Geometry concur A circle intersects side $BC$, $CA$, $AB$ of $\triangle ABC$ in points $L$ and $L'$, $M$ and $M'$, $N$ and $N'$. show that $AL$, $BM$, $CN$ concur if and only if $AL'$, $BM'$, $CN'$ concur(handle the case when ratios are negative)
Since if $AL$, $BM$, $CN$ concur, then $\frac{BL}{LC}\frac{CM}{MA}\frac{AN}{NB}=1$, need to prove $\frac{BL'}{L'C}\frac{CM'}{M'A}\frac{AN'}{N'B}=1$, I can't figure out the relation ship between these two function. Someone has hint? Thanks
 A: By the "Power of a Point" Theorems, we have
$$\begin{align}
|AN| |AN^\prime| &= |AM^\prime| |AM| &(= \text{power of pt } A \text{ wrt } \bigcirc\;)\\
|BL| |BL^\prime| &= |BN^\prime| |BN| &(= \text{power of pt } B \text{ wrt } \bigcirc\;)\\
|CM| |CM^\prime| &= |CL^\prime| |CL| &(= \text{power of pt } C \text{ wrt } \bigcirc\;)
\end{align}$$
where (in Ceva-like fashion) the lengths of oppositely-directed segments are given opposite signs. (This takes into consideration whether each of points $A$, $B$, $C$ is inside or outside of the circle.)
Multiplying the three left-hand sides and the three right-hand sides, and then dividing appropriately, gives
$$\frac{|BL||CM||AN|}{|CL||AM||BN|} = \frac{|CL^\prime| |AM^\prime| |BN^\prime|}{|BL^\prime||CM^\prime||AN^\prime|}$$
which, in more Ceva-friendly form (replacing $|CL|$ with $-|LC|$, etc, but then noticing that these sign changes cancel in the full equation), becomes
$$\frac{|BL||CM||AN|}{|LC||MA||NB|} = \frac{|CL^\prime| |AM^\prime| |BN^\prime|}{|L^\prime B||M^\prime C||N^\prime A|}$$
If the fraction on one side of the equation reduces to $1$, then so does the fraction on the other side. By Ceva's Theorem, the concurrency of one triplet of cevians is linked to the concurrency of the other triplet.
A: In the following picture, you can see that the triangles $\Delta NBL'$ and  $\Delta N'BL$ are similar and we have:
$$
\frac{BL'}{BN'}=\frac{BN}{BL}
$$
In a similar way, you have: 
$$
\frac{AN'}{AM'}=\frac{AM}{AN}, \frac{CM'}{CL'}=\frac{CL}{CM}
$$
And therefore: 
$$
\frac{AN'}{AM'}\frac{BL'}{BN'}\frac{CM'}{CL'}=\frac{AM}{AN}\frac{BN}{BL}\frac{CL}{CM}
$$
This means that if the first set of segments are concurrent, the second set are also concurrent.

A: Just a comment. (I didn't have enough rep - 4 points missing, lol)
This stuff is known as cyclocevian conjugacy and, as Darij Grinberg has proven,
the cyclocevian conjugate of a point P is the 
isotomic conjugate
of the homothety at centroid (k=-2) image 
of the isogonal conjugate
of the homothety at centroid (k=-1/2) image 
of the isotomic conjugate
of the point P.
( see Link )
That's useless, but funny.
