\[\boxed{\mathbf{1063.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AB = c;\]
\[AC = b;\]
\[AD - биссектрисса;\]
\[\angle A = \alpha.\]
\[\mathbf{Найти:}\]
\[AD - ?\]
\[\mathbf{Решение.}\]
\[1)\ S_{\text{ABC}} = S_{\text{ABD}} + S_{\text{ADC}};\]
\[\frac{1}{2}ab \bullet \sin\alpha =\]
\[= \frac{1}{2}c \bullet AD \bullet \sin\frac{\alpha}{2} + \frac{1}{2}b \bullet AD \bullet \sin\frac{\alpha}{2};\]
\[ab \bullet \sin\alpha =\]
\[= AD\left( c \bullet \sin\frac{\alpha}{2} + b \bullet \sin\frac{\alpha}{2} \right).\]
\[2)\ AD = \frac{ab \bullet \sin\alpha}{\sin\frac{\alpha}{2}(c + b)} =\]
\[= \frac{2ab \bullet \sin\frac{\alpha}{2} \bullet \cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}(c + b)} =\]
\[= \frac{2ab \bullet \cos\frac{\alpha}{2}}{c + b}.\]
\[\ Ответ:\ \frac{2ab \bullet \cos\frac{\alpha}{2}}{c + b}.\]