It has been posted before : http://www.mathlinks.ro/Forum/viewtopic.php?t=31693 Kostas Vittas.

Let $ABCD$ be the bicentric quadrilateral and let $E, F, G, H$ be the points of tangency of the incircle $(I)$ with $\overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}$, respectively. Let $A', B', C', D'$ be the midpoints of $\overline{HE}, \overline{EF}, \overline{FG}, \overline{GH}$, respectively. Let $I, O$ be the center of the circumcircle, incircle, respectively, and denote $X \equiv AC \cap BD.$ By Brianchon's Theorem applied to "hexagons" $AEBCGD$ and $BFCDHA$, it follows that $X \in EG$ and $X \in FH$, respectively. Now, consider the inversion with pole $I$ about $\odot (EFGH).$ It is clear that $E, F, G, H$ are their own inverses, and $A, B, C, D$ are sent to $A', B', C', D'$, respectively. Note that from the midpoint conditions, we have $A'B', \; C'D' \parallel FH$ and $B'C', \; D'A' \parallel EG.$ Therefore, $A'B'C'D'$ is a parallelogram, but is also cyclic (because $ABCD$ was cyclic). Hence, $A'B'C'D'$ must in fact be a rectangle. By the parallel lines, it follows that $EG \perp FH.$ Furthermore, note that the image $O'$ of $O$ under inversion lies on the line $IO.$ Therefore, it suffices to prove that $I, O', X$ are collinear. To do so, we use complex numbers. WLOG let $\odot (EFGH)$ be the unit circle, and let $FH$ be parallel to the real axis. From $EG \perp FH$, it follows that $e = \tfrac{1}{g}$ and $h = -\tfrac{1}{f}.$ By the intersection of chords formula, we get \[x = \frac{eg(f + h) - fh(e + g)}{eg - fh} = \frac{f + h + e + g}{2}.\] Because the center of a rectangle is also it's center of mass, we derive \[o' = \frac{a' + b' + c' + d'}{2} = \frac{\frac{h + e}{2} + \frac{e + f}{2} + \frac{f + g}{2} + \frac{g + h}{2}}{4} = \frac{e + f + g + h}{4}.\] It is then clear that $\frac{x}{o'} \in \mathbb{R}$, as required. $\square$

Inversion around $(I) $ (incircle of the quadrilateral) reduces the problem to the following lemma: Let $ABCD $ be a cyclic quadrilateral with circumcenter $O $ such that $AC\perp BD $. Then the anticenter of $ABCD $ lies on $OP $, where $P\equiv AC\cap BD $. Let $M $ and $N $ be the midpoints of $AC $ and $BD $ respectively. The proof now follows immediately as $OMPN $ is a rectangle ($\implies MN $ bisects $OP $).

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