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.
Best Answer
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.