We will calculate $\angle O_{1} E A$. Since triangle $O_{1} E D$ is isosceles with apex $O_{1}$ and since $A E F D$ is a cyclic quadrilateral, we have

$$\angle O_{1} E A=\angle O_{1} E D-\angle A E D=90^{\circ}-\frac{1}{2} \angle E O_{1} D-\angle A F D .$$

Furthermore,

$$\begin{aligned} \frac{1}{2} \angle E O_{1} D & =\angle E S D \quad \text { central angle-circumference angle theorem on } \Gamma_{1} \\ & =\angle C S D-\angle C S E \\ & =\angle C S D-\angle C D E \quad \text { circumference angle theorem in cyclic quadrilateral } C S D E \\ & =\angle F C D-\angle S D C-\angle C D E \quad \text { exterior angle theorem in } \triangle C S D \\ & =\angle F C D-\angle S D E \\ & =\angle F C D-\angle A D E \quad \\ & =\angle F C D-\angle A F E . \quad \text { circumference angle theorem in cyclic quadrilateral } A E F D \end{aligned}$$

From this, together with (7), we get

$$\angle O_{1} E A=90^{\circ}-\angle F C D+\angle A F E-\angle A F D$$

The line $O_{1} O_{2}$ is perpendicular to $C D$ and bisects the segment $C D$, so it is the perpendicular bisector of $C D$. Therefore, $F$ lies on the perpendicular bisector of $C D$, which implies that triangle $C D F$ is isosceles with apex $F$. Thus,

$$\angle F C D=90^{\circ}-\frac{1}{2} \angle C F D=90^{\circ}-\angle A F D .$$

Combined with (8), this gives

$$\angle O_{1} E A=\angle A F D+\angle A F E-\angle A F D=\angle A F E$$

Now, by the tangent-circumference angle theorem on $\Gamma_{3}$, it follows that $O_{1} E$ is tangent to $\Gamma_{3}$.