we do a little trigbashing let $\alpha = \angle{LCD} = \angle{LI_CD} = \angle{LCF}$ if F is foot of altitude from C to AB, because $CF \parallel DI_C$, and $\beta = \angle{LCB} = \angle{LCA}$ then $\frac{DI_C}{CI_C} = \frac{1}{2\cos\alpha} = \frac{QI_C}{CI_C} = \sin\beta \implies 2\sin\beta\cos\alpha = 1$ note that by right triangles, if $R = PQ \cap CI_C$ then $\frac{CR}{I_CR} = \frac{1}{\tan^2\beta} \implies$ it suffices to show that $\frac{CR-I_CR}{2 I_CR} = \frac{CL}{LI_C}$, or $\frac{CR-I_CR}{2 I_CR} = \frac{CF}{CD}$ by the angle bisector theorem, so it suffices to show $\frac{\frac{1}{\tan^2\beta}-1}{2}=\cos{2\alpha}$. since $\cos\alpha = \frac{1}{2\sin\beta}$, $\cos{2\alpha} = 2\cos^2\alpha-1 = \frac{1-2\sin^2\beta}{2\sin^2\beta}$ so the equation becomes $$\frac{1}{\tan^2\beta}-1=\frac{1-2\sin^2\beta}{\sin^2\beta}$$$$\frac{\cos^2\beta-\sin^2\beta}{\sin^2\beta} = \frac{1-2\sin^2\beta}{\sin^2\beta}$$which is clearly true.