Proof Referring to the figure, draw a line from $D$, tangent to $\omega$, and the line intersects $AB$, $AC$, $BC$ at points $M$, $N$, $K$ respectively.

Since
$$\angle KDE = \angle AEK = \angle EFC,$$
we know $MK \parallel CG$.
By Newton's theorem, the lines $BN$, $CM$, $DE$ are concurrent.
By Ceva's theorem, we have
$$\frac{BE}{EC} \cdot \frac{CN}{NA} \cdot \frac{AM}{MB} = 1. \qquad \textcircled{1}$$
From Menelaus' theorem,
$$\frac{BK}{KC} \cdot \frac{CN}{NA} \cdot \frac{AM}{MB} = 1. \qquad \textcircled{2}$$
① ÷ ②, we have
$$BE \cdot KC = EC \cdot BK,$$
thus
$$BC \cdot KE = 2EB \cdot CK. \qquad \textcircled{3}$$
Using Menelaus' theorem and ③, we get
$$1 = \frac{CB}{BE} \cdot \frac{ED}{DF} \cdot \frac{FG}{GC} = \frac{CB}{BE} \cdot \frac{EK}{CK} \cdot \frac{FG}{GC} = \frac{2FG}{GC}.$$
So $CF = GF$.