Let $\alpha, \beta, \gamma, \theta$ denote half the measures of $\widehat{AB}, \widehat{AD}, \widehat{DC}, \widehat{CB}$ respectively, and let $M$ be the midpoint of minor arc $\widehat{BC}$. Furthermore, let $p = \overline{AB} \cap \overline{CD}$, so $J$ is the $P$-excenter of $\triangle PBC$.
It suffices to show that $$[ICM] + [CMJ] = [ICJ] \iff \frac{\sin \angle MCJ}{CI} + \frac{\sin \angle MCI}{CJ} = \frac{\sin \angle ICJ}{CM}.$$The rest is a (surprisingly short) computation.
By the Law of Sines and some angle chasing we may rewrite
\begin{align*}
CI&= BC \cdot \frac{\sin \frac{\gamma}2}{\cos \frac{\beta+\theta}2} \\
CJ &= BC \cdot \frac{\cos \frac{\gamma+\beta}2}{\cos \frac{\theta-\beta}2} \\
CM &= \frac{BC}{2 \cos \frac{\theta}2},
\end{align*}and we also have
\begin{align*}
\sin \angle MCJ &= \sin \frac{\gamma}2 \\
\sin \angle MCI &= \sin \frac{\theta+\alpha}2 \\
\sin \angle ICJ &= \cos \frac{\beta}2.
\end{align*}Now notice that $\cos \frac{\beta+\gamma}2 = \sin \frac{\alpha+\theta}2.$ Then the relation is equivalent to $$\cos \frac{\theta+\beta}2 + \cos \frac{\theta-\beta}2 = 2 \cos \frac{\theta}2 \cos \frac{\gamma}2$$after much cancellation, which is obvious by sum to product.