In diamond ABCD, point E is on diagonal AC, point F is on the extension of BC, EF = EB, EF and CD intersect at point G How to prove that triangle EGC is similar to triangle DGF after connecting DF

In diamond ABCD, point E is on diagonal AC, point F is on the extension of BC, EF = EB, EF and CD intersect at point G How to prove that triangle EGC is similar to triangle DGF after connecting DF

It is proved that in diamond ABCD, BC = CD, ∠ BCE = ∠ DCE, CE = CE
So △ BCE ≌ △ DCE (SAS)
So ∠ EBC = ∠ EDC,
And EF = EB,
So, f = EBC,
So ∠ f = EDC,
So △ DEG ∽ CFG
So eg / CG = dg / GF
Eg * GF = CG * GD