\[Схематический\ рисунок.\]
\[Дано:\]
\[BD - биссектриса\ \angle B;\]
\[AB = 12\ см;\]
\[BC = 15\ см;\]
\[AC = 18\ см.\]
\[Найти:\]
\[\text{BD.}\]
\[Решение:\]
\[1)\ В\ \mathrm{\Delta}ABC:\]
\[\frac{\text{AD}}{\text{CD}} = \frac{\text{AB}}{\text{BC}} = \frac{12}{15} = \frac{4}{5}\text{\ \ \ }\]
\[AD = \frac{4}{5}CD;\]
\[AC = AD + CD =\]
\[= \frac{4}{5}CD + CD = 18\]
\[\frac{9}{5}CD = 18\ \ \ \]
\[CD = 18 \cdot \frac{5}{9} = 10\ см;\]
\[AD = \frac{4}{5} \bullet 10 = 8\ см.\]
\[432\cos{\angle A} = 243\ \ \ \]
\[\cos{\angle A} = \frac{9}{16}.\]
\[2)\ В\ \mathrm{\Delta}ABD:\]
\[= 144 + 64 - 2 \bullet 12 \bullet 8 \bullet \frac{9}{16} =\]
\[= 208 - 108 = 100;\ \ \ \]
\[BD = 10\ см.\]
\[Ответ:\ \ 10\ см.\]