My solution:
Let $S,T$ be circles through $(A,B),(A,C)$, respectively; with centers $X,W$. Then $BC$ is their common external tangent. Also $I$, the intersection of $BC$ and $XW$; is their external center of similitude. So letting the second point of intersections of $IA$ with $S,T$ be $R,U$; $RB$ is parallel to $AC$. Then
$\angle URB = \angle UAC = \angle UCK$, for $K$ any point in the ray through $C$ opposite to ray $CB$. So
$BCUR$ is cyclic. Then $IU.IR =IC.IB$....(1)
Now consider power of point $I$ wrt $S,T$. Then
$IA.IR = IB^2$ and $IA.IU =IC^2$. Multiplying and using (1), we get $IB.IC = IA^2$. So $V$ coincides with $I$.