Estonian Mathematical Olympiad

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 $$\begin{array}{rl}\angle GDA& =\frac{180^{\circ }-\angle GAD}{2}=90^{\circ }-\frac{\angle GAD}{2}=90^{\circ }-\angle GCD\\ & =90^{\circ }-\angle ECD=90^{\circ }-\gamma .\end{array}$$

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 $$\begin{array}{rl}\angle GDA& =\frac{180^{\circ }-\angle GAD}{2}=90^{\circ }-\frac{\angle GAD}{2}=90^{\circ }-\angle GCD\\ & =90^{\circ }-\angle ECD=90^{\circ }-\gamma .\end{array}$$ 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.