HELP Maths question

Since AB//DF//EG,
angle ABC = angle EDC = angle GEC
angle BAC = angle DFC = angle EGC
angle ACB = angle FCD = angle GCE
Therefore triangle ABC, FDC and GEC are similar (AAA)
AB/BC = FD/DC = GE/EC
AB/(BD + DE + EC) = FD/(DE + EC) = EG/EC
FD = AB(DE + EC)/(BD + DE + EC)
EG = AB(EC)/(BD + DE + EC)
Therefore FD + EG
= AB(DE + EC)/(BD + DE + EC) + AB(EC)/(BD + DE + EC)
= AB(DE + EC + EC)/(BD + DE + EC)
Since BD = EC (given)
= AB(DE + BD + EC)/(BD + DE + EC)
= AB

Draw a point M on AB such that MD //AC, therefore, AMDF is a parallelogram, so DF = AM ..........(1)
For triangle MBD and triangle GEC
BD = EC ( given)
Angle BMD = angle BAF ( corr. angle MD//AF)
But angle BAF = angle EGC ( corr. angle AB//EG)
therefore, angle BMD = angle EGC
Angle MDB = angle GCE ( corr. angle AC//MD)
Therefore triangle MBD congruent triangle GEC (AAS)
so MB = GE .............(2)
(1) + (2) we get
AM + MB = AB = DF + GE.