Let $X=K_2\cap EL, J_1=FX\cap IB$ By Pascal theorem at$IBEXFC$, $BE\cap CF,L,J_1$ is collinear But, $BE\parallel CF$ Hence, $LJ_1\parallel BE\parallel LJ\Rightarrow J=J_1$ QED (Note that $E$Is Symmetrical point of $B$ to $AI$)

Notice that $\angle IJL=\frac{\angle C}{2}$, but $\angle EFI=\angle EBI=\angle ECI=\frac{\angle C}{2}$, similarly we can get $\angle ILJ=\angle IEF=\frac{\angle B}{2}$, so $\triangle ILJ \sim \triangle IEF$, now there's an obvious spiral similarity that takes $LJ$ to $EF$, and it's well-known that $D=LE \cap JF$ lies on $(IEF)$.

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