Solution:

First we prove (already using the notation appropriate to the problem) the following lemma: The perpendicular bisector of a side $BII'$ and the angle bisector through the third vertex $D$ intersect on the circumcircle of every non-isosceles triangle $BI'D$. ("South Pole Theorem")

Proof of the lemma: Let $M$ be the center of the circumcircle of $BI'D$ and $S$ the intersection point of the angle bisector with the circumcircle different from $D$. Then $\square BMI I'$, being a central angle, is twice as large as $\square BD I'$ and $\square SD I'$ is half as large as $\square BD I'$. But since the perpendicular bisector of $BI'$ bisects both arcs of the circumcircle belonging to this chord, it must pass through $S$.

Corollary: Let the other intersection point of the perpendicular bisector of $BI'$ with the circumcircle be $E$. Then, by the lemma, $D$ lies on the Thales circle over $ES$ and we have $\square EDS = 90^\circ$. Thus $DE$ is the external bisector of $\square BDI'$. The converse also holds: if for the point $E$ on the external bisector of $\square BDI'$ we also have $BE = EI'$, then $E$ lies on the circumcircle of $BI'D$.

For the main proof we denote the reflection of $I$ by $I'$ and the second intersection point of $AI$ with the circumcircle of the triangle $ABD$ by $D'$. Since $AD'$ is the angle bisector of $\square BAD$, $D'$ lies at the midpoint of the arc $BD$ and so $DD' = BD' = CD'$ holds.

Using the inscribed angle theorem over the chord $AD$ as well as suitable angle bisectors, we obtain $\square DD'E = \square DD'A = \square DBA = \square CBI = \square ICB$. Since $D'$, because of the equal distances, is the circumcenter of the triangle $BCD$, we have $\square EDD' = 90^\circ - \square D'DC = \square CBD = \square CBI$, so, because $\square CBI = \square ICB$, the triangles $ED'D$ and $IBC$ are similar. From this it follows that $\frac{BC}{CI} = \frac{BC}{CI} = \frac{DD'}{D'E} = \frac{BD'}{D'E}$.

Moreover, $\square I'CB = \square ACB + \square I'CA = \square ACB + \square ACI = \square ACB + \square CBD = \square BDA = \square BD'E$ holds, and therefore the triangles $BCI'$ and $BD'E$ are similar; likewise $BCD'$ and $BI'E$ are similar triangles, since under the spiral similarity about $B$ that carries $BCI'$ into $BD'E$, $CD'$ is carried into $I'E$.

But since $BCD'$ is isosceles, we also have $BE = EI'$.

Now $DE \perp AC$, and by the corollary $E$ lies on the circumcircle of the triangle $BI'D$.