\]
\[\textbf{а)}\ AB = \frac{AD \bullet BC}{\text{CD}} = \frac{7,5 \bullet 9}{4,5} =\]
\[= 15\ см.\]
\[\textbf{б)}\ DC = \frac{AD \bullet BC}{\text{AB}} = \frac{20 \bullet 16}{30} =\]
\[= 10\frac{2}{3}.\]
\[\mathbf{Ответ:}\mathbf{а)}\ 15\ см;б)\ 10\frac{2}{3}.\]