Предмет: Алгебра, автор: Nellaaaa

УМОЛЯЮ РЕШИТЕ Диагонали AC и BD равнобокой трапеции ABCD пересекаются в точке O. Известно, что AD:BC=5:3. Окружность ω с центром O, проходящая через вершины A и D, пересекает продолжение основания BC за точку B в точке K. Оказалось, что BK=BO. Найдите отношение основания AD к радиусу окружности ω.

antonovm: 7 : 9

Ответы

Автор ответа: antonovm

Ответ:

7 : 9

Объяснение:

Приложения:

antonovm: Может быть стоит добавить : AD/AO = 5x/5y = x/y = 7/9

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