Решебник по геометрии 9 класс Мерзляк Задание 107

\[Схематический\ рисунок.\]

\[Дано:\]

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

\[AM - медиана;\]

\[\angle MAB = \alpha;\]

\[\angle MAC = \beta;\]

\[AM = m.\]

\[Найти:\]

\[AB;\ AC.\]

\[Решение.\]

\[1)\ Дополнительное\ построение:\]

\[MK = AM;\ \ \ \]

\[K \in AM.\]

\[2)\ Четырехугольник\ ABKC:\]

\[AM = MK;\ \ \]

\[BM = MC.\]

\[Следовательно:\]

\[ABKC - параллелограмм.\]

\[Отсюда:\]

\[\angle BAC = \angle BAK + \angle CAK = \alpha + \beta;\]

\[CK = AB;\ \ \ \]

\[\angle ACK = 180{^\circ} - \angle BAC.\]

\[\sin{\angle ACK} = \sin{\angle BAC} =\]

\[= \sin(\alpha + \beta).\]

\[3)\ В\ \mathrm{\Delta}ACK:\]

\[\angle AKC = 180{^\circ} - \angle ACK - \angle CAK =\]

\[= \angle BAC - \angle CAK = \alpha;\]

\[AK = 2AM = 2m.\]

\[\frac{\text{AK}}{\sin{\angle ACK}} = \frac{\text{CK}}{\sin{\angle CAK}} =\]

\[= \frac{\text{AC}}{\sin{\angle AKC}};\]

\[AB = CK = \frac{AK \bullet \sin{\angle CAK}}{\sin{\angle ACK}} =\]

\[= \frac{2m\sin\beta}{\sin(\alpha + \beta)};\]

\[AC = \frac{AK \bullet \sin{\angle AKC}}{\sin{\angle ACK}} =\]

\[= \frac{2m\sin\alpha}{\sin(\alpha + \beta)}.\]

\[Ответ:\ \ \frac{2m\sin\beta}{\sin(\alpha + \beta)};\ \frac{2m\sin\alpha}{\sin(\alpha + \beta)}.\]