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.