由於 $AB$ 為該圓在 $B$ 的切線,可得
$\begin{array}{ll}
\angle ABD = \angle ACB & \text{(同弓形上的圓周角)}
\end{array}$
考慮 $\Delta ABD$ 及 $\Delta ACB$,
$\begin{array}{ll}
\angle BAD = \angle CAB & \text{(公共角)} \\
\angle ABD = \angle ACB & \text{(已證)}
\end{array}$
$\begin{array}{rll}
\angle ADB & = 180^\circ – \angle BAD – \angle ABD & \text{($\Delta$的內角和)} \\
& = 180^\circ – \angle CAB – \angle ACB & \text{(已證)} \\
& = \angle ABC & \text{($\Delta$的內角和)}
\end{array}$
所以,$\Delta ABD\sim \Delta ACB$ (AAA)。由此,可得
$\begin{array}{rcl}
\dfrac{\Delta ABD\text{ 的面積}}{\Delta ACB\text{ 的面積}} & = & \left(\dfrac{AD}{AB}\right)^2 \\
& = & \left( \dfrac{1}{2} \right)^2 \\
& = & \dfrac{1}{4}
\end{array}$
所以,可得
$\begin{array}{rcl}
\dfrac{\Delta ABD\text{ 的面積}}{\Delta BCD\text{ 的面積}} & = & \dfrac{1}{4-1} \\
& = & \dfrac{1}{3}
\end{array}$