Show that $PQ$ bisects the line joining the Incenters Given a cyclic quadrilateral $ABCD$ such that $I_1, I_2$ denote the incenters of $\Delta ADC, \Delta ABC$ respectively and $P, Q$ are the midpoints of the arcs $AB,CD$, then show that $PQ$ bisects $I_1I_2$.
I have tried doing it. All I could find out is that if $K\equiv PQ\cap CD$, then, $MI_2KI_1$ forms a parallelogram and that completes the proof. 
But I can't show how it forms a parallelogram. 
 A: Let $R$, $S$ the midpoints of the arcs $BC$, $AD$, respectively, and $I_3$, $I_4$ the incenters of $\triangle BDC$, $\triangle ABD$. It's not difficult to show that $I_1I_3I_2I_4$ is a rectangle: Note that $PQ \perp RS$ by chasing angles and see that are respectively parallel to $I_4I_2$, $I_1I_3$ and $I_4I_1$, $I_2I_3$). In fact, in this angles chasing you'll realize  that $\triangle PI_2I_4$, $\triangle RI_2I_3$, $\triangle QI_3I_1$ and $\triangle SI_4I_1$ are isosceles. Then $PQ$ is the perpendicular bissector of $I_4I_2$, then this line pass through the midpoint of $I_1I_2$. 
A: Tricky. I slightly changed the notation for the midpoints of arcs:

We have that $AI_2, CI_2, AI_1, CI_1$ go through the midpoint of some arc, hence $M_{AB}M_{CD}, M_{AD} M_{BC}$ and $I_1 I_2$ are concurrent by Pascal's theorem. We also have $M_{AB}M_{CD}\perp M_{AD} M_{BC}$, so by setting $X=M_{AB}M_{CD}\cap M_{AD} M_{BC}$ it is not difficult to prove $I_1 X = I_2 X$ through some trigonometry. On the other hand, if you already know that $I_1,I_2,I_3,I_4$ are the vertices of a rectangle the other proof is far superior.
