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 $\triangle OMG\implies NO=NM$.