\[\boxed{\mathbf{644.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[\angle BAD = \angle DAC;\]
\[AB = 14\ см;\]
\[BC = 20\ см;\]
\[AC = 21\ см.\]
\[\mathbf{Найти:}\]
\[BD - ?\]
\[DC - ?\]
\[\mathbf{Решение.}\]
\[1)\ AH - высота,\ общая\ для\ \]
\[\mathrm{\Delta}\text{ABD\ }и\ \mathrm{\Delta}ACD:\]
\[\frac{S_{\text{ABD}}}{S_{\text{ACD}}} = \frac{\text{BD}}{\text{CD}}.\]
\[2)\ \angle BAD = \angle DAC:\]
\[\frac{S_{\text{ABD}}}{S_{\text{ACD}}} = \frac{AB \bullet AD}{AD \bullet AC} = \frac{\text{AB}}{\text{AC}}.\]
\[3)\frac{\text{AB}}{\text{AC}} = \frac{\text{BD}}{\text{CD}}\]
\[AC \bullet BD = AB \bullet CD\]
\[\frac{\text{BD}}{\text{DC}} = \frac{\text{AB}}{\text{AC}}.\]
\[4)\ Пусть\ BD = x;\]
\[DC = BC - x = 20 - x.\]
\[5)\frac{x}{20 - x} = \frac{14}{21}\]
\[\frac{x}{20 - x} = \frac{2}{3}\]
\[3x = 2(20 - x) = 40 - 2x\]
\[5x = 40\]
\[x = 8\ см.\]
\[3)\ DC = 20 - 8 = 12\ см.\]
\[\mathbf{Ответ:}BD = 8\mathbf{\ см};DC = 12\ см.\]