\[Схематический\ рисунок.\]
\[Дано:\]
\[AM = 3BM;\]
\[CK = 3BK;\]
\[AC = 16\ см.\]
\[Найти:\]
\[MK \parallel AC;\]
\[\text{MK.}\]
\[Решение.\]
\[1)\ Отметим\ точки\ \text{E\ }и\ F:\]
\[AE = BE;\ \ \ \]
\[CF = BF.\]
\[2)\ \mathrm{\Delta}ABC:\]
\[EF - средняя\ линия;\]
\[EF \parallel AC;\ \ \]
\[EF = \frac{1}{2}AC = 8.\]
\[3)\ \mathrm{\Delta}EBF:\]
\[AB = AM + BM = 4BM;\]
\[BM = \frac{1}{4}AB;\ \ \]
\[EB = \frac{1}{2}AB;\]
\[BC = BK + CK = 4BK;\]
\[BK = \frac{1}{4}BC;\ \ \ \]
\[BF = \frac{1}{2}BC;\]
\[BM = \frac{1}{2}EB;\ \ \ \]
\[BK = \frac{1}{2}\text{BF.}\]
\[MK - средняя\ линия:\]
\[MK = \frac{1}{2}EF = 4\ см.\]
\[Отсюда:\]
\[MK \parallel EF \parallel AC.\]
\[Что\ и\ требовалось\ доказать.\]
\[Ответ:\ \ 4\ см.\]