Since $AB$ is the tangent to the circle at $B$, then we have
$\begin{array}{ll}
\angle ABD = \angle ACB & \text{($\angle$s in alt. segment)}
\end{array}$
Consider $\Delta ABD$ and $\Delta ACB$,
$\begin{array}{ll}
\angle BAD = \angle CAB & \text{(common $\angle$)} \\
\angle ABD = \angle ACB & \text{(proved)}
\end{array}$
$\begin{array}{rll}
\angle ADB & = 180^\circ – \angle BAD – \angle ABD & \text{($\angle$ sum of $\Delta$)} \\
& = 180^\circ – \angle CAB – \angle ACB & \text{(proved)} \\
& = \angle ABC & \text{($\angle$ sum of $\Delta$)}
\end{array}$
Therefore, $\Delta ABD\sim \Delta ACB$ (AAA). Hence, we have
$\begin{array}{rcl}
\dfrac{\text{area of }\Delta ABD}{\text{area of }\Delta ACB} & = & \left(\dfrac{AD}{AB}\right)^2 \\
& = & \left( \dfrac{1}{2} \right)^2 \\
& = & \dfrac{1}{4}
\end{array}$
Therefore, we have
$\begin{array}{rcl}
\dfrac{\text{area of }\Delta ABD}{\text{area of }\Delta BCD} & = & \dfrac{1}{4-1} \\
& = & \dfrac{1}{3}
\end{array}$
