Solution:

We denote the second point of intersection of $CF$ and $K$ by $H$ and the point of intersection of $AB$ and $g$ by $X$. Since $AB \perp g$, it suffices to show that $GH \parallel g$. This is precisely the case when $\angle AGH = \angle AFD$. The inscribed angle theorem and vertical angles give $\angle AGH = \angle ACH = \angle DCF$; hence it remains to show $\angle DCF = \angle AFD$. Since the triangles $DFC$ and $ADF$ share an angle at $D$, it only remains to prove their (opposite-sense) similarity. This holds precisely when $\frac{|DC|}{|DF|} = \frac{|DF|}{|DA|}$,
i.e. $|DF|^2 = |DA| \cdot |DC|$. However, by the tangent-secant theorem we have $|DE|^2 = |DA| \cdot |DC|$, so it only remains to show $|DE| = |DF|$, which is equivalent to $\angle DEF = \angle EFD$.

Since $\angle FXB = \angle AEB = 90^\circ$, $AEXF$ is a cyclic quadrilateral and we have $\angle EFX = \angle EAB$. The equality of the tangent-chord angles gives $\angle DEF = \angle EAB$, from which $\angle DEF = \angle EFD$ follows.