Question

AB=CB,AB=CD and EF bisects BD at G. Prove that G is midpoint of EF

Answer 1

If AB = CB and AB = CD then AD can have any length ,
so we can not answer this question .

I assumed that AD = CB and AB = CD .
So ABCD becomes a parallelogram , therefore AB//CD .

Think △DEG and △BFG .
AB//CD so ∠GDE = ∠GBF , and clearly ∠DGE = ∠BGF .
And G bisects BD so DG = BG .
Therefore , △DEG and △BFG are congruent .
So EG = FG , G is the mid-point of EF .