\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[AD = \frac{4}{7}AC;\]
\[AE = \frac{4}{7}AB;\]
\[BC = 21\ см.\]
\[Найти:\]
\[\text{DE.}\]
\[Решение.\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}ADE - второй\ признак:\]
\[\angle BAC = \angle DAE;\]
\[\frac{\text{AB}}{\text{AE}} = \frac{7}{4};\ \ \ \]
\[\frac{\text{AC}}{\text{AD}} = \frac{7}{4}.\]
\[Отсюда:\]
\[\frac{\text{BC}}{\text{DE}} = \frac{\text{AB}}{\text{AE}} = \frac{7}{4}\text{\ \ \ }\]
\[DE = \frac{4}{7}BC = 12\ см.\]
\[Ответ:\ \ 12\ см.\]