\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[BMKN - квадр;\]
\[\angle B = 90{^\circ};\]
\[BM = 6\ см;\]
\[AB = 10\ см.\]
\[Найти:\]
\[\text{CN.}\]
\[Решение.\]
\[1)\ BMKN - квадрат:\]
\[BN = MK = BM = 6\ см;\]
\[BM \parallel KN;\ \ \ \]
\[BN \parallel MK.\]
\[2)\ \mathrm{\Delta}AMK\sim\mathrm{\Delta}ABC:\]
\[MK \parallel BC.\]
\[Значит:\]
\[AM = AB - BM = 4\ см.\]
\[3)\ \frac{\text{AM}}{\text{AB}} = \frac{\text{MK}}{\text{BC}}\text{\ \ }\]
\[\frac{4}{10} = \frac{6}{\text{BC}}\text{\ \ \ }\]
\[BC = 15\ см.\]
\[CN = BC - BN = 9\ см.\]
\[Ответ:\ \ 9\ см.\]