Invert with respect to $O$ with power $\frac{R^2}2$, where $R$ is the circumradius of $ABCD$. The line $OP$ turns into itself, while the lines $O_1O_3,O_2O_4$ turn into the circles $(OMP),(ONQ)$, where $M,N,P,Q$ are the midpoints of $AB,BC,CD,DA$ respectively. It hence suffices to prove that those two circles meet again on $OP$. If $T=AB\cap CD,\ S=AD\cap BC$, then $(OMP)$ is the circle with diameter $OT$, so it passes through the foot $X$ of the perpendicular through $O$ to $ST$. In the same way we prove that $(ONQ)$ passes through $X$. Since, as is well-known, the perpendicular from $O$ to $ST$ is precisely $OP,\ X\in OP$ as well, and we're done.

There is another interesting problem which is quite similar to this one: http://www.mathlinks.ro/Forum/viewtopic.php?p=462779

tangents to the circle at $A,B,C,D$ intersect at $O_1',O_2',O_3',O_4'$ and $O,O_i,O_i'$ are collinear.from brianchen theorem $O_3'O_1'$ and $O_2'O_4'$ intersect at $P$ and we are done.

See here for the generalization of the problem: http://www.mathlinks.ro/viewtopic.php?t=187214

shobber wrote: The centre of the circumcircle of quadrilateral $ ABCD$ is $ O$ and $ O$ is not on any of the sides of $ ABCD$. $ P = AC \cap BD$. The circumecentres of $ \triangle{OAB}$, $ \triangle{OBC}$, $ \triangle{OCD}$ and $ \triangle{ODA}$ are $ O_1$, $ O_2$, $ O_3$ and $ O_4$ respectively. Prove that $ O_1O_3$, $ O_2O_4$ and $ OP$ are concurrent. Denote by $ R_2$, $ R_4$ the reflection points of $ O$ in $ O_2$, $ O_4$, respectively. We have $ R_2B$, $ R_2C$, $ R_4A$, $ R_4D$ are all tangent to $ (O)$. Call $ Q$ the intersection of $ AD$ and $ BC$. We know that $ P$ lies on the polar line of $ Q$ with respect to $ (O)$, so we conclude that $ R_2$, $ R_4$, $ P$ are collinear. It follows $ O_2O_4$ passes through the middle point of the segment $ OP$. Similarly, $ O_1O_3$ also passes through the mid-point of segment $ OP$. Hence, $ O_1O_3$, $ O_2O_4$, and $ OP$ are concurrent at the mid-point of the segment $ OP$. $ \square$

shobber wrote: The centre of the circumcircle of quadrilateral $ ABCD$ is $ O$ and $ O$ is not on any of the sides of $ ABCD$. $ P = AC \cap BD$. The circumecentres of $ \triangle{OAB}$, $ \triangle{OBC}$, $ \triangle{OCD}$ and $ \triangle{ODA}$ are $ O_1$, $ O_2$, $ O_3$ and $ O_4$ respectively. Prove that $ O_1O_3$, $ O_2O_4$ and $ OP$ are concurrent.

Image not found Solution. Let $ A'\in O_{1}O_{4}\cap OA, \ \ B'\in O_{1}O_{2}\cap OB,\ \ C'\in O_{2}O_{3}\cap OC, \ \ D'\in O_{3}O_{4}\cap OD.$ So these points are the middlepoints of segments $ OA, \ OB, \ OC$ and $ OD$ respectively. $ \Longrightarrow OA' = OB' = OC' = OD'.\ (*)$ Using the radical axis argument we get that: $ OA'\perp O_{1}O_{4}, OB'\perp O_{1}O_{2}, OC'\perp O_{2}O_{3}$ and $ OD'\perp O_{3}O_{4}. \ (**)$ From relations $ (*)$ and $ (**)$ we obtain that the quadrilateral $ O_{1}O_{2}O_{3}O_{4}$ is tangent (the center of its incircle is $ O$). Newton's Theorem applied in $ O_{1}O_{2}O_{3}O_{4}$ yieds: $ O_{1}O_{3}, O_{2}O_{4}, A'C', B'D'$ concur. Now it's easy to see that line $ A'C'$ meets $ [OP]$ in its middlepoint and so does line $ B'D'.$ Thus lines $ O_{1}O_{3} \ , O_{2}O_{4} \ , A'C' \ , B'D'$ and $ OP$ concur in the middlepoint of $ [OP].$

where can i find what`s newton theorem? :

b555 wrote: where can i find what`s newton theorem? : What he means by Newton's Theorem is Theorem 2 in Circumscribed quadrilaterals revisited (or you can find it at http://www.mathlinks.ro/Forum/viewtopic.php?t=19549 ). I don't know if it was known to Newton, but usually people mean a different fact (also about circumscribed quadrilaterals) by "Newton's theorem". darij

Lemma: $O_1O_3$ bisects $OP$. Proof: Let $AB$ and $CD$ meet at $E$, and $AD$ and $BC$ meet at $F$. Let $FP$ meet $OE$ at $X$. By Brokard's Theorem, $FP$ is the polar of $E$ w.r.t. $(O)$, so $EA \cdot EB = ED \cdot EC = EX \cdot EO$. Therefore, $OXAB$ and $OXDC$ are cyclic, so $X$ is the second intersection of $(O_1)$ and $(O_3)$. Then $O_1O_3$ is the perpendicular bisector of $OX$, so $O_1O_3 \parallel FX$ and $O_1O_3$ passes through the midpoint of $OX$; hence, it also bisects $OP$, as desired. Similarly, we can get that $O_2O_4$ bisects $OP$, so $OP$, $O_1O_3$, and $O_2O_4$ are concurrent, as desired.

I see nobody posted this solution,so i will do that! Just observe that the circumcenter of APD,O,the circumcenter of BPC and P make a parallelogram(angle-chasing).After that we use aforementioned lemma(we proved that) and problem is finished

It's easy problem. Show that O1 P O3 O parallelogram.

Problem can be generalized by taking point $O$ as any point on the plane. Here is the general form which is also true: $ABCD$ circumscribed about a circle with center $O$ and diagonals $AC,BD$ intersect at $P$. $Q$ is an arbitrary point on the plane and $X,Y,Z,W$ are the circumcenters of triangles $ABQ,BCQ,CDQ,DAQ$ respectively. Show that $XZ,YW,OP$ are concurrent.

GuvercinciHoca wrote: Problem can be generalized by taking point $O$ as any point on the plane. Here is the general form which is also true: $ABCD$ circumscribed about a circle with center $O$ and diagonals $AC,BD$ intersect at $P$. $Q$ is an arbitrary point on the plane and $X,Y,Z,W$ are the circumcenters of triangles $ABQ,BCQ,CDQ,DAQ$ respectively. Show that $XZ,YW,OP$ are concurrent. Solution: Let $(BCQ)\cap (ADQ)=R,Q$ and $(ABQ)\cap (DCQ)=S,Q$. It's easy to show that $XZ\cap YW=O'$ is the circumcenter of $\triangle QRS$. Thus, we want to prove $O,P,O'$ are collinear. Let $E,F$ be the radical centers of $(ABQ),(DCQ),(ABCD)$ and $(ADQ),(BCQ),(ABCD)$ respectively. $E\in QS \quad , F\in QR \implies EF$ is radical axis of $(O),(O')$. So $OO'\perp EF$. On the other hand we have $OP\perp EF$ by brocards theorem $\implies O,P,O'$ are collinear.

GuvercinciHoca wrote: Problem can be generalized by taking point $O$ as any point on the plane. Here is the general form which is also true: $ABCD$ circumscribed about a circle with center $O$ and diagonals $AC,BD$ intersect at $P$. $Q$ is an arbitrary point on the plane and $X,Y,Z,W$ are the circumcenters of triangles $ABQ,BCQ,CDQ,DAQ$ respectively. Show that $XZ,YW,OP$ are concurrent. I think its very easy question We will just move the point $D$ on the $(ABC)$ Then we can see that $deg(Z)=deg(W)=deg(P)=1$ Let $L=XZ \cap YW$ then $deg(L)\leq 2 $ But if we take $D=B$ $\implies$ $XZ=YW$ because of coincide lemma $deg(L)\leq 1$ We want to prove that $O-P-L$ collinear.So its enough to show that for $0+1+1+1$ cases of point $D$ Obviously for points $D$ :$D=A$ ,$D=C$ and $AD\parallel BC$ satisfies the condition.$\blacksquare$

sevket12 wrote: GuvercinciHoca wrote: Problem can be generalized by taking point $O$ as any point on the plane. Here is the general form which is also true: $ABCD$ circumscribed about a circle with center $O$ and diagonals $AC,BD$ intersect at $P$. $Q$ is an arbitrary point on the plane and $X,Y,Z,W$ are the circumcenters of triangles $ABQ,BCQ,CDQ,DAQ$ respectively. Show that $XZ,YW,OP$ are concurrent. I think its very easy question We will just move the point $D$ on the $(ABC)$ Then we can see that $deg(Z)=deg(W)=deg(P)=1$ Let $L=XZ \cap YW$ then $deg(L)\leq 2 $ But if we take $D=B$ $\implies$ $XZ=YW$ because of coincide lemma $deg(L)\leq 1$ We want to prove that $O-P-L$ collinear.So its enough to show that for $0+1+1+1$ cases of point $D$ Obviously for points $D$ :$D=A$ ,$D=C$ and $AD\parallel BC$ satisfies the condition.$\blacksquare$ Why... Just why