How prove this $\angle PK_{3}B=\angle BK_{3}Q$ Question:

three circles $K_{1},K_{2},K_{3}$ are tangent to each other,(Circles $K_{1}$ and $K_{2}$ are externally tangent at a point $T$),Denote by $L_{1}$ is the exterior common tangent of the circles $K_{1},K_{2}$,meets $K_{3}$ in $P,Q$,and Denote by $L_{2}$ is the common tangent at $T$ meet $K_{3}$ in $B$.

show that:

$$\angle PK_{3}B=\angle BK_{3}Q$$


maybe this is famous thereom?  if someone have see it?can you link .Thank you
 A: It is just the angle-preservation property of the circular inversion: we just need to show that the tangent to $K3$ in $B$ is always parallel to $L1$.
Let $U$ be the projection of $T$ on $L1$ and $K4$ be the circle with center $T$ and radius $TU$.
Let $P^{-1}$ denotes the circular inversion of $P$ with respect to $K4$.
$K1^{-1}$ and $K2^{-1}$ are two lines, both parallel to $L2$, so $K3^{-1}$ is congruent to $L1^{-1}$ (it is just a translated copy of $L1^{-1}$ along the direction of $K1^{-1}$).
This gives that for any possible $K3$, the diameter of $K3^{-1}$ is constant, and the tangent $\tau$ to $K3^{-1}$ in $B^{-1}$ always has the same direction. In particular, $\tau$ and $L1$ cut $L2$ with opposite angles; moreover, the center of the circle $K5$ through $T$ that is tangent to $\tau$ lies in the intersection of $TU$ with the line through $B^{-1}$ orthogonal to $\tau$.
By inverting a second time, this gives that the tangent to $K3$ in $B$ is parallel to $L1$, as wanted.

(In this figure, $K3$ does not meet $L1$ in order to avoid an even messier configuration. 
We just provided a visual evidence that the tangent in $B$ to $K3$ is parallel to $L1$.)
