So messy..
Basically let < G be θ
Then < DOB = 2θ
We know triangles DOC & BOC are congruent so < DOC is θ
< DFC = θ as FC || GB so corresponding angles
Since < DXC = < DOC, D, X, O and C are concylic points since angles in the same segment are equal
then < OXC = < ODC as angles in the same...