Решебник по геометрии 9 класс Мерзляк Задание 109

\[Схематический\ рисунок.\]

\[Дано:\]

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

\[AA_{1},\ BB_{1},\ CC_{1} - высоты\ \mathrm{\Delta}ABC;\]

\[AA_{1} \cap BB_{1} = H.\]

\[Доказать:\]

\[R_{\text{AHB}} = R_{\text{BHC}} = R_{\text{AHC}} = R_{\text{ABC}}.\]

\[Доказательство.\]

\[1)\ В\ \mathrm{\Delta}ABC:\]

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

\[\sin{\angle C} = \sin(\angle A + \angle B);\]

\[R_{\text{ABC}} = \frac{\text{AB}}{2\sin{\angle C}} = \frac{\text{AB}}{2\sin(\angle A + \angle B)}.\]

\[2)\ В\ \mathrm{\Delta}AHB:\]

\[\angle AA_{1}B = 90{^\circ};\ \ \ \]

\[\angle BAH = 90{^\circ} - \angle B;\]

\[\angle BB_{1}A = 90{^\circ};\ \ \ \]

\[\angle ABH = 90{^\circ} - \angle A;\]

\[\angle AHB = 180{^\circ} - \angle BAH - \angle ABH;\]

\[\angle AHB = \angle A + \angle B.\]

\[R_{\text{AHB}} = \frac{\text{AB}}{2\sin{\angle AHB}} =\]

\[= \frac{\text{AB}}{2\sin(\angle A + \angle B)}.\]

\[3)\ Аналогично:\]

\[R_{\text{ABC}} = R_{\text{BHC}} = \frac{\text{BC}}{2\sin(\angle B + \angle C)};\]

\[R_{\text{ABC}} = R_{\text{AHC}} = \frac{\text{AC}}{2\sin(\angle A + \angle C)}.\]

\(Что\ и\ требовалось\ доказать.\)