Решебник по геометрии 9 класс Мерзляк Задание 176

Авторы:
Год:2022
Тип:учебник

Задание 176

\[1)\ A_{1}A_{2}\ldots A_{n} - многоугольник;\]

\[O - центр\ окружности:\]

\[OA_{1} = OA_{2} = \ldots = OA_{n} = R,\ \ \ \]

\[A_{1}A_{2} = A_{2}A_{3} = \ldots = A_{n}A_{1};\]

\[\mathrm{\Delta}A_{1}OA_{2} = \mathrm{\Delta}A_{2}OA_{3} =\]

\[= \ldots = \mathrm{\Delta}A_{n}OA_{1};\]

\[\angle OA_{1}A_{2} = \angle OA_{2}A_{1} =\]

\[= \ldots = \angle OA_{n}A_{1};\]

\[\angle A_{1}A_{2}A_{3} = \angle A_{2}A_{3}A_{4} =\]

\[= \ldots = \angle A_{n}A_{1}A_{2}.\]

\[Ответ:\ \ да.\]

\[2)\ ABCD - прямоугольник;\ \ \ \]

\[AB \neq BC:\]

\[\angle A = \angle B = \angle C = \angle D = 90{^\circ};\]

\[\angle A + \angle C = \angle B + \angle D;\]

\[ABCD - вписанный.\]

\[Ответ:\ \ нет.\]

\[3)\ A_{1}A_{2}\ldots A_{n} - многоугольник;\]

\[O - центр\ окружности;\]

\[OH_{1}\bot A_{1}A_{2};\ \ \]

\[OH_{2}\bot A_{2}A_{3},\ \ldots;\]

\[OH_{n}\bot A_{n}A_{1}.\]

\[OH_{1} = OH_{2} = \ldots = OH_{n} = R;\]

\[A_{1}A_{2} = A_{2}A_{3} = \ldots = A_{n}A_{1};\]

\[\mathrm{\Delta}OH_{1}A_{2} = \mathrm{\Delta}OH_{2}A_{2};\ \ \ \]

\[\angle OA_{2}H_{1} = \angle OA_{2}H_{2};\]

\[\mathrm{\Delta}OA_{1}A_{2} = \mathrm{\Delta}OA_{2}A_{3};\ \ \ \]

\[OA_{1} = OA_{3};\]

\[OA_{1} = OA_{2} = \ldots = OA_{n};\]

\[\mathrm{\Delta}A_{1}OA_{2} = \mathrm{\Delta}A_{2}OA_{3} = \ldots = \mathrm{\Delta}A_{n}OA_{1};\]

\[\angle OA_{1}A_{2} = \angle OA_{2}A_{1} = \ldots = \angle OA_{n}A_{1};\]

\[\angle A_{1}A_{2}A_{3} = \angle A_{2}A_{3}A_{4} =\]

\[= \ldots = \angle A_{n}A_{1}A_{2}.\]

\[Ответ:\ \ да.\]

\[4)\ A_{1}A_{2}\ldots A_{n} - многоугольник;\]

\[O - центр\ окружности;\]

\[OH_{1}\bot A_{1}A_{2};\ \ \ \]

\[OH_{2}\bot A_{2}A_{3},\ \ldots;\ \]

\[OH_{n}\bot A_{n}A_{1}.\]

\[OH_{1} = OH_{2} = \ldots = OH_{n} = R;\]

\[\angle A_{1}A_{2}A_{3} = \angle A_{2}A_{3}A_{4} =\]

\[= \ldots = \angle A_{n}A_{1}A_{2};\]

\[\mathrm{\Delta}OH_{1}A_{2} = \mathrm{\Delta}OH_{2}A_{2};\ \ \]

\[\angle OA_{2}H_{1} = \angle OA_{2}H_{2} = \frac{1}{2}\angle A_{2};\]

\[\angle OA_{1}H_{1} = \angle OA_{2}H_{1} = \angle OA_{2}H_{2} =\]

\[= \ldots = \angle OA_{n}H_{1};\]

\[\mathrm{\Delta}A_{1}OA_{2} = \mathrm{\Delta}A_{2}OA_{3} = \ldots = \mathrm{\Delta}A_{n}OA_{1};\]

\[OA_{1} = OA_{2} = \ldots = OA_{n}.\]

\[Ответ:\ \ да.\]

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