The line $OC$ perpendicular to the radius $AC$ of the circle $\alpha$ is tangent to this circle. Hence we get $\angle GDC = 180^\circ - \angle OCG = 180^\circ - \angle OCE$. From the equality of inscribed angles, we get $\angle CDF = \angle CEF = \angle CEO$. Since $OC = OE$, we finally get $\angle CEO = \angle OCE$. In conclusion,
$$\angle GDF = \angle GDC + \angle CDF = 180^\circ - \angle OCE + \angle CEO = 180^\circ.$$

Therefore, points $G$, $D$, and $F$ are collinear.

Solution 2:

We express the value of angle $GDF$ as the sum of the values of angles $GDA$, $ADO$, and $ODF$ (Fig. 35). Let $\angle DCE = \gamma$.
Firstly, note that $AD = AG$. Using this and the relationship between central and inscribed angles in circle $\alpha$, we get
$$\begin{aligned} \angle GDA &= \frac{180^\circ - \angle GAD}{2} = 90^\circ - \frac{\angle GAD}{2} = 90^\circ - \angle GCD \\ &= 90^\circ - \angle ECD = 90^\circ - \gamma. \end{aligned}$$

By symmetry, $\angle ADO = \angle ACO = 90^\circ$.
Next, note that $OD = OF$. Using this and the equality of inscribed angles,

$$\angle GDF = \angle GDA + \angle ADO + \angle ODF = (90^\circ - \gamma) + 90^\circ + \gamma = 180^\circ.$$

Therefore, points $G$, $D$, and $F$ are collinear.

Solution 3:

We express the value of angle $GDF$ as the sum of the values of angles $GDA$, $ADE$, and $EDF$ (Fig. 36). Let $\angle DCE = \gamma$.
Firstly, note that $AD = AG$. Using this and the relationship between central and inscribed angles in circle $\alpha$, we get
$$\begin{aligned} \angle GDA &= \frac{180^\circ - \angle GAD}{2} = 90^\circ - \frac{\angle GAD}{2} = 90^\circ - \angle GCD \\ &= 90^\circ - \angle ECD = 90^\circ - \gamma. \end{aligned}$$
By symmetry, $\angle ADO = \angle ACO = 90^\circ$. The line $AD$, which is perpendicular to the radius $OD$ of the circle $\omega$, is tangent to circle $\omega$. Therefore, $\angle ADE = \angle ECD = \gamma$. Since $EF$ is a diameter of circle $\omega$, $\angle EDF = 90^\circ$. In conclusion, $\angle GDF = \angle GDA + \angle ADE + \angle EDF = (90^\circ - \gamma) + \gamma + 90^\circ = 180^\circ$. Therefore, points $G$, $D$, and $F$ are collinear.