Let $AB=BC=AC=2x$. Since $AB=BC=AC$, then $\Delta ABC$ is an equilateral triangle and $\angle BAC = 60^\circ$. Since $BE=CE=x$, then $\angle AEB=90^\circ$ and $\angle BAE = 30^\circ$. By applying the Pythagoras Theorem to $\Delta ABE$, we have
$\begin{array}{rcl}
AE^2 & = & AB^2 – BE^2 \\
AE & = & \sqrt{(2x)^2 – x^2} \\
& = & \sqrt{3}x
\end{array}$
Since $AD=DE$, then $AD=DE=\dfrac{\sqrt{3}}{2}x$. By applying the Pythagoras Theorem to $\Delta BED$, we have
$\begin{array}{rcl}
BD^2 & = & BD^2 + DE^2 \\
BD & = & \sqrt{x^2 +(\dfrac{\sqrt{3}}{2}x)^2} \\
& = & \dfrac{\sqrt{7}}{2}x
\end{array}$
By applying the sine law to $\Delta ABD$, we have
$\begin{array}{rcl}
\dfrac{AD}{\sin \theta} & = & \dfrac{BD}{\sin 30^\circ} \\
\sin \theta & = & \dfrac{AD\sin30^\circ}{BD} \\
& = & \dfrac{\sqrt{3}}{2}x \times \dfrac{1}{2} \div \dfrac{\sqrt{7}}{2}x \\
& = & \dfrac{\sqrt{3}}{2\sqrt{7}} \\
& = & \dfrac{\sqrt{3}}{2\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} \\
& = & \dfrac{\sqrt{21}}{14}
\end{array}$
