By the tangent-chord angle theorem we have $\angle CAF = \angle CBA$. Since $p$ is the reflection of the line $BC$ over the line $AB$, we have $\angle CBA = \angle ABD$. The points $A, B, E, C$ are concyclic, so $\angle ABD = \angle ACE$. Hence, $\angle CAF = \angle ACE$, and the line $CE$ is parallel to the line $FD$.

We would like to show that the line $GB$ is also parallel to $FD$. Using the tangent-chord angle theorem one more time we see that $\angle DAB = \angle ACB$. Also, $\angle CBA = \angle ABD$, so the triangles $BAD$ and $BCA$ are similar, having two congruent angles. This implies that $\frac{|AD|}{|AB|} = \frac{|CA|}{|CB|}$. Since $F$ is the reflection of the point $D$ over the point $A$, we have $|AD| = |FA|$. So, $\frac{|FA|}{|AB|} = \frac{|CA|}{|CB|}$, or $\frac{|FA|}{|CA|} = \frac{|AB|}{|CB|}$. We also have $\angle CAF = \angle CBA$, so the triangles $FAC$ and $ABC$ are similar as well. This implies that $\angle AFC = \angle BAC = \angle BGC$ and the line $GB$ is parallel to $FD$.