Let $AB$ and $CD$ intersect at point $F$, and let the midpoints of segments $EF$, $AD$, $BC$ be $X$, $Y$, $Z$ respectively.
By Gauss's theorem, the three points $X$, $Y$, $Z$ are collinear (the Newton line). The line originally containing $X$, $Y$, $Z$ passes through point $F$; applying a scaling by factor $2$ centered at point $E$ will make this line pass through point $I$. Therefore $YZ \parallel FI$.
Let the midpoints of $AC$, $BD$ be $U$, $V$ respectively. Since $FI$ is the angle bisector of $\angle DFA$, $YZ$ is also the angle bisector of $\angle UYV$. But $YZ$ bisects segment $UV$ ($UYVZ$ is a parallelogram), so $UYVZ$ is a rhombus, which gives $YU = YV$. Therefore $AB = 2YU = 2YV = CD$.