Вопрос школьника по предмету Геометрия
отрезки ВЕ и DС равны и пересикаются в точке К так что СК=DK ВК=КЕ найдите DEK усли СВЕ= 60
Ответ учителя по предмету Геометрия
1) ΔCBK = ΔEDK, т.к CK = DK, BK = EK, ∠BKC = ∠DKE (вертикальные). 2) Т.к ΔCBK = ΔEDK, то ∠CBK = ∠DEK = 60°. Ответ: 60°.
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