Problem 1133

Without loss of generality, we can assume that $AC \geq AB$. We denote $N$ as the intersection point of the external bisector of the angle $\widehat{BAC}$ with $\Omega$ and $S$ as the intersection point of the internal bisector with $\Omega$. We place a point $D$ on the ray $[AB)$ that does not belong to the segment $[AB]$. Since $(AN)$ is the perpendicular to $(AS)$ passing through $A$, we have $\widehat{NAC} = 90^{\circ} - \widehat{CAS} = 90^{\circ} - \frac{1}{2} \widehat{BAC}$ and $\widehat{NAD} = 180^{\circ} - 90^{\circ} - \widehat{SAB} = 90^{\circ} - \frac{1}{2} \widehat{BAC}$. Therefore, $\widehat{NAD} = \widehat{NAC}$, so $(AN)$ is the bisector of the angle $\widehat{CAD}$.

By the inscribed angle theorem, we have $\widehat{NAC} = \widehat{NBC}$. On the other hand, $\widehat{NCB} = 180^{\circ} - \widehat{BAN} = \widehat{NAD}$. Since $\widehat{NAD} = \widehat{NAC}$, we find that $\widehat{NBC} = \widehat{NCB}$, so the triangle $BNC$ is isosceles. Therefore, the point $N$ is equidistant from the points $B$ and $C$, so $N$ lies on the perpendicular bisector of the segment $BC$. Thus, the intersection point of the perpendicular bisector of the segment $[BC]$ and the external bisector of the angle $\widehat{BAC}$ is actually $N$, which we defined as belonging to the circumcircle of triangle $ABC$.

Therefore, the intersection point of the perpendicular bisector of the segment $[BC]$ and the external bisector of the angle $\widehat{BAC}$ indeed lies on the circumcircle of triangle $ABC$.