\[Схематический\ рисунок.\]
\[Дано:\]
\[ABCD - прямоугольник;\]
\[OH\bot AD;\]
\[OH = 7\ см.\]
\[Найти:\]
\[\text{AB.}\]
\[Решение.\]
\[1)\ ABCD - прямоугольник:\]
\[AB\bot AD;\ \ \ \]
\[BO = DO.\]
\[2)\ По\ теореме\ Фалеса:\]
\[OH\bot AD;\ \ \ \]
\[AB\bot AD;\ \ \ \]
\[OH \parallel AB;\]
\[\frac{\text{AH}}{\text{BO}} = \frac{\text{DH}}{\text{DO}};\ \ \]
\[\frac{\text{AH}}{\text{DH}} = \frac{\text{BO}}{\text{DO}} = 1.\]
\[3)\ В\ \mathrm{\Delta}ADB:\]
\[AH = DH;\ \ \ \]
\[BO = DO.\]
\[OH - средняя\ линия:\]
\[AB = 2OH = 14\ см.\]
\[Ответ:\ \ 14\ см.\]