\[Схематический\ рисунок.\]
\[Дано:\]
\[CM = 15\ см;\]
\[CK = 12\ см;\]
\[AC = 20\ см;\]
\[BC = 25\ см;\]
\[AB = 30\ см.\]
\[Найти:\]
\[\text{MK.}\]
\[Решение.\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}MKC - второй\ признак:\]
\[\angle ACB = \angle MCK;\]
\[\frac{\text{CM}}{\text{BC}} = \frac{15}{25} = \frac{3}{5};\]
\[\frac{\text{CK}}{\text{AC}} = \frac{12}{20} = \frac{3}{5}.\]
\[Отсюда:\]
\[\frac{\text{MK}}{\text{AB}} = \frac{\text{CM}}{\text{BC}} = \frac{3}{5}\]
\[MK = \frac{3}{5}AB = 18\ см.\]
\[Ответ:\ \ 18\ см.\]