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1) Пусть ABCD - трапеция, ВС и AD основания трапеции, O - точка пересечения диагоналей.
2) Рассмотрим треугольники BOC и AOD:
<ВОС = <AOD (вертикальные)
<CBO = <ODA (накрестлежащие при параллельных прямых AD и BC)
<BCO = <OAD (накрестлежащие при параллельных прямых AD и BC)
Отсюда, треугольники BOC и AOD подобны
3) Но высоты в этих треугольниках равны ( по условию) ⇒ коэффициент подобия равен 1, т. е. треугольники равны ⇒ BC = AD
4) По признаку параллелограмма (противоположные стороны равны и параллельны) четырехугольник является не трапецией, а параллелограммом.