Given points $A$, $B$, $C$ and $D$ lying on a circle and lines $BD$ and $AC$ intersecting at $F$, prove lines are parallel

Question

The diagram shows the points $A$, $B$, $C$ and $D$ lying on a circle. $AC$ and $BD$ intersect at point $F$. $EG$ is tangent to the circle at point $C$. $AD$ is produced to meet the tangent at point $E$. Given that $\angle{ABC} = \angle{BFC}$ prove that $BD \parallel EG$.

In part 1, we need to prove lines are parallel, which means we can show angles $\angle{AFB} = \angle{FCB}$ (corresponding angles), $\angle{AFB} = \angle{FCE}$ (alternate angles), $\angle{AFB} + \angle{CFB} = 180^{\circ}$ The problem is line segment $CA$ is not a radius (it's not mentioned in the question) so we can not say $CA$ is perpendicular. Then how do we find those angles and equate it ?? I also tried to apply exterior angle property on triangle on triangle $\triangle AFB$ ($\angle{BFC} = \angle{FAB} + \angle{FBA}$) but didn't get anything.

Answer 1

Since $\angle ABC=\angle BFC$, and $\angle BCA$ is common to both triangles, then$$\triangle ABC\sim\triangle BFC$$and$$\angle CAB=\angle FBC$$

And by the alternate segment theorem$$\angle CAB=\angle BCG$$Therefore$$\angle FBC=\angle BCG$$and hence$$BD\parallel EG$$

Answer 2

Since $\angle BFC=\frac{\overset{\huge\frown}{AD}+\overset{\huge\frown}{BC}}{2}$ and $\angle ABC=\frac{\overset{\huge\frown}{AD}+\overset{\huge\frown}{DC}}{2}$ where measure of arcs are read from the center and $\angle BFC=\angle ABC$, we have $\overset{\huge\frown}{BC}=\overset{\huge\frown}{DC}$ which means that $\angle BDC=\angle DCE$ and $DB\parallel EG$.