Question

如图,正方形ABCD对角线AC,BD交于点O,P是BD上任意一点,过点P作PE⊥AB于E,PF⊥AD于点F,连接OE,O

Answer 1

证明：

∵正方形ABCD

∴AO＝BO,∠AOB＝∠BAD＝90,∠ABD＝∠DAC＝45

∵PE⊥AB

∴PE＝BE

∵PF⊥AD

∴矩形AEPF

∴AF＝PE

∴AF＝BE

∴△AOF≌△BOE

∴∠AOF＝∠BOE

∵∠BOE+∠AOE＝∠AOB＝90

∴∠AOF+∠AOE＝90

∴∠EOF＝90

∴OE⊥OF