Step-by-step explanation:

Given;

AB = CB

AB = CD

EF BISECTS BD at G. So, DG = BG;

To Prove : G is the midpoint of EF or GE = GF.

Construction: Join A and C with a straight line.

Proof : In △ BAC;

 ∠CAB =  ∠ACB (angles opposite to equal sides are equal) (i)

In △ GDC and △GBC;

GD = GB. (given)

DC = BC. (given)

GC = GC. (common)

Therefore, △ GDC ≅ △GBC. (by SSS criteria of congruence)

HENCE,  ∠DGC =  ∠BGC (C.P.C.T)

∠DCG =  ∠BCG (C.P.C.T) (II)

In △ GBC and △GBA;

∠CAB =  ∠ACB (from eq. (i))

∠AGB =  ∠DGC (vertically opposute angle)

AB = BC (given)

Therefore, △ GBC ≅ △GBA (by AAS criteria of congruence)

Hence, GC =  AG (C.P.C.T) (iii)

∠BAG =  ∠BCG (C.P.C.T) (iv)

From equation (ii) and (iv);

∠DCG =  ∠BAG. (v)

∠BAG = ∠GAF (vi)

∠DCG = ∠GCE. (VII)

Now, in △CGE and △AGF;

∠GAF =  ∠GCE. (from equation (v), (vi) and (vii))

AG = GC. (from equation (iii))

∠AGF =  ∠CGE. (vertically opposite angle)

Therefore, △CGE ≅ △AGF (by ASA criteria of congruence)

Hence, GE = GF. (by C.P.C.T)

That's all.