Let $D$ be an arbitrary point on the side $BC$ of a given triangle $ABC$ and let $E$ be the intersection of $AD$ and the second external common tangent of the incircles of triangles $ABD$ and ACD. As $D$ assumes all positions between $B$ and $C$, prove that the point $E$ traces an arc of a circle. Click here for the proof.
Proof: Mark various tangent points as the above. We have $JK=ML$. $$\because JK=KE+JE=EN+EO=EN+EN+NO=2EN+NO$$ and $$ ML=LD+MD=DO+DN=DO+DO+NO=2DO+NO$$ Therefore $EN=DO$, $KE=DL$, and $JE=MD$.
We have $$2AE=AC-CM-DL+AB-BL-MD$$ $$=AB+AC-(CM+MD+DL+BL)=AB+AC-BC$$ Therefore $$AE=\dfrac{AB+AC-BC}{2}$$
which implies that the length of $AE$ is a constant for $\triangle{ABC}$, and $E$ is on an arc of a circle, as shown below, where $P$ and $Q$ are tangent points of the in-circle of $\triangle{ABC}$ on $AC$ and $AB$ respectively.
