17. (12 points) As shown in Figure 10, given that the diameter of a circle is equal to the height of equilateral $\triangle ABC$, the circle is tangent to sides $AB$ and $BC$ at points $D$ and $E$, respectively, and intersects side $AC$ at points $F$ and $G$. Find the measure of $\angle DEF$.
17. As shown in Figure 15, draw a perpendicular from point $E$ to $BC$ intersecting the circle at point $H$ and $AC$ at point $O$. Connect $AH$ and $DH$, and draw $AM \perp BC$, with the foot of the perpendicular at $M$.

Since $E$ is the point of tangency, $EH$ must pass through the center of the circle, i.e., $EH$ is a diameter. Therefore, $DH \perp DE$. Also, since $D$ and $E$ are points of tangency,
$$
BD = BE \text{.}
$$

Given $\angle B = 60^{\circ}$, we know that $\triangle DBE$ is an equilateral triangle, so $\angle BDE = \angle BAC = 60^{\circ}$. Hence, $DE \parallel AC$ and $DH \perp AC$.

From the given information, $AM = EH$. Also, $AM \parallel EH$, so quadrilateral $AMEH$ is a rectangle, which means $AH \perp HE$, i.e., $AH$ is a tangent. Therefore, $AD = AH$, and $AC$ perpendicularly bisects $DH$, so $AC$ must pass through the center of the circle. Thus, the intersection point $O$ of $AC$ and $EH$ is the center of the circle. Consequently, $OE = OF$.
Since $\angle COE = 90^{\circ} - \angle C = 30^{\circ}$, we have,
$$
\begin{array}{l}
\angle OEF = 75^{\circ} . \\
\text{Also, } \angle DEO = \angle EOC = 30^{\circ}, \text{ so} \\
\angle DEF = 30^{\circ} + 75^{\circ} = 105^{\circ} .
\end{array}
$$