The measure of arc $\widehat{DE}$ is equal to the measure of $\angle DGE$. Since $\triangle DGE$ is an isosceles triangle, $\angle DGE = \angle DEG$, so $\angle DGE = \angle DEG = 60^\circ$ as well. The sum of the angle measures in quadrilateral $DAEB$ is $360^\circ$, so we have \begin{align*}
\angle DAB + \angle DEA + \angle DEG + \angle DGE &= 360^\circ
\\\Rightarrow\qquad \angle DAB + \angle DEA + 60^\circ + 60^\circ &= 360^\circ
\\\Rightarrow\qquad \angle DAB + \angle DEA &= 180^\circ
\end{align*} So $\triangle DAE$ is a triangle with an angle sum of $180^\circ$. Therefore, $DE$ is a straight line segment, so $\widehat{DE}$ is the entire circumference of circle $G$, or $\boxed{24}$ units.
$\triangle{DEF}$ is inscribed in circle $G$. In circle $G$, $\angle DFE = 60^\circ$. If the circumference of circle $G$ is $24$ units, what is the length of $\widehat{DE}$? Express your answer as a common fraction.
1 answer