ГДЗ по геометрии 9 класс Атанасян ФГОС Задание 643

\[\boxed{\mathbf{643.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}ABC;\]

\[BD - биссектриса\ \angle B;\]

\[\textbf{а)}\ BC = 9\ см;\]

\[AD = 7,5\ см;\]

\[DC = 4,5\ см;\]

\[\textbf{б)}\ AB = 30;\]

\[AD = 20;\]

\[BC = 16.\]

\[\mathbf{Найти:}\]

\[\textbf{а)}\ AB - ?\]

\[\textbf{б)}\ DC - ?\]

\[\mathbf{Решение.}\]

\[1)\ BH - высота,\ общая\ для\ \]

\[\mathrm{\Delta}\text{ABD\ }и\ \mathrm{\Delta}BCD:\ \]

\[\frac{S_{\text{ABD}}}{S_{\text{BCD}}} = \frac{\text{AD}}{\text{CD}}.\]

\[2)\ \angle ABD = \angle DBC:\]

\[\frac{S_{\text{ABD}}}{S_{\text{BCD}}} = \frac{AB \bullet BD}{BD \bullet BC} = \frac{\text{AB}}{\text{BC}}.\]

\[3)\ \frac{\text{AD}}{\text{CD}} = \frac{\text{AB}}{\text{BC}}\]

\[AD \bullet BC = CD \bullet AB\]

\[\frac{\text{AB}}{\text{AD}} = \frac{\text{BC}}{\text{CD}}.\]

\[\textbf{а)}\ AB = \frac{AD \bullet BC}{\text{CD}} = \frac{7,5 \bullet 9}{4,5} =\]

\[= 15\ см.\]

\[\textbf{б)}\ DC = \frac{AD \bullet BC}{\text{AB}} = \frac{20 \bullet 16}{30} =\]

\[= 10\frac{2}{3}.\]

\[\mathbf{Ответ:}\mathbf{а)}\ 15\ см;б)\ 10\frac{2}{3}.\]