\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[BDEK - ромб;\]
\[AB = 10\ см;\]
\[BC = 15\ см.\]
\[Найти:\]
\[\text{BD.}\]
\[Решение.\]
\[1)\ BDEK - ромб:\]
\[KB = BD = DE = EK;\]
\[EK \parallel BD;\ \ \ \]
\[ED \parallel BK.\]
\[2)\ \mathrm{\Delta}AKE\sim\mathrm{\Delta}ABC:\]
\[EK \parallel BC.\]
\[Значит:\]
\[AK = AB - BK = 10 - EK.\]
\[3)\ \frac{\text{AK}}{\text{AB}} = \frac{\text{AE}}{\text{AC}} = \frac{\text{EK}}{\text{BC}};\ \ \ \]
\[\frac{\text{AK}}{10} = \frac{\text{EK}}{15}\]
\[15AK = 10EK\]
\[3(10 - EK) = 2EK\]
\[30 - 3EK = 2EK\]
\[5EK = 30\ \ \ \]
\[EK = 6\ см.\]
\[Ответ:\ \ 6\ см.\]