This is non-bary bashable

Let $P,Q,R,S$ be pairwise intersections of internal angle bisectors of $(\widehat{A},\widehat{B}),$ $(\widehat{B},\widehat{C}),$ $(\widehat{C},\widehat{D}),$ $(\widehat{D},\widehat{A})$ and let $P',Q',R,S'$ be pairwise intersections of external angle bisectors of $(\widehat{A},\widehat{B}),$ $(\widehat{B},\widehat{C}),$ $(\widehat{C},\widehat{D}),$ $(\widehat{D},\widehat{A}).$ $P$ and $P'$ become incenter and excenter of the triangle bounded by $DA,AB,BC$ $\Longrightarrow$ $A,B,P,P'$ are concyclic and likewise $B,C,Q,Q'$ are concyclic $\Longrightarrow$ $\widehat{BPP'}=\widehat{BAP'}=\widehat{BCQ}=\widehat{BQ'Q}$ and similarly $\widehat{BP'P}=\widehat{BQQ'}$ $\Longrightarrow$ $\triangle BPP' \sim \triangle BQQ'$ are directly similar $\Longrightarrow$ $\frac{BP'}{BQ}=\frac{BP}{BQ'} \Longrightarrow \frac{\overline{BP} \cdot \overline{BQ}}{\overline{BP'} \cdot \overline{BQ'}}=-1 \Longrightarrow \frac{\text{pow}(B,\odot (I))}{\text{pow}(B,\odot (J))}=-1.$ Similarly, the ratio of the powers of $A,C$ and $D$ WRT $\odot(I)$ and $\odot(J)$ is $-1$ $\Longrightarrow$ $(O),(I),(J)$ are coaxal and $O$ divides $\overline{IJ}$ in the ratio $-1,$ i.e. $O$ is midpoint of $\overline{IJ},$ as desired.

Let $P,Q,R,S$ be pairwise intersections of internal angle bisectors of $(\widehat{A},\widehat{B}),$ $(\widehat{B},\widehat{C}),$ $(\widehat{C},\widehat{D}),$ $(\widehat{D},\widehat{A})$ and let $P',Q',R',S'$ be pairwise intersections of external angle bisectors of $(\widehat{A},\widehat{B}),$ $(\widehat{B},\widehat{C}),$ $(\widehat{C},\widehat{D})$, $(\widehat{D},\widehat{A})$. Let $E = \overline{AB}\cap\overline{CD}$ and $F = \overline{AD}\cap\overline{BC}$. Let $K$, $K'$, $L$ and $L'$ be the projection of $P$ and $P'$ onto $\overline{AB}$, and $Q$ and $Q'$ onto $\overline{BC}$, respectively. WLOG $AB > CD$. Note that $ABPP'$ are concyclic, since,$\angle P'AP = 90^{\circ} = P'BP$. Similarly for $BCQQ'$, $CDRR'$ and $ADSS'$. Let $I' = \overline{QL} \cap \overline{PK}$. Now, \[\angle PQI' = \angle BQL = 90^{\circ} - \tfrac12\angle B = \angle KPB = \angle QPI'\]Thus, $I'P = I'Q$ and $\angle PIQ = \angle B$. \[\angle PIQ = 2\angle PRQ = 2\angle R'RC = 2\angle R'DC = 180^{\circ} - \angle D = \angle B\]Therefore, $I \equiv I'$. Similarly, $J = P'K' \cap P'L'$. Now note that $K$ and $K'$ are isotomic conjugates w.r.t. $\overline{AB}$ (since, $P$ and $P'$ are the incentre and $F$-excentre of $\triangle FAB$.) Thus, the result follows. (Also, for the case where $AB = CD$, replace $F$ in the above proof with $E$ and for the case where $ABCD$ is a rectangle, the three circles are concenctric)

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let $I_1,I_2,I_3,I_4$ the intersections of the in-bisectors wrt $A$ and $B$ ,$B$ and $C$,$C$ and $D$,$D$ and $A$. similarly $J_1,J_2,J_3,J_4$ the intersections of the ex-bisectors.Remark that $I_1,J_1$ are the incenter and excenter of $ ABK_1$,the triangle formed by $ AB,BC,AD$, if $L_1$ s the midpoint of the arc $AB$ of its circumcircle not containing $K_1$ then $L_1$ is the midpont of $I_1J_1$ thus $BA$-bisector is the midline of $h_1,h'_1$ (the altitudes of $I_1,J_1$ to $AB$ ) idem we deduce that the $BC$-bisector is the midline of $h_2,h'_2$ ( the altitudes of $I_2,J_2$ to $BC$ ) .Moreover angle chase leads to $I_2I_4AB,ABJ_2J_4$ are cyclic so Brahmagupta 's yields $I_1I$ is the altitude line of $I_1AB$ and $J_1J$ is the altitude line of $J_1AB$ hence $I\in h_1,J\in h'_1$ similarly we get $I\in h_2,J\in h'_2$ therefore $O$ is the midpoint of $IJ$ RH HAS

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