Take d as the center of symmetry, point a' as the symmetry point of point A, and ABA'c as a parallelogram.

Take the symmetry points e' and f' of e and f on A'C and ABC respectively.

It is easy to prove that EFE'F' is a parallelogram.

Consider triangle BFP and CEP.

∠ABP=∠ACP，∠BFP=∠CEP = 90

The triangular BFP is similar to CEP.

PF/PE = BF/CE = BF'/CE '

∠EPF = ∠ECF'= 180 - ∠A

The triangle of EPF is similar to that of ECF.

FEP = "European Community"

∠EFF' = ∠PEC = 90

The parallelogram EFEF is a rectangle,

Therefore, DE=DF = half the diagonal of the rectangular EFEF.