First, we will proof some lemma:
Lemma 1: Suppose a line $l$ pass $A,B,C,D$ four points, let $B',C',D'$ be the image when doing inversion of $A$. Then $(A,C;B,D)$ is a harmonic division if and only if $C'$ is the midpoint of $B'D'$.
Proof: Let $A$ be zero, and $B,C,D$ has the coordinate $b,c,d$. So we have $(A,C;B,D)$ is a harmonic division if and only if $(c-b)d=b(d-c)$, if and only if $1/b+1/d=2/c$, if and only if $C'$ is the midpoint of $B'D'$.
Then we use inversion to finish your proof:
We use inversion w.r.t. the incircle of $ABCD$. Let four tangent points of $AD,DC,CB,BA$ to be $Q,R,S,T$. Let $E,F,G,H$ be the midpoint of $QT,TS,SR,RQ$. So $E,F,G,H$ are images of $A,B,C,D$ respectively. Furthermore, let $K$ be the intersection of $EG$ and $HF$, so $K$ is the image of $P$. Let $L$ to be the image of $O$. Let the circumcircle of $HIF$ and $EIG$ be $J$. So we only need to proof, by lemma 1, $JK=KL$.
Now we seek for another description of $L$. Notice that $O$ is the intersection of the perpendicular bisector of $BD$ and the perpendicular bisector of $AC$. Let $HIFU$ be the harmonic quadrilateral, so $L$ is on the circle passing $I$, $U$, and orthogonal to the circumcircle of $HIF$. So we know that $\angle ULI=90^\circ-\angle UJI$. Similarly, let $V$ be the point such that $EIGV$ is the harmonic quadrilateral, so $\angle VLI+\angle VGI=90^\circ$.
Notice that $I,J,K,L$ are collinear* (this is because $O,X,I,P$ collinear), and $K$ is the intersection of $IK$ and circumcircle of $FIH$. So $JU\parallel HF$. Similarly, $VJ\parallel EG$. Since $K$ is the midpoint of $EG$ and $VJ\parallel EG$, we have $KJ=KV$. Similarly, $KJ=KU$. So $K$ is the circumcenter of $UJV$. Since $I,J,L$ are collinear, and also $\angle VLI+\angle VGI=90^\circ$ and $\angle ULI=90^\circ-\angle UJI$, we have $\angle JUL=\angle JVL=90^\circ$. So $J,V,L,U$ are on the same circle, and since $\angle JVL=90^\circ$, $JL$ is the diameter. Since $K$ is the circumcenter of $UJV$, we have $KJ=KL$. So the claim is proved.
- If we don't know this is collinear, we can first using radical axis of circumcircles of $HIF$, $EIG$ and $EFGH$ to proof $I,J,K$ collinear, then we can proof that $U,I,V$ collinear (this can be derived from $\angle EIH+\angle HIG=180^\circ$). Then proof $K$ to be the circumcircle of $JUV$. Then, we let $L'$ to be the intersection of the perpendicular line of $JU$ passing $U$ and the perpendicular line of $JV$ passing $V$. We have $J,U,L',V$ on the same circle and $I,J,K,L'$ are collinear. Then we have $\angle UL'K=90^\circ-\angle UJK=\angle ULK$ and $\angle VL'K=90^\circ-\angle VJK=\angle VLK$, so $L=L'$. So $I,J,K,L$ are collinear,
Best Answer
False. Let $S = \{(-2,0),(-1,1/4),(1,0),(2,0)\}$. Then $m = (1,0)$ (via mathematica, or by showing $(1,0)$ is a local minimum by analyzing small perturbations). Let $s = (1,0)$ and $s' = (1,1/4)$. Now, by symmetry and uniqueness of median, the new median, $m'$, has $x$-coordinate $0$. So $|m-m'| \ge 1$.