Concurrency of lines formed by pair of circles joining pairwise.

Question

How can i prove that lines $AD,EB,CF$ are concurrent ?

My attempt

Considering $\Delta ACB$ I've got the condition that $\cfrac {CP \cdot BQ \cdot AR}{PB \cdot AQ \cdot RC}=1 $ ,but I don't see how can I apply this condition since I don't see how to get those ratios.

Playing with a weaker version of the problem,namely with 3 circles where two of them are congruent ,I see that the above condition is applicable since every of the above lengths,like $CP$ and $RC$ , are in ratio $1$ ,therefore it's easy to prove the concurrency of lines.So it's obvious that I am not looking for the right pattern,the kind of pattern which is applicable to the general situation where none of the circles are congruent.

One usefull observation I can make is that any line which connects the centers of any two circles is perpendicular to the line which connects the endpoints of the areas in common between the two circles.To be clear if you observe the above diagram $OO_1$ is always perpendicular to $AD$ .

That's the relevant part of my effort,now I am trying just to "add" some random line to see if i can stab at something usefull.

Answer 1

You can prove that using the following :

Let $O_1$ be the circle on which $A,F,D,C$ exist. Also, let $O_2$ be the circle on which $C,F,B$. Let $X$ be the intersection point between $AD$ and $CF$. And let $E'(\not =B)$ be the intersection point between $O_2$ and $BX$. Then, four points $E',A,B,D$ are on a circle.

This is because we have $$\text{$\triangle{CXE'}$ and $\triangle{BXF}$ are similar}$$ $$\text{$\triangle{CXD}$ and $\triangle{AXF}$ are similar}$$ $$\text{$\triangle{E'XA}$ and $\triangle{DXB}$ are similar}$$

Answer 2

This is actually a well known theorem called the radical axis theorem.

Define the radical axis of two circles as the locus of points which has equal power with respect to both circles. ( Show that this locus is a line. To do this show this, take a point $P$ which has equal power with respect to both the circles and draw the tangents to both the circles from point $P$. From the right triangles so formed use the pythagorean theorem to deduce that the perpendicular from the point $P$ to the segment joining both the centres ($O_1O_2$) of the circles is always the same.)

Let $\Gamma_1,\Gamma_2$ and $\Gamma_3$ be $3$ circles. Let $l_1$ be the radical axis of $\Gamma_1$ and $\Gamma_2$, $l_2$ be the radical axis of $\Gamma_1$ and $\Gamma_3$ and $l_3$ be the radical axis of $\Gamma_2$ and $\Gamma_3$. Let $R=l_1\cap{}l_2$ then $R$ has equal power with respect to all the three circles. This implies that $R\in{l_3}$. Thus $l_1,l_2$ and $l_3$ are concurrent.

Note that the radical axis of two intersecting circles is the line passing through the two intersection points.