see here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2014625&sid=2359cf55efbf13451fe8d818ec1e676d#p2014625

Here is my solution(I don't know if it has been posted before):- Some trivial observations are as follows:- 1. $ M_1M_3\perp M_2M_4 $ and $ I_1I_3\perp I_2I_4 $. 2. $ M_2M_4 $ is the perpendicular bisector of $ I_1I_3 $ and similar for $ M_1M_3 $. We have to prove that the center of homothety of $ I_1I_3 $ and $ M_1M_3 $ is same as the center of homothety of $ I_2I_4 $ and $ M_2M_4 $. To prove this take $ I_1X\perp AB, I_3Y\perp CD $ where $ X,Y $ lie on $ AB,CD $ respectively. $ Z=I_1X\cap I_3Y $. Note that, since $ \frac {AX}{XB}=\frac {DY}{YC} $, so $ Z,K,O $ are collinear. So both the centers of homothety lies on $ OK $. Suppose, $ M,M' $ are the midpoints of $ M_1M_3 $ and $ M_2M_4 $. $ I,I' $ are the midpoints of $ I_1I_3,I_2I_4 $. But note that, if $ J=M_1M_3\cap M_2M_4 $, then $ KIJI' $ and $ JMOM' $ are rectangles. So $ IM,I'M',OK $ are concurrent. So the two centers of homothety are the same. So done. Edited.

RSM wrote: $ KIJI' $ and $ KMOM' $ are rectangles. Maybe this is a typo. $KIJI'$ and $JMOM'$ are rectangles. However, I think you showed a nice solution, RSM I used lengthy and boring calculation to solve this...

Dear Mathlinkers, a synthetic proof can be seen on http://perso.orange.fr/jl.ayme vol. 20 Forme et mouvement p. 5 Sincerely Jean-Louis

Official solution by ARMO-2010: We have points $(A,I_1, M_2)$, $ (A,I_4,M_3) $, $(B,I_1,M_4)$, $(B,I_2,M_3)$, $(C,I_2,M_1)$, $(C,I_3,M_4)$, $(D,I_3,M_2)$, $(D,I_4,M_1)$ are lie one line. So, $I_2I_4\| M_2M_4$, $I_1I_3\|M_1M_3$ and from $M_1M_3\bot M_2M_4$ $ \Rightarrow $ $I_1I_3\bot I_2I_4$. Let $ A'=DM_1\cap BM_4$, $ B'=AM_2\cap CM_1 $, $ C'=BM_3\cap DM_2$ and $ D'=AM_3\cap CM_4 $. We have $A'B'C'D'$ is restangle and $A'B'\| I_2I_4\| M_2M_4\| C'D'$, $B'C'\| I_1I_3\| M_1M_4\| A'D'$. Let $P=M_1I_1\cap M_3I_3$, $PI_2\cap M_1M_3=X$ and $PI_2\cap I_1I_3=Y$ and $PI_2\cap B'C'=Z$. Then \[ \frac{I_1Y}{I_3Y}=\frac{M_1X}{M_3X}=\frac{B'Z}{C'Z} .\]Hence, $M_2,I_2,P$ are collinear. Analoguously, $M_4,I_4,P$ are collinear. Hence, $M_1I_1, M_2I_2, M_3I_3, M_4I_4 $ are intersect at one point $P$. This is a hard problem in 'Uzbekistan NO', but old. I think the official solution is fabulous (by autor the problem PAVEL KODJEVNIKOV).