Extend FP intersection AD to point g.

Ideas:

1) according to the square, we get: AC bisecting angle BAD,

So angle BAC= angle DAC=45 degrees.

2) provable: PE=PG, PF=GD.

Because: EPF angle = PGD angle =90 degrees.

So: triangle EPF is equal to triangle PGD.

So: EF=DP