\[\boxed{\mathbf{353}.еуроки - ответы\ на\ пятёрку}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\]
\[AB - отрезок;\]
\[C \in AB;\]
\[AC\ :BC = 2\ :1;\]
\[D \in AC;\]
\[AD\ :CD = 3\ :2.\]
\[Найти:\]
\[AD\ :DB.\]
\[Решение:\]
\[Пусть\ AB = x;\ \ AC = \frac{2}{3}x;\ \ \]
\[BC = \frac{1}{3}\text{x.}\]
\[AD = \frac{3}{5}AC = \frac{3}{5} \cdot \frac{2}{3}x = \frac{2}{5}x;\]
\[BD = AB - BD = x - \frac{2}{5}x = \frac{3}{5}x;\]
\[AD\ :BD = \frac{2}{5}x\ :\frac{3}{5}x = 2\ :3.\]
\[Ответ:AD\ :DB = 2\ :3.\]