\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[\angle AKM = \angle ACB.\]
\[Найти:\]
\[\text{BM.}\]
\[Решение.\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}AMK - первый\ \]
\[признак:\]
\[\angle ACB = \angle AKM;\]
\[\angle BAC = \angle MAK.\]
\[Отсюда:\]
\[\frac{\text{AB}}{\text{AM}} = \frac{\text{AC}}{\text{AK}}\]
\[AB = \frac{AM \bullet AC}{\text{AK}};\]
\[BM = AB - AM =\]
\[= \frac{AM \bullet AC}{\text{AK}} - AM.\]
\[Ответ:\ \ \frac{AM \bullet AC}{\text{AK}} - AM.\]