\[Схематический\ рисунок.\]
\[Дано:\]
\[AB = 8\ см;\]
\[BC = 12\ см;\]
\[AC = 16\ см;\]
\[CD = 9\ см.\]
\[Найти:\]
\[\text{BD.}\]
\[Решение.\]
\[\mathrm{\Delta}ABC\sim\mathrm{\Delta}BDC - второй\ признак:\]
\[\angle BCD = \angle ACB;\]
\[\frac{\text{BC}}{\text{AC}} = \frac{12}{16} = \frac{3}{4};\]
\[\frac{\text{CD}}{\text{BC}} = \frac{9}{12} = \frac{3}{4}.\]
\[Отсюда:\]
\[\frac{\text{BD}}{\text{AB}} = \frac{\text{BC}}{\text{AC}} = \frac{3}{4}\]
\[BD = \frac{3}{4}AB = 6\ см.\]
\[Ответ:\ \ 6\ см.\]