Let $ U, V$ and $ W$ be the centers of the circles $ \Gamma_{A}, \Gamma_{B}$ and $ \Gamma_{C}$ respectively. The pairs of lines $ EF$ and $ VW$, $ FD$ and $ WU$, and $ DE$ and $ UV$ meet at $ X, Y$ and $ Z$ respectively (one or all three points may be at infinity, but the argument below works in those cases too). By applying Menelaus to triangle $ UVW$ and transversals $ \over arrow{E - F - X}, \over arrow{F - D - Y}$ and $ \over arrow{D - E - Z}$ yields that $ X, Y$ and $ Z$ are the intersection points of the common external tangents to $ \Gamma_{B}$ and $ \Gamma_{C}$, $ \Gamma_{C}$ and $ \Gamma_{A}$, and $ \Gamma_{A}$ and $ \Gamma_{B}$ respectively. Now by Monge theorem we have that the points $ X, Y$ and $ Z$ are collinear (see here http://mathworld.wolfram.com/MongesCircleTheorem.html ). Consequently, the triangles $ ABC$ and $ DEF$ are perspective, therefore the lines $ AD, BE$ and $ CF$ are concurrent (by Desargues' Theorem).

we don`t really have to use monge`s theorem. the incircle of triangle $\triangle UVW$ touches it`s sides at $D,E,F$. $X$ is in fact the harmonical conjugate of $D$ wrt $V,W$, and it can be proven using very basic geometry facts that $X$ lies on $EF$ and $BC$. now, the fact that $X,Y,Z$ are collinear is another menelaus, or, speaking from projective geometry`s point of view, note that $X$ is the pole of $UD$ wrt the incircle of $UVW$. since $UD,VE,WF$ are concurrent, it follows that $X,Y,Z$ are collinear. note : some of the points $X,Y,Z$ may be points at infinity, but the arguments above remain true.

maky wrote: note that $ X$ is the pole of $ UD$ wrt the incircle of $ UVW$. since $ UD,VE,WF$ are concurrent, it follows that $ X,Y,Z$ are collinear. why X is the pole of UD wrt the incircle of UVW?? and how menelaus yields that X,Y,Z are the intersections of tangents?? i am really a beginner of projective geom, but i really want to understand the problem.

Let $P,Q,R$ be centers of the circles $\omega_a, \omega_b, \omega_c$. $E$ is the insimilicenter of $w_a,w_c$ , $F$ is the insimilicenter of $w_a,w_b$, and $QR\cap BC= P_1$ (say) is the exsimilicenter of $\omega_b,\omega_c$. So by Monge's theorem $E,F,P_1$ are collinear. So $EF \cap BC = P_1$, similarly we have $ED \cap AB = R_1$ and $FD \cap AC = Q_1$, where $P_1, Q_1, R_1$ are exsimilicenters of $\omega_b,\omega_c$ and $\omega_a, \omega_c$ and $\omega_a, \omega_b$ respectively. Again by Monge's theorem $P_1,Q_1,R_1$ are collinear. So $\triangle ABC$ and $\triangle DEF$ are perspective from a line, so they must be perspective from a point. So $AD,BE,CF$ are concurrent.

I think it's sufficient to use three-center theorem on circles $Wa,Wb,Wc$,then by Desargues (in triangle DEF and ABC) it's trivial.

Let $ X=DE \cap AC$ and define $Y,Z$ similarly. By Monge, the internal centers of $(w_c,w_b),(w_b,w_a)$ and the external center of $(w_a,w_c)$ are collinear. That is, $Y$ is the external center of $(w_a,w_c)$. By Monge, the external centers of $(w_a,w_b),(w_c,w_a),(w_b,w_c)$ are collinear. That is, $XYZ$ are collinear. Thus by Desargues, $AD, BE, CF$ are concurrent.

Let $X$, $Y$, $Z$ be the exsimilicenteres of $\omega _b$ and $\omega _c$, $\omega _a$ and $\omega _c$, $\omega _a$and $\omega _b$. From Monge's theorem,$X$, $Y$, $Z$ are colliniar,and because the exsimilicenter, insimilicenter and the centers of two circles are colliniar, we get that triangles $ABC$ and $DEF$ are perspective.So,by Desargues' theorem, the conclusion follows.

Notice $X=\overline{AB}\cap\overline{DE}$ is the exsimilicenter of $\omega_a$ and $\omega_b$ by 2-1 Monge. Similarly define $Y$ and $Z.$ By Monge, $X,Y,$ and $Z$ are collinear. Hence, $\triangle ABC$ and $\triangle DEF$ are perspective and Desargues finishes. $\square$

Suppose that $X, Y, Z$ be external homothetic center of $(\omega_b, \omega_c); (\omega_c, \omega_a); (\omega_a, \omega_b),$ respectively. Then It's easy to see that $X \in BC, Y \in CA, Z \in AB$. By Monge - D'Alembert theorem for $\omega_a, \omega_b, \omega_c,$ we have $X, Y, Z$ are collinear. Applying Desargues theorem for $\triangle ABC, \triangle DEF,$ we have $AD, BE, CF$ concur