如图所示,设PF⊥CD, ∵BP=FP, 由翻折变换的性质可得BP=B′P, ∴FP=B′P, ∴FP⊥CD, ∴B′,F,P三点构不成三角形, ∴F,B′重合分别延长AE,DC相交于点G, ∵AB平行于CD, ∴∠BAG=∠AGC, ∵∠BAG=∠B′AG,AGC=∠B′AG, ∴GB′=AB′=AB=5, ∵PB′(PF)⊥CD, ∴PB′∥AD, ∴△ADG∽△PB′G, ∵Rt△ADB′中,AB′=5,AD=3, ∴DB′=4,DG=DB′+B′G=4+5=9, ∴△ADG与△PB′G的相似比为9:5, ∴AD:PB′=9:5, ∵AD=3, ∴PB′=,即相等距离为. |