In a cyclic pentagon $ABCDE$, prove $\frac{a}{\sin(B+E)}=\frac{d}{\sin(C+E)}$

Question

I've found this question online without a solution, I have tried myself and I couldn't go further than what I've mentioned in the picture.

Question:

My attempt

So I simplified the question into proving $\frac{a}{\sin \beta} = \frac{d}{\sin \alpha}$. It seems to me that the sine law will be used in some way, but I don't know how.

Answer 1

Let $\hat{CD}=2\alpha,$ $\hat{DE}=2\beta,$ $\hat{AE}=2\gamma,$ $\hat{AB}=2\delta$, $\hat {BC}=2\epsilon$ and $R$ be a radius of the circumcircle.

Thus, we need to prove that $$\frac{2R\sin\alpha}{\sin(\alpha+\beta+\gamma+\alpha+\epsilon+\delta)}=\frac{2R\sin\delta}{\sin(\beta+\gamma+\delta+\alpha+\epsilon+\delta)}$$ or $$\frac{\sin\alpha}{\sin(\alpha+540^{\circ})}=\frac{\sin\delta}{\sin(\delta+540^{\circ})}$$ or $$-1=-1$$ and we are done!