Show by using sine rule I'm a bit stuck on how to answer this question that was in a past paper (there are no answers to the paper). I'm completely lost so any help is greatly appreciated!

 A: This question appeared in the 2019 Extension 2 HSC Mathematics exam. One possible solution proceeds as follows:
We can apply the sine rule in both $\triangle BCD$ and $\triangle ABC$ to obtain
$$\frac{a}{\sin{(\angle CBD)}} = \frac{e}{\sin{(\angle BDC)}}$$
and
$$ \frac{d}{\sin{(\angle ACB)}} = \frac{e}{\sin{(\angle BAC)}}$$
respectively.
Noting that $\angle BDC = \angle BAC$ since both angles are subtended by the arc BC, we find that
\begin{equation*}
\frac{a}{\sin{(\angle CBD)}} = \frac{d}{\sin{(\angle ACB)}}.\tag*{(1)}
\end{equation*}
Now observe that since ABDE is a cyclic quadrilateral, the opposite angles are supplementary so $\angle ABD = 180^{\circ} - E.$ Considering the angle sum at $B$ then gives
\begin{align*}
\angle CBD &= B - \angle ABD \\
&= B + E - 180^{\circ}.
\end{align*}
Similarly, $\angle ACD = 180^{\circ} - E$ since AEDC is a cyclic quadrilateral, and so
\begin{align*}
\angle ACB &= C - \angle ACD \\
&= C + E - 180^{\circ}.
\end{align*}
Placing these into $(1)$ and using the identity $\sin{(\theta - 180^{\circ})} = - \sin{\theta}$ gives the required result.
