Предмет: Геометрия, автор: ek1800

Рівнобічну трапецію вписано в коло, центр якого належить одній з основ. Кут між діагоналями трапеції, протилежний її бічній стороні, дорівнює 56°. Знайдіть кути трапеції.

Ответы

Автор ответа: siestarjoki

Равные хорды стягивают равные дуги

AB=CD => ∪AB=∪CD

Угол между хордами равен полусумме высекаемых дуг

APB =(∪AB+∪CD)/2 =∪AB =56

ACB =∪AB/2 =56/2 =28

ACD =90 (опирается на диаметр AD)

BCD =28+90 =118° =ABC

CDA =180-118 =62° (внутренние односторонние при параллельных) =BAC

Приложения:

sofiiia882: а какое ДАНО?

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