[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.656131964867349, xmax = 27.072735601891157, ymin = -11.846767671002189, ymax = 7.785867598733776; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw(circle((8.5,-3.6), 6), linewidth(1.2) + dtsfsf); draw((3.872968729519343,0.2197619849925414)--(12.859879926796904,0.5220682944261639), linewidth(1.2)); draw((10.029893586558202,2.201674380194745)--(8.633575671841877,-7.570915139709353), linewidth(1.2)); draw((8.366424328158123,0.37091513970935264)--(8.633575671841877,-7.570915139709353), linewidth(1.2)); draw((12.859879926796904,0.5220682944261639)--(13.127031270480654,-7.419761984992542), linewidth(1.2)); draw((3.7762431908187954,3.0952025553175107)--(4.1401200732030965,-7.722068294426164), linewidth(1.2)); draw((4.1401200732030965,-7.722068294426164)--(13.127031270480654,-7.419761984992542), linewidth(1.2)); draw((10.029893586558202,2.201674380194745)--(10.089527824311496,0.42887777528675747), linewidth(1.2)); draw((3.7762431908187954,3.0952025553175107)--(12.859879926796904,0.5220682944261639), linewidth(1.2)); draw((3.7762431908187954,3.0952025553175107)--(10.029893586558202,2.201674380194745), linewidth(1.2)); draw((3.679517652118248,5.97064312564248)--(12.859879926796904,0.5220682944261639), linewidth(1.2)); draw((3.679517652118248,5.97064312564248)--(3.7762431908187954,3.0952025553175107), linewidth(1.2)); draw((3.7762431908187954,3.0952025553175107)--(8.366424328158123,0.37091513970935264), linewidth(1.2)); /* dots and labels */ dot((8.5,-3.6),dotstyle); label("$O$", (8.579063440232295,-3.4107188874479926), NE * labelscalefactor); dot((3.872968729519343,0.2197619849925414),dotstyle); label("$A$", (3.9459931700881117,0.41156408542095635), NE * labelscalefactor); dot((12.859879926796904,0.5220682944261639),dotstyle); label("$B$", (12.941871277951401,0.7204354367639018), NE * labelscalefactor); dot((8.366424328158123,0.37091513970935264),linewidth(4pt) + dotstyle); label("$M$", (8.443932224019756,0.5273908421745609), NE * labelscalefactor); dot((8.633575671841877,-7.570915139709353),linewidth(4pt) + dotstyle); label("$K$", (8.714194656444834,-7.426046454906282), NE * labelscalefactor); dot((10.029893586558202,2.201674380194745),dotstyle); label("$P$", (10.104115737488089,2.399923409691167), NE * labelscalefactor); dot((3.7762431908187954,3.0952025553175107),linewidth(4pt) + dotstyle); label("$Q$", (3.8494708727934412,3.249319625884267), NE * labelscalefactor); dot((10.089527824311496,0.42887777528675747),linewidth(4pt) + dotstyle); label("$H$", (10.162029115864891,0.5853042205513631), NE * labelscalefactor); dot((4.1401200732030965,-7.722068294426164),linewidth(4pt) + dotstyle); label("$D$", (4.216255602513189,-7.561177671118821), NE * labelscalefactor); dot((13.127031270480654,-7.419761984992542),linewidth(4pt) + dotstyle); label("$C$", (13.212133710376479,-7.271610779234809), NE * labelscalefactor); dot((10.059710705434846,1.3152760777407515),linewidth(4pt) + dotstyle); label("$E$", (10.142724656405957,1.4733093556623311), NE * labelscalefactor); dot((3.679517652118248,5.97064312564248),dotstyle); label("$A'$", (3.7529485754987704,6.164293004183314), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution(with amar_04): Firstly we'll give a defination of few points. Let $\omega$ be the given circle. Let $AQ \cap \omega=D \neq A$. Let $DK \cap \omega=C\neq D$ and let $QB \cap PH=E$. and finally let $A'$ be the reflection of $A$ over $Q$. Observe that $C$ is the $A$-antipode w.r.t $\omega$ and similiarly $D$ is the $B$-antipode w.r.t $\omega \implies ABCD$ is a rectangle. Now $PH \parallel AA'$. Consider a homothety centred at $B$ s.t. $PH \mapsto AA'$ under this homothety. Note that $Q$ is the midpoint of $AA'$ be defination. So we just need to show that under this homothety $P \mapsto A'$ or that $B-P-A'$ are collinear. Now note that by thales we have $MQ \parallel A'B$. So if we are able to prove $MQ \parallel PB$ we would be done. Now we can restate the problem as follows. Restated problem wrote: Let $ABCD$ be a rectangle and $P$ be any point on $\odot(ABCD)$. Let $M,K$ be the midpoint of $CD,AB$ respectively. Let the line perpendicular to $PK$ at $K$ intersect $AD$ at $Q$. Then prove that $QM \parallel PB$. For this let $PQ \cap \odot(ABCD)=T$ . Note that $TO \cap PK=Y \in \odot(ABCD) \implies MTKY$ is a parallelogram $\implies MT \equiv MY \parallel PK \implies \angle PTM=90^\circ \implies TMAQ$ cyclic. Now let $TP \cap AB=X$. Note that $\angle XPB=\angle TAB \equiv \angle TAM=\angle TQM \implies MQ \parallel PB$. Done $\blacksquare$.

Thank you for recall own problem. Here is an extension for this. Let $M$ be the midpoint of the chord $AB$ of a circle centered at $O$. Point $K$ is symmetric to $M$ with respect to $O$, and points $P$, $Q$ are chosen arbitrarily on the circle. $R$ is on arc $PQ$ such that $\frac{RP}{RQ}=\frac{AP}{AQ}$. Let $N$ be the intersection of the line perpendicular to $AB$ through $A$ and the line perpendicular to $RK$ through $R$. Let $H$, $L$ be the projection of $P$, $Q$ onto $AB$ respectively. Prove that $BN$ passes through intersection of $PL$ and $QH$. When $P=Q$, we have original problem.

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I'm really honored when this problem of mine was selected in the Sharygin Geometry Olympiad Final Round. The Sharygin Geometry Olympiad has always been the largest and most prestigious Geometry competition in the world. I am grateful to the people who created this contest. Another generalization for this problem, you can see here.