Let $ AI, $ $ BJ, $ $ CI, $ $ DJ $ cuts $ \odot (O) $ again at $ A^*, $ $ B^*, $ $ C^*, $ $ D^* $, respectively. Clearly, $ (A^*, C^*), $ $ (B^*, D^*) $ are symmetry WRT $ O $. Since $ AI $ $ \cap $ $ BJ, $ $ AC^* $ $ \cap $ $ BD^*, $ $ IC^*$ $ \cap $ $ JD^* $ lie on the bisector of $ \angle (AD, BC) $, so from Desargues' theorem ($ \triangle AIC^* $ and $ \triangle BJD^* $) we get $ C^*, $ $ D^*, $ $ P $ are collinear. Similarly, we can prove $ R $ $ \in $ $ A^*B^* $, so we get $ A^*B^* $ $ \parallel $ $ C^*D^* $ $ \parallel $ $ OM $. Analogously, we can prove $ B^*C^* $ $ \parallel $ $ D^*A^* $ $ \parallel $ $ ON $, so we conclude that $ OM $ $ \perp $ $ ON $.

Define $A^*,B^*,C^*,D^*$ as same as above. Let $R^*\equiv A^*B^*\cap DC, S^*\equiv AD\cap C^*B^*$ and $Q^*,P^*$ analogously. Applying Pascal's Theorem for $B^*C^*CDAA^*,B^*BCDD^*A^*$ yields that $R^*,S^*,I$ and $R^*,J,Q^*$ are collinear, similarly we get $P^*,Q^*,J$ and $P^*,I,S^*$ are collinear, it follows that $I,J,Q^*,S^*\in \overline{P^*R^*}.$ i.e. $CD,A^*B^*,IJ$ are concurrent. Now the problem is equivalent to: $ABCD$ is a given square and $O$ is the center, let $S,Q$ be points on $\overline{BC},\overline{AD}$ respectively, $R\equiv AB\cap SQ,P\equiv CD\cap SQ.$ Let $M,N$ be the midpoints of $\overline{PR},\overline{QS}.$ Prove that $OM\perp ON.$ [asy][asy] size(7cm); pathpen=black; pointpen=black; pointfontpen=fontsize(10pt); void b(){ pair A=D("A",dir(45),dir(45)); pair B=D("B",dir(135),dir(135)); pair C=D("C",dir(-135),dir(-135)); pair D=D("D",dir(-45),dir(-45)); pair S=D("S",3*B/5+2*C/5,dir(45)); pair Q=D("Q",4*D/5+A/5,dir(120)); pair P=D("P",extension(S,Q,C,D),dir(-45)); pair R=D("R",extension(S,Q,A,B),dir(135)); pair O=D("O",origin,N); pair M=D("M",midpoint(P--R),dir(-90)); pair N=D("N",midpoint(S--Q),dir(-90)); D(R--P); D(B--R); D(D--P); D(A--B--C--D--cycle); } b(); pathflag=false; b(); [/asy][/asy] which is very easy to prove that $OM\parallel AB$ and $ON\parallel AD,$ so $OM\perp ON.\square$

Lemma. $A, B, C, D$ are points such that $AB \parallel CD$. Let the midpoint of $AC, BD$ be $M, N$. Then $MN \parallel AB \parallel CD$. Proof of Lemma. We have $d(M,AB) = \frac{1}{2} d(C, AB) = \frac{1}{2} d(D, AB) = d(N, AB)$ which is enough. $\blacksquare$ Now denote $E = AD \cap BC$. Let $AI, BJ, CI, DJ$ cut the circle $O$ again at $A', B', C', D'$. Trivially $A'C'$ and $B'D'$ are diameters of $O$ - so $A'B'C'D'$ is actually a rectangle. First, $AI \cap BJ$ is the intersection of the bisectors of $\angle EAB, \angle EBA$, so this is the incenter of $\triangle EAB$. Second, $IC' \cap JD'$ is the intersection of the bisectors of $\angle ADC, \angle DCB$. Notice that they are the external angles of $\angle EDC, \angle ECD$. Therefore, this point is the $E$-excenter of $\triangle EDC$. Finally, for $AC' \cap BD'$, as $\angle BAC' = \frac{1}{2}(180-\angle DAB)$, we have $AC'$ be the external angle bisector of $\angle EAB$. Similarly, $BD'$ is the external angle bisector of $\angle EBA$, so $AC' \cap BD'$ is the $E$-excenter of $\triangle EAB$. Therefore, $AI \cap BJ$, $IC' \cap JD'$, and $AC' \cap BD'$ all lie on the angle bisector of $\angle AEB$, so they are colinear. By Desargues's Theorem for $\triangle AIC'$ and $\triangle BJD'$, we have $AB, IJ, C'D'$ concurrent. This implies $C', D', P$ colinear, and similarly $A', B', Q$ are colinear as well. Now by our Lemma, we have $OM \parallel A'B' \parallel C'D'$ and $ON \parallel B'C' \parallel D'A'$. Now $\angle A'B'C' = 90 \implies A'B' \perp B'C' \implies OM \perp ON$, so we have our conclusion. $\blacksquare$