过D作DE⊥BC交BC于E,过A作AF⊥BC交BC于F∵AB = AC,∠BAC = 90°∴2AF = BC∵AD//BC∴AF = DE∴2DE = BC∵BC = BD∴2DE = BD∵DE⊥BC∴∠DBC = 30°∵BD =BC∴∠BDC = ∠BCD = 75°∴∠DOC = ∠OBC + ∠OCB = 30° + ...
几何证明梯形如图,梯形ABCD,AN平行于BC,AB=AC,∠BAC=90°,BD=BC,求证 CO=CD
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