It's a very nice problem. We invert wrt $O$ and get this equivalent problem: A line $\ell$ cuts $OA,OB,OC$ in $P,Q,R$ respectively. The same line cuts the lines $BC,CA,AB$ in $K,L,M$ respectively. Show that the circles $(OPK),(OQL),(ORM)$ also concur in a point different from $O$. From Desargue's theorem we find that the mapping $P\leftrightarrow K,\ Q\leftrightarrow L,\ R\leftrightarrow M$ is an involution on the line $\ell$. At the same time, we know that a pencil of circles (the set of circles passing through two fixed points determines an involution on any line (which cuts them, of course; if a line cuts two of them then it cuts all of them-this is easy to prove). Assume the circles $(OPK),(OQL)$ also intersect at $X$. The circle $(OXR)$ must cut $\ell$ at $M$ because the involution is already determined by the pairs $P\leftrightarrow K,\ Q\leftrightarrow L$, so the circle $(OXR)$ is the same as the circle $(ORM)$, so the circle $(ORM)$ passes through $X$, Q.E.D. Do you happen to know who proposed it?

Dear Grobber can you explain more?????? Wht's involution? What's Desargue theorem? I will be thankful if you answer me

An involution is a homographic transformation (which conserves the cross-ratio of four points) from the set of points of a line to the same set, such that $f(f(P))=P$ for all points $P$ of the line. Desargue's theorem (not the one that's so well-known, but another one): A line $\ell$ intersects opposite pairs of sides of a complete quadrilateral in pairs of points which correspond to each-other in an involution (on the line $\ell$).

Thank you very much.

Let O be a point in the plane of the triangle ABC A circle T which passes throught O intersects the second time the lines OA,OB,OC in P,Q,R and also intersects the second time the circles (BOC),(COA),(AOB) in K,L,M. Prove that PK,QL,RM are concurrent I have solution: Let H be the center of T E,F,G are centers of (BOC),(COA),(AOB) We have PK,QL,RM are concurrent <=> LP.QM.RK=PM.QK.RL <=> sinAOMsinBOKsinCOL=sinAOLsinBOMsinCOK <=> sinHGFsinHEGsinHFE=sinHFGsinHGEsinHEF Because EH,FH,GH are concurrent so applying the Ceva sin we have done

Dear Mathlinkers, an article concerning the “’Du cercle des huit points au cercle des six pieds or the theorem on the six pedals’’ has been put on my website with interesting examples. http://perso.orange.fr/jl.ayme vol. 9 p. 8 You can use Google translator Sincerely Jean-Louis

Can someone explain to me why we have this "PK,QL,RM are concurrent <=> LP.QM.RK=PM.QK.RL"