Let $I$ be the center of the incircle of triangle $ABC$ and let $A', B'$ and $C'$ be the symmetrics of $I$ with respect to the lines $(BC), (CA)$ and $(AB)$ respectively. The circumcircle of $A'B'C'$ passes through $B$. Find $\widehat{A B C}$.
Let $r$ be the radius of the inscribed circle. Then $I A^{\prime}=I B^{\prime}=I C^{\prime}=2 r$, so the circumcircle of triangle $A^{\prime} B^{\prime} C^{\prime}$ is the circle centered at $I$ with radius $2 r$. Since $B$ is on this circle, we have $I B=2 r$. Let $\beta=\widehat{A B C}$ and $H$ be the orthogonal projection of $I$ onto $[A B]$. Then $\widehat{A B I}=\frac{\beta}{2}$ and in the right triangle $I H B$ we have $\sin (\beta / 2)=\frac{r}{2 r}=1 / 2$ so $\beta / 2=30^{\circ}$ and $\beta=60^{\circ}$.