\[\boxed{\mathbf{692.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AB = 10\ см;\]
\[BC = 12\ см;\]
\[CA = 5\ см.\]
\[\mathbf{Найти:}\]
\[AD,\ AE,\ EB,\]
\[BF,\ FC,\ CD.\]
\[\mathbf{Решение.}\]
\[1)\ По\ свойству\ касательныъ\ к\ \]
\[окружности:\ \]
\[AD = AE;EB = BF;FC = CD.\]
\[2)\ Пусть\ EB = x;\ \ \ AE = 10 - x;\ \ \]
\[FC = 1 - x:\]
\[AC = AD + DC = AE + FC\]
\[5 = 10 - x + 12 - x;\]
\[2x = 17\]
\[x = 8,5.\]
\[EB = BF = 8,5\ см.\]
\[3)\ AE = AD = 10 - 8,5 = 1,5\ см.\]
\[4)\ CD = FC = 12 - 8,5 = 3,5\ см.\]
\[Ответ:CD = FC = 3,5\ см;\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ AE = AE = 1,5\ см;\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ EB = BF = 8,5\ см.\ \]