\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[AB = 21\ см;\]
\[AC = 42\ см;\]
\[BC = 28\ см;\]
\[BM = 8\ см;\]
\[BK = 6\ см.\]
\[Найти:\]
\[\text{KM.}\]
\[Решение.\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}KBM - второй\ признак:\]
\[\frac{\text{AB}}{\text{BK}} = \frac{21}{6} = \frac{7}{2};\]
\[\frac{\text{BC}}{\text{BM}} = \frac{28}{8} = \frac{7}{2};\]
\[\angle ABC = \angle KBM - вертикальные.\]
\[Отсюда:\]
\[\frac{\text{AC}}{\text{KM}} = \frac{\text{AB}}{\text{BK}} = \frac{7}{2}\]
\[KM = \frac{2}{7}AC = 12\ см.\]
\[Ответ:\ \ 12\ см.\]