解:(1)PD与EF垂直且相等;
证明:过P作PM⊥CD,PN⊥AD,
∵AC是正方形对角线,
∴PM=PF,PE=PN,
∵PM⊥CD,PN⊥AD,
∴PNDM为矩形,
∴PN=DM,
∴PE=PN=DM,
∵PM=PF,PE=PN=DM,∠PMD=∠FPE=90°,
∴△PMD≌△FPE,
∴PD=EF,∠PEF=∠PDM,∠DPM=∠EFP,
延长DP与EF相交于G,
∵∠DPM=∠EPG,∠PEF=∠PDM
∴∠EPG+∠PEF=90°,
∴EF⊥PD;
(2)画图,(1)中所猜想的结论仍然成立.
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