\[Схематический\ рисунок.\]
\[Дано:\]
\[P_{\text{ABC}} = 30\ см;\]
\[AE\ :BE = 3\ :2;\]
\[CF = 5\ см.\]
\[Найти:\]
\[AB;\ BC;\ AC.\]
\[Решение.\]
\[1)\ Окружность:\]
\[AE = AK;\ \ \ \]
\[BE = BF;\ \ \]
\[CF = CK.\]
\[2)\ В\ \mathrm{\Delta}ABC:\]
\[P_{\text{ABC}} = AB + BC + AC = 30\]
\[AE + BE + BF + CF + CK + AK = 30\]
\[AE + BE + BE + CF + CF + AE = 30\]
\[AE + BE + CF = 15\]
\[\frac{3}{2}BE + BE + 5 = 15\]
\[2,5BE = 10\ \ \]
\[BE = 4\ см.\]
\[AE = \frac{3}{2} \bullet 4 = 3 \bullet 2 = 6\ см.\]
\[AB = AE + BE = 6 + 4 = 10\ см.\]
\[BC = BF + CF = 4 + 5 = 9\ см.\]
\[AC = AK + CK = 6 + 5 = 11\ см.\]
\[Ответ:\ \ 9\ см;\ 10\ см;\ 11\ см.\]