Extending AC, let QF pass Q and be perpendicular to AC, and let congruent triangles's judgment theorem be from angle A= angle QCF, angle PEA= angle QFC=90 degrees, and PA=QC. It is concluded that all triangles PEA are equal to triangle QFC, so AE = CFQF=PE, so all triangles PED are equal to triangle QFD, and ED = FD = ED+DC+CF = ED+DC+AE =1,so the length of DE is 1/2.