(1)AB=CD, 理由是:过O作OE⊥AB于E,OF⊥CD于F,连接OB、OD, ∵∠APM=∠CPM,∠APM=∠BPN,∠CPM=∠DPN, ∴∠BPN=∠DPN, ∵OE⊥AB,OF⊥CD, ∴OE=OF, 在Rt△BEO和Rt△DOF中,OF=OE,OD=OB,由勾股定理得:BE=DF, ∵OF⊥CD,OE⊥AB, OF、OE过O, ∴由垂径定理得:CD=2DF,AB=2BE, ∴AB=CD.
(2)AB=CD成立, 证明:过O作OE⊥AB于E,OF⊥CD于F,连接OB、OD, ∵∠APM=∠CPM, ∴OE=OF, 在Rt△BEO和Rt△DOF中,OF=OE,OD=OB,由勾股定理得:BE=DF, ∵OF⊥CD,OE⊥AB, OF、OE过O, ∴由垂径定理得:CD=2DF,AB=2BE, ∴AB=CD. |