\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[AE \parallel BF \parallel CM \parallel DK;\]
\[AB = 25\ см;\]
\[BC = 20\ см;\]
\[CD = 35\ см;\]
\[EK = 48\ см.\]
\[Найти:\]
\[EF;\ FM;\ MK.\]
\[Решение:\]
\[1)\ AD = AB + BC + CD =\]
\[= 25 + 20 + 35 = 80\ см.\]
\[2)\ По\ теореме\ Фалеса:\]
\[\frac{\text{EF}}{\text{AB}} = \frac{\text{FM}}{\text{BC}} = \frac{\text{MK}}{\text{CD}} = \frac{\text{EK}}{\text{AD}} = \frac{3}{5};\]
\[EF = \frac{3}{5} \bullet AB = 15\ см;\]
\[FM = \frac{3}{5} \bullet BC = 12\ см;\]
\[MK = \frac{3}{5} \bullet CD = 21\ см.\]
\[Ответ:\ \ 15\ см;\ 12\ см;\ 21\ см.\]