China National Team Selection Test

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 $$\frac{BE}{EC}\cdot \frac{CN}{NA}\cdot \frac{AM}{MB}=1.\stackrel{◯}{1}$$ From Menelaus' theorem, $$\frac{BK}{KC}\cdot \frac{CN}{NA}\cdot \frac{AM}{MB}=1.\stackrel{◯}{2}$$ ① ÷ ②, we have $$BE\cdot KC=EC\cdot BK,$$ thus $$BC\cdot KE=2EB\cdot CK.\stackrel{◯}{3}$$ 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 $CF=GF$.