Proof. Note that the incenter, $I$ is determined by the intersection of the internal angle bisectors of the triangle. We will consider the $A$-excircle, i.e. the excircle whose center, let's call it $X$, is determined by the intersection of the external angle bisectors at vertices $B$ and $C$. Note that $X$ lies on the line $AI$ as well.

Let $\angle BAC = \alpha$, $\angle ABC = \beta$ and $\angle ACB = \gamma$. Furthermore, let $AX \cap (ABC) = M$, so we need to show that $MI = MX$. We claim that $BICX$ is a cyclic quadrilateral centered at $M$. Firstly, we have $\angle BAM = \angle CAM = \frac{\alpha}{2}$, so $MB = MC$. Next, we show that $\triangle MCI$ is isosceles. Indeed, we have $\angle MIC = \angle IAC + \angle ACI = \frac{\alpha}{2} + \frac{\gamma}{2} = \angle BAM + \angle BCI = \angle BCM + \angle BCI = \angle MCI$.
Since $MB = MC = MI$, $M$ is the circumcenter of $\triangle BCI$, it remains to show that $X ∈ (BCI)$. We have $\angle BIC = 180° - \frac{\beta}{2} - \frac{\gamma}{2}$. On the other hand, $\angle BXC = 180° - \angle CBX - \angle BCX = 180° - \frac{180° - \beta}{2} - \frac{180° - \gamma}{2} = \frac{\beta}{2} + \frac{\gamma}{2}$.
Therefore, $\angle BIC + \angle BXC = 180°$, i.e. $X ∈ (BCI)$ and $(BCI)$ is centered at $M$, so $MI = MX$, as desired.