\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[AF = 1,5\ м;\]
\[AC = 39\ м;\]
\[EH = 3\ м;\]
\[AG = 1,8\ м.\]
\[Найти:\]
\[\text{BD.}\]
\[Решение.\]
\[1)\ AFHG - прямоугольник:\]
\[FH = AG = 1,8\ м.\]
\[2)\ ACDG - прямоугольник:\]
\[CD = AG = 1,8\ м.\]
\[3)\ В\ \mathrm{\Delta}ABC:\]
\[EF = EH - FH = 1,2\ м.\]
\[EF\bot AC;\ \ \ \]
\[BC\bot AC;\]
\[EF \parallel BC.\ \ \ \]
\[Следовательно:\]
\[\mathrm{\Delta}AEF\sim\mathrm{\Delta}ABC.\]
\[4)\ \frac{\text{BC}}{\text{EF}} = \frac{\text{AC}}{\text{AF}}\text{\ \ \ }\]
\[BC = \frac{EF \bullet AC}{\text{AF}}\]
\[BC = \frac{1,2 \bullet 39}{1,5} = 31,2\ м.\]
\[BD = BC + CD = 33\ м.\]
\[Ответ:\ \ 33\ м.\]