Let $ABCD$ be a convex quadrilateral, such that $ A$, $ B$, $C$, and $D$ lie on a circle, with $\angle DAB < \angle ABC$. Let $I$ be the intersection of the bisector of $\angle ABC$ with the bisector of $\angle BAD$. Let $\ell$ be the parallel line to $CD$ passing through point $I$. Suppose $\ell$ cuts segments $DA$ and $BC$ at $ L$ and $J$, respectively. Prove that $AL + JB = LJ$.