How to prove the concurrence of three lines? 
Let quadrilateral $ABCD$ inscribe in circle whose center is $O$ and let $AC$ and $BD$ intersect at $P$,while $Q$ be a point on the straight line $OP$.The circumcenters of $\Delta QAB$, $\Delta QBC$, $\Delta QCD$, $\Delta QDA$ are $O_1$, $O_2$, $O_3$, $O_4$ respectively.Then how to prove the concurrence of the straight lines $O_1 O_3$, $O_2 O_4$, $OP$?($O$ is not coincide with $P$)

I'm sorry that I can't provide any useful ideas.(I have tried to violently calculate through trigonometric functions,but it only made me crazy)
And I'm very sorry for my possible wrong grammars and strange expressions because I'm still an English learner and my first language is very different from the English system.
I'd appreciate it if someone could share his ideas about the question.
 A: Here's an outline of my coordinate proof.


Lemma. For $\overline{RR'}$ a diameter of $\bigcirc O$ with radius $r$, and $\overline{AA}$ a chord through $P$ with $p:=|OP|$, and $\alpha :=\angle RR'A$ and $\alpha':=\angle R'RA'$,
$$\frac{r-p}{r+p}=\frac{(r-p)/|PA|}{(r+p)/|PA|}\quad\underbrace{\overbrace{\;\qquad=\qquad\;}^{\triangle PRA'\sim\triangle PAR'}}_{\triangle PR'A'\sim\triangle PAR}\quad\frac{|RA'|/|R'A|}{|R'A'|/|RA|}=\frac{|RA|/|R'A|}{|R'A'|/|RA'|}=\frac{\tan\alpha}{\tan\alpha'} \tag1$$

With this, and the Inscribed Angle Theorem, we can coordinatize chords $\overline{AA'}$ and $\overline{BB'}$ passing through $P$ as (abusing notation so that $\operatorname{cis}\theta=(\cos\theta,\sin\theta)$)
$$
A  = r \operatorname{cis} 2\alpha \qquad 
A' = r \operatorname{cis}(2\alpha'-\pi) \qquad
B  = r \operatorname{cis} 2\beta \qquad 
B' = r \operatorname{cis}(2\beta'-\pi)
\tag2$$
such that
$$\frac{\tan\alpha}{\tan\alpha'}=\frac{r-p}{r+p}=\frac{\tan\beta}{\tan\beta'}\tag3$$
With $P=(p,0)$ and $Q=(q,0)$, we find after a bit of symbol-crunching that the circumcenters of $\triangle QAB$ and $\triangle QA'B'$ are
$$\begin{align}
K\phantom{'} &:= \phantom{-}\frac{r^2 - q^2}{2((r - q) + \tan\alpha \tan\beta (r + q))} 
\;\left( 1 - \tan\alpha \tan\beta, \tan\alpha + \tan\beta\right) \tag4 \\[8pt]
K' &:= - \frac{r^2 - q^2}{2 ((r + q) + \tan\alpha' \tan\beta' (r - q))} 
\;\left( 1 - \tan\alpha' \tan\beta', \tan\alpha' + \tan\beta'\right) \tag5
\end{align}$$
Then, $\overleftrightarrow{KK'}$ meets $\overleftrightarrow{OP}$ (aka, the $x$-axis) at

$$K_\star := \left(\frac{p (r^2-q^2)}{2 (r^2-p q)}, 0\right) \tag6$$

This is independent of $\alpha$, $\beta$, $\alpha'$, and $\beta'$. That is, it's defined solely by $P$ and $Q$ along a diameter of $\bigcirc O$.
Consequently, not only does $K_\star$ also lie on the line through the circumcenters of $\triangle QA'B$ and $\triangle QAB'$ (thus completing the proof), but it lies on the line through the circumcenters of any pair of "opposite" triangles determined by $Q$ and two chords through $P$.
In particular, taking $\overline{AA'}\perp\overline{PQ}$, and $B=A$ and $B'=A'$, we have the case where the circumcircles are internally tangent to $\bigcirc O$. This gives us a construction of $K$ as the intersection of $\overline{OA}$ with the perpendicular bisector of $\overline{QA}$ (likewise for $K'$), and we see that $K_\star$ is the projection of this point onto $\overline{PQ}$:

Be that as it may ...
The independence of $K_\star$ seems like it's trying to tell me something (ideally, something that would have helped me avoid coordinatizing the problem), but I'm not sure what it is. I may need to come back to it with fresh eyes at a later date.
A: Here's a slick inversive proof.
Let $(QAB)$ meet $(QCD)$ again at $X$, let $(QBC)$ meet $(QDA)$ again at $Y$. It suffices to show that the centre of $(QXY)$ (which is $O_1O_3\cap O_2O_4$) lies on $OP$.
First, note that $(QAC)$ and $(QBD)$ meet again on line $OP$. Indeed, let $(QAC)$ meet $OP$ again at $R$. Then $PQ\cdot PR=PA\cdot PC=PB\cdot PD$, which implies that $R$ lies on $(QBD)$ too.
Now invert centre $Q$ with arbitrary radius, and denote inverses with $'$. The line $OP$ is fixed, and $ABCD$ maps to another cyclic quadrilateral $A'B'C'D'$, whose center $J$ still lies on $OP$. Since $(QAC)$ and $(QBD)$ meet on $OP$, we know that $R'=A'C'\cap B'D'$ is on $OP$.
Circles $(QAB)$, $(QCD)$ map to lines $A'B'$, $C'D'$ respectively, so $X'=A'B'\cap C'D'$. Similarly, $Y'=B'C'\cap D'A'$.
But by Brokard's theorem, we know that $JR'\perp X'Y'$, i.e. $OP\perp X'Y'$. This implies that the centre of $(QXY)$ lies on $OP$, as desired.
A: My friend has given his answer.I think I should present it.
Let $E=O_1O_3\cap O_2O_4$,and $F=AB\cap CD$,$G=AD∩BC$.Let $\Gamma$ be a circle whose centre is $E$,radius is $EQ$.(If $E$ is coincide with $Q$,then let $\Gamma$ be a null circle.)And $\Gamma$ meets $\bigcirc O_1$ again at $K$.
Then $FB\cdot FA=FC\cdot FD$,which implies that  $F$ is  on the radical axis of $\bigcirc O_1$ and $\bigcirc O_3$.So $FQ\perp O_1O_3$,which means $FQ\perp O_1E$.So $F$ is on the radical axis of $\bigcirc O_1$ and $\Gamma$.Then we have that $F,K,Q$ are on the same line,which implies that $FK\cdot FQ=FB\cdot FA$.Thus $F$ is on the radical axis of $\bigcirc O$ and $\Gamma$.Similarly,$G$ is on the radical axis of $\bigcirc O$ and $\Gamma$.So the straight line $FG$ is the radical axis of $\bigcirc O$ and $\Gamma$.Therefore,$OE\perp FG$.
And by Brokard's theorem,we know that $OP\perp FG$.Thus,$O,E,P$ are collinear.
So,the staight lines $O_1O_3$,$O_2O_4$ and $OP$ are concurrent at $E$.
