\[Схематический\ рисунок.\]
\[Дано:\]
\[\angle E = \angle K;\]
\[DE = 2,5 \bullet MK;\]
\[EF = 2,5 \bullet KN;\]
\[DF - MN = 30\ см.\]
\[Найти:\]
\[DF;\ MN.\]
\[Решение.\]
\[\mathrm{\Delta}DEF\sim\mathrm{\Delta}MKN - второй\ признак:\]
\[\frac{\text{DE}}{\text{MK}} = \frac{\text{EF}}{\text{KN}} = 2,5;\]
\[\angle E = \angle K.\]
\[Отсюда:\]
\[\frac{\text{DF}}{\text{MN}} = \frac{\text{DE}}{\text{MK}} = 2,5\]
\[DF = 2,5 \bullet MN.\]
\[2,5 \bullet MN - MN = 30\]
\[1,5 \bullet MN = 30\]
\[MN = 20\ см.\]
\[DF = 2,5 \bullet 20 = 50\ см.\]
\[Ответ:\ \ 20\ см;\ 50\ см.\]