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В правильной пирамиде SABCD постройте сечение, проходящее через диагональ AC основания и параллельное боковому ребру SD.
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Пусть К - середина SB. Тогда ОК - средняя линия треугольника SBD ⇒ OK ║ SD.KAC - искомое сечение.Док-во: АС ⊂ КАС, SD║ OK⇒SD║KAC
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Автор:
jaidyntgxe - 6 лет назад
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B. she gave a bad acting performance.
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