Решебник по геометрии 9 класс Мерзляк Задание 103

\[Схематический\ рисунок.\]

\[Дано:\]

\[\mathrm{\Delta}ABC - равнобедренный;\]

\[AD - биссектриса\ \angle A;\]

\[AC = a;\ \]

\[\angle B = \alpha.\]

\[Найти:\]

\[\text{AD.}\]

\[Решение.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

\[\angle A = \angle C = \frac{180{^\circ} - \angle B}{2} = 90{^\circ} - \frac{\alpha}{2};\]

\[\angle BAD = \angle CAD = \frac{1}{2}\angle A = 45{^\circ} - \frac{\alpha}{4};\]

\[\sin{\angle C} = \cos\frac{\alpha}{2}.\]

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

\[AB = \frac{AC \bullet \sin{\angle C}}{\sin{\angle B}} = \frac{a\cos\frac{\alpha}{2}}{\sin\alpha}.\]

\[2)\ В\ \mathrm{\Delta}ABD:\]

\[\angle BDA = 180{^\circ} - \angle BAD - \angle B =\]

\[= 180{^\circ} - \left( 45{^\circ} + \frac{3\alpha}{4} \right);\]

\[\sin{\angle BDA} = \sin\left( 45{^\circ} + \frac{3\alpha}{4} \right).\]

\[\frac{\text{AB}}{\sin{\angle BDA}} = \frac{\text{AD}}{\sin{\angle B}}\]

\[AD = \frac{AB \bullet \sin{\angle B}}{\sin{\angle BDA}} =\]

\[= \frac{a\cos\frac{\alpha}{2}}{\sin\left( 45{^\circ} + \frac{3\alpha}{4} \right)}.\]

\[Ответ:\ \ \frac{a\cos\frac{\alpha}{2}}{\sin\left( 45{^\circ} + \frac{3\alpha}{4} \right)}.\]