19th Turkish Mathematical Olympiad

We will show that $EF$ is a common tangent of the circumcircles of $AGF$ and $BGE$. Note that it is enough to show that $\angle GBE=\angle GEF$ and $\angle GAF=\angle GFE$.

Let $H$ be the point of intersection of $AB$ and the line passing through $F$ parallel to $BC$.

Then $AHF\sim ABC$ and $\frac{AF}{AC}=\frac{HF}{BC}$. On the other hand, it is given that $\frac{AF}{AC}=\frac{EC}{BC}$. Therefore, $HF=EC=ED$ and hence $HFCE$ is a parallelogram and $HFDE$ is a rectangle.

Since $ACDG$ is a cyclic quadrilateral and $FC\parallel HE$, we have $\angle BGD=\angle ACB=\angle HED$. Then $H,E,D,G$ are cyclic. Consequently, $H,G,D,E,F$ are on the circle of diameter $[HE]$. Then $\angle BGE=90^{\circ }$ and $\angle GBE=90^{\circ }-\angle GED=\angle GEF$. Since $AGDC$ and $GFED$ are cyclic quadrilaterals $\angle BAC=180^{\circ }-\angle GDE=\angle GFE$.