Решебник по геометрии 9 класс Мерзляк Задание 102

\[Схематический\ рисунок.\]

\[Дано:\]

\[BD - биссектриса\ \angle B;\]

\[\angle A = \alpha;\ \]

\[\angle C = \gamma;\]

\[AC = b.\]

\[Найти:\]

\[\text{BD.}\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\angle B = 180{^\circ} - \angle A - \angle C =\]

\[= 180{^\circ} - \alpha - \gamma;\]

\[\sin{\angle B} = \sin(180{^\circ} - \alpha - \gamma) =\]

\[= \sin(\alpha + \gamma);\]

\[\frac{\text{AB}}{\sin{\angle C}} = \frac{\text{AC}}{\sin{\angle B}}\]

\[AB = \frac{AC \bullet \sin{\angle C}}{\sin{\angle B}} = \frac{b\sin\gamma}{\sin(\alpha + \gamma)};\]

\[\angle DBC = \frac{1}{2}\angle B = 90{^\circ} - \frac{1}{2}(\alpha + \gamma).\]

\[2)\ В\ \mathrm{\Delta}BDC:\]

\[\angle BDA = \angle DBC + \angle ACB;\]

\[\angle BDA = 90{^\circ} - \frac{1}{2}(\alpha + \gamma) + \gamma =\]

\[= 90{^\circ} - \frac{\alpha - \gamma}{2};\]

\[\sin{\angle BDA} = \cos\frac{\alpha - \gamma}{2}.\]

\[3)\ В\ \mathrm{\Delta}ABD:\]

\[\frac{\text{AB}}{\sin{\angle BDA}} = \frac{\text{BD}}{\sin{\angle A}}\]

\[BD = \frac{AB \bullet \sin{\angle A}}{\sin{\angle BDA}} =\]

\[= \frac{b\sin\alpha\sin\gamma}{\sin(\alpha + \gamma)\cos\frac{\alpha - \gamma}{2}}.\]

\[Ответ:\ \ \frac{b\sin\alpha\sin\gamma}{\sin(\alpha + \gamma)\cos\frac{\alpha - \gamma}{2}}.\]