It is easy to prove that O1OO2G is a parallelogram So GN=NO If we can prove OM⊥PQ then the problem is solved => all we have to do is to prove M is the midpoint of XY => we need XP = YQ ∠YCB+∠XAD=∠YGB=∠XGD=∠PAD=∠XAD+∠XAP => ∠YCB=∠XAP => ∠XAD=∠QCY PX/sin∠PAX=AP/sin∠AXY PX=AP/sin∠AXY* sin∠PAX=AP/sin∠GBY*sin∠YCB=AP/GY*BY YQ/sin∠QCY=YC/sin∠YQC YQ= YC/sin∠YQC* sin∠QCY=YC/sin∠CAD*sin∠XAD=YC/CD*XD AP/GY*BY=AP/GD*XD YC/CD*XD PX/YQ=(AP/GD)/ (YC/CD) But we have AP/GD= sin∠PGA/ sin∠CAD YC/CD= sin∠YDC/ sin∠CYD ∠CAD=∠CYD ∠PGA=∠YDC=∠YBC => PX=YQ => M is the midpoint of XY => OM⊥PQ DONE!

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I think the idea for this problem isn’t new. We can solve this problem by using a well know lemma (It’s similar a problem in IMO 1977): Let (O) and (O’) have common AB, a line throughs A, intersect two circles resepective at M, N. We have MN always pass a fixed point T, T is midpoint of CD ( C in (O), D in (O’) and CD is perpendicular to AB at B)

I think the idea for this problem isn’t new. We can solve this problem by using a well know lemma (It’s similar a problem in IMO 1977): Let $(O)$ and $(O^')$ have common $AB$, a line throughs $A$, intersect two circles resepective at $M, N$. We have $MN$ always pass a fixed point $T$, $T$ is midpoint of $CD$ ( $C$ in $(O)$, $D$ in $(O^')$ and $CD$ is perpendicular to $AB$ at $B$)

Can't we draw ${XN }$ ⊥${QP}$ then try to prove ${GX=MX}$ ${\rightarrow}$ ${MO\parallel XN}$ to get it ?Dear Mathlinkers ? By this way I find that :when ${H}$ is not on ${AB \cap CD}$ we also have :${MN=ON}$[See the figure ,I think the conclusion is right ]

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By angle chasing we have ∠P’AX’=∠Y’CB P’X’=AP’*sin∠P’AX’/ sin∠AX’Y’=AP’* sin∠Y’CB/ sin∠AX’Y’=AP’*BY’/AY’ Y’Q’=CY’* sin∠Y’CQ’/ sin∠Y’Q’C=CY’* sin∠X’AD/ sin∠DBC=CY’*X’D/CD P’X’/Y’Q’= (AP’*BY’/AY’) / (CY’*X’D/CD) Notice that ⊿X’CD∽⊿Q’CY’ => CD/CY’=CX’/CQ’ => P’X’/Y’Q’=( AP’ *BY’* CX’)/( CQ’*X’D*AY’) Again notice ⊿BCQ’∽⊿DAP’ => AP’/CQ’=AD/BC => P’X’/Y’Q’=(AD*BY’* CX’)/( BC *X’D*AY’) It is well-known if BD,AC and X’Y’ are concurrent in a circle , then we have AD*BY’* CX’= BC *X’D*AY’ (It can be proved by sin law) Thus we acquire P’X’=Y’Q’=> M’is the midpoint of X’Y’ => OM’⊥ X’Y’ From the above , we have O1OO2G is a parallelogram So GN=NO Which means M’N=NO

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we can apply this result (these points are different from those of the problem): let $XYZW$ a parallelogram and let $l$ be a line through $X$. consider the intersections $P,Q$ of the circles of center $Y$ and $W$ of radius $XY$ and $XW$, resp, with $l$. then $Z$ lies on the perpendicular bisector of $PQ$. proof: let $l$ be the x-axis, $X = (0,0)$, $Y = (a,b)$, $W = (c,d)$, then, $P = (2a,0)$, $Q = (2c,0)$ and $Z = (a + c,b + d)$, and we're done... applying the lemma to parallelogram $OO_1GO_2$ and any line $l$, it follows that $OP = OQ$... this implies that $OM$ is perpendicular to $l$... this implies that triangle $OMG$ is right-angled, and we conclude that $ON = MN$... (there's no need for $H$ to be on $l$)

This well-known result came from Russia 1995, Romanian TST 1996 and others : Let $T$ be the second intersection of $(O_1)$ and $(O_2)$ then $HO\perp GT$ and $H,T,O$ are collinear. Since $OO_2\perp BC, O_1G\perp BC$ then $OO_2//O_1G$, similarly we get $GO_1OO_2$ is a parallelogram thus they intersect at the midpoint $N$ of $OG$. We only need to show that the perpendicular bisector of $PQ$ passes through $O. (*)$ Let $HO\cap (O_1)=\{J\}. \cap (O_2)=\{F\}. OO_2//GJ$ then $OO_2$ is midline parallel to $GJ$ of triangle $GJF.$ It follows that $O$ is the midpoint of $JF$ and $(*)$ is right.

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As mentioned above $N$ is the midpoint of $OG$. Let $(GAB),(GCD)$ intersect at $M'$, then since $AB,M'G,CD$ are concurrent, $M'$ lies on $HG$. It is well known that $\angle OM'G=90^{\circ}$, and from here, we see that $M'$ is the midpoint of $PQ\implies M\equiv M'$. Hence $N$ is circumcenter in right triangle $\triangle OMG\implies NO=NM$.