\[Схематический\ рисунок.\]
\[Дано:\]
\[BD - биссектриса\ \angle B;\]
\[AB = 28\ см;\]
\[BC = 20\ см;\]
\[AC = 36\ см.\]
\[Найти:\]
\[AD;\ CD.\]
\[Решение.\]
\[AC = AD + CD = 36\ см.\]
\[BD - биссектриса\ \angle B:\]
\[\frac{\text{AD}}{\text{AB}} = \frac{\text{CD}}{\text{BC}}\text{\ \ \ }\]
\[AD = \frac{AB \bullet CD}{\text{BC}}\]
\[AD = \frac{28 \bullet CD}{20} = 1,4CD;\]
\[1,4CD + CD = 36\]
\[2,4CD = 36\ \ \ \]
\[CD = 15\ см.\]
\[AD = 1,4 \bullet 15 = 21\ см.\]
\[Ответ:\ \ AD = 21\ см;\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ CD = 15\ см.\]