. Let $\omega$ be the circumcircle of the triangle $C E F$ and let $H$ be the second point of intersection of the circle $\omega$ with the line $C G$ (Figure 8). Assume also, without loss of generality, that $A C < B C$. (If $A C = B C$, the whole problem becomes trivial due to symmetry.) Then the points $G, H, B$ lie on the same side of the line $C F$ and the vertex $A$ lies on the opposite side. The points $E, F, H, C$ lie on the circle $\omega$ while the points $E$ and $D$ are symmetric with respect to the line $\mathrm{CH}$. Hence

$$\angle E F C = \angle E H C = \angle C H D.$$

The line $A C$ is tangent to the incircle of $A B C$ at $E$, so we have $\angle G D F = \angle E D F = \angle A E F = 180^{\circ} - \angle C E F$. Now, using the circle $\omega$, we see that $\angle C E F = 180^{\circ} - \angle C H F = 180^{\circ} - \angle G H F$. Combining the above relations we conclude that $\angle G D F = \angle G H F$, so that the points $G, F, H, D$ lie on a circle. Therefore $\angle C H D = \angle G H D = \angle G F D$, which together with $(*)$ proves the assertion of the problem.

Remark. This problem is trivial for those who rely on the following known result: a symmedian through one of the vertices of a triangle passes through the point of intersection of the tangents to the circumcircle at the other two vertices (http://www. cut-the-knot.org/Curriculum/Geometry/Symmedian.shtml\#explanation). Applying this result to triangle $D E F$ and the symmedian through $F$ gives that the symmedian coincides with $F C$. Now use the definition of symmedian.