cyclic quadrilateral $ABCD$ is given prove $OM \geq ON$ Given a cyclic quadrilateral, $ABCD$ inscribed into a circle with center $O$. let $M$ be the intersections between two diagonals of $ABCD$. $E,F,G,H$ are midpoints of $AB,BC,CD,DA$ . let $N$ be the intersections between $EG$ and $FH$ . prove $OM  \geq ON$
my attempt:
let $P, Q$ be the midpoints of $BD, AC$ I tried to prove $\frac{1}{2}EF\geq ON$ and  $OM\geq\frac{1}{2}EF$ for the first one I used $m_a^2=\frac{2b^2+2c^2-a^2}{4}$ formula, therefore, we will have:
$b^2=OF^2=\frac{4r^2-AC^2}{4}$
$c^2=OE^2=\frac{4r^2-BD^2}{4}$
hence suffice to show $4EF^2\geq 8r^2-AC^2-BD^2$ 
but I don't know how to continue.
also, there is a second solution :
Let $P,Q$ be midpoints of $AC,BD$. $OP$ and $AC$ respectively $OQ$ and $BD$ are perpendicular. $OPMQ$ is cyclic and $N$ is the midpoint of segment $PQ$. Now this means  $N$ lies inside the circle $OPQ$, whose diameter id $OM$ so $ON\le{OM}$ 
but unfortunately, it is not true for all time it depends on the sequence of $Q, M, P, O$ if we have $QPMO$, not $QMPO$ it won't be a cyclic quadrilateral.
please share your ideas in comments and post an answer even if your solution isn't complete. thanks!
 A: 
Construct $P,Q$ as the midpoints of $AC$ and $BD$ respectively.
My proof consists of two parts: First, we prove that $P,N,Q$ are on the same line, and $N$ is always the midpoint of $PQ$. Then, we prove that $OPMQ$ is cyclic, and $OM$ is the diameter. This will conclude our proof because in any circle the maximum distance between two points is the diameter.
Proof of the first part: $N$ is the midpoint of $PQ$
Let $N'$ be the intersection of $PQ$ and $EG$. Since $EQ = \frac{1}{2}\cdot AD = PG$ and $EQ\| PG$, we. see that $EQGP$ is a parallelogram, and hence $EN' = GN'$.
In the original construction of $N$, we have $EH = \frac{1}{2}\cdot BD = FG$ and $EH\|FG$ which implies $EFGH$ is a parallelogram. Hence, because  $N$ is the intersection of the two diagonals, we must have $EN = GN$.
There can only be one midpoint of $EG$. Since both $N$ and $N'$ are  midpoints, we conclude that $N=N'$.
Proof of the second part: $OPMQ$ is cyclic and $OM$ is a diameter
Since $P$, $Q$ are the midpoints of chords, it follows that $OP\perp AC$ and $OQ \perp BD$. Hence, $OPMQ$ is cyclic and $OM$ is a diameter of the circle.
Final Remarks
Since $OM$ is a diameter and $N$ is a point in the circle $OPMQ$, it follows that $OM \geq ON$.
