Don't touch my circles [externally;)]

[sowmitra]

Circles $S_2$ and $S_3$ touch circle $S_1$ externally at $C$ and $D$. Another circle $S_4$ touches $S_2$ and $S_3$ externally at $E$ and $F$. Centres of $S_1, S_2, S_3, S_4$ are $U, A, B, V$.
$AD\cap BC=U_0$, and, $AF\cap BE=V_0$.
Prove that, $ UU_0, VV_0,$ and, $AB$ are concurrent.

[asif e elahi]

Circle $S_{1}$ touches $S_{2}$ and $S_{3}$ at point $C$ and $D$ respectively.By Monge De Alembert's theorem $CD$ passes through the exsimilicentre of circles $S_{2}$ and $S_{3}$. Let the exsimilicentre is $P$.So $CD$ passes through $P$.Similarly $EF$ passes through $P$.Again $P$ lies on $AB$.
Let $UU_{0}\cap AB=Q$.In$\bigtriangleup UAB$; $UQ,AD,BC$ concurrent.So by Sheva's theorem,
$\frac{UC}{AC}\times \frac{AQ}{BQ}\times \frac{BD}{DU}=1$
Again $P,C,D$ are collinear.So by Mellenau's theorem,
$\frac{UC}{AC}\times \frac{AP}{BP}\times \frac{BD}{DU}=1$
Dividing these equations we get
$\frac{AP}{BP}=\frac{AQ}{BQ}$
Let $R=VV_{0}\cap AB$.Similarly we can prove
$\frac{AP}{BP}=\frac{AR}{BR}$
So $\frac{AQ}{BQ}=\frac{AP}{BP}=\frac{AR}{BR}$
or $\frac{AQ}{BQ}=\frac{AR}{BR}$
This implies $P\equiv Q$
So $UU_{0},VV_{0},AB$ are concurent.

[sowmitra]

I think it will be $R \equiv Q$
Anyway, Nice solution...