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(1)证明:在正方形ABCD中,AB=AD,∠BAE=∠D=90°,
∴∠DAF+∠BAF=90°,
∵AF⊥BE,
∴∠ABE+∠BAF=90°,
∴∠ABE=∠DAF,
∵在△ABE和△DAF中,
,
∴△ABE≌△DAF(ASA),
∴AF=BE;
(2)解:MP与NQ相等.
理由如下:如图,过点A作AF∥MP交CD于F,过点B作BE∥NQ交AD于E,
∵AB∥CD,AD∥BC,
∴四边形AMPF与四边形BNQE是平行四边形,
∴AF=PM,BE=NQ,
∵在正方形ABCD中,AB=AD,∠BAE=∠D=90°,
∴∠DAF+∠BAF=90°,
∵AF⊥BE,
∴∠ABE+∠BAF=90°,
∴∠ABE=∠DAF,
∵在△ABE和△DAF中,
,
∴△ABE≌△DAF(ASA),
∴AF=BE;
∴MP=NQ.