The perpendicular bisector of the bisector $BL$ of the triangle $ABC$ intersects the bisectors of its external angles $A$ and $C$ at points $P$ and $Q$, respectively. Prove that the circle circumscribed around the triangle $PBQ$ is tangent to the circle circumscribed around the triangle $ABC$.
Define $T=BB \cap AC$. Well-known $TB=TL$ so $T$ lies on $PQ$. Let $I_A, I_C$ be excenters so that $I_AACI_C$ is cyclic. By Reim's $PQCA$ is cylic, so now we have $TB^2 = TA \cdot TC = TP \cdot TQ$ and so $(PBQ)$ is tangent to $(ABC)$.
Claim 1: $PQCA$ is cyclic
Proof: Angle chasing
Let $T$ be an intersection of $AC$ and tangent at $B$ to $\odot(ABC)$
Claim 2: $TB=TL$
Proof: Angle chasing
From Claim 2 it follows that $T$ lies on perpendicular bisector of $BL$. As a result $TB, AC$ and perpendicular bisector of $BL$ are concurrent. From Power of Point it follows that:
$$ TA \cdot TC = TP \cdot TQ = TB^2$$This implies that $TB$ is tangent to $\odot (PBQ)$. Therefore $\odot (ABC)$ and $\odot (PBQ) $ are tangent to each other at point $B$ and we are done.