<Polygon-2024-269>, hint

Let AD^BE=H, CF^AD=I, CF^BE=K

Since [ABCD]=[AFED] and [BCDE] =[BAFE] --->[ABH]=[DEH]

and let [HIK]=s [AHB]=x [DEKI]=x-s, [CDI]=y [AFKH]=y-s [EFK]=z [BCIH]=z-s

Since [ABCD]=AFED]<--> x+y-s+z=z-s+y+x-s  Hence s=0

Thus AD, BE, CF concur at one point O.

Let OA=a, OB=b, OC=c, OD=d, OE=e, OF=f

AB+BC+CD=AF+FE+ED and BC+CD+DE=EF+FA+AB--->AB=DE

Similarly BC=EF  CD=AF. [OAB]=[ODE]---> ab=de bc=ef, cd=af

Consider quadrilateral ABDE  ab=de <---> a/e=d/b triangles AOE and DOB are similar

Hence ABDE is a parallelogram OA=OD, OB=OE

Thus O is the center of symmetry.