A mathematical solid geometry problem!

Take the midpoint of PC as h and connect FH.EH.

Because CD is perpendicular to P 1D, BA is perpendicular to P 1d (the nature of origami).

CD//AB and BC//AD, so the quadrilateral ABCD is a parallelogram, then AB=CD,

In triangular PCD, F is the midpoint of PD and H is the midpoint of PC, so FH is parallel and equal to 1/2CD (midline theorem), while AE= 1/2AB and AB are parallel and equal to CD. So FH is parallel and equal to AE, then the quadrilateral AEHF is a parallelogram, so AF is parallel to EH, EH is in the plane PEC, but AF is not in the plane PEC.