\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[\angle ACB = \angle DFE;\]
\[FE = 8,4\ м;\]
\[AB = 2\ м;\]
\[CB = 2,4\ м.\]
\[Найти:\]
\[\text{DE.}\]
\[Решение.\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}DEF - первый\ \]
\[признак:\]
\[\angle ACB = \angle DFE;\]
\[\angle ABC = \angle DEF = 90{^\circ}.\]
\[Отсюда:\]
\[\frac{\text{DE}}{\text{AB}} = \frac{\text{FE}}{\text{BC}}\]
\[\frac{\text{DE}}{2} = \frac{8,4}{2,4}\]
\[DE = 7\ м.\]
\[Ответ:\ \ 7\ м.\]