The argument hinges on the claim below:

Claim. The lines $AE$ and $BI$ are perpendicular; similarly, $AF$ and $CI$ are perpendicular

*Proof.* Let $\alpha = \angle BAC$, $\beta = \angle CAB$ and $\gamma = \angle ACB$. Let $DX$ cross the circle $ADC$ again at $D'$. Note that $\angle ECX = \angle ACX - \angle ACE = 90^\circ - \beta/2 - \gamma/2 = \alpha/2 = \angle DAC = \angle DD'C = \angle XD'C$. As $\angle CXE = \angle CXD'$, triangles $XCD'$ and $XEC$ are similar, so $XD' \cdot XE = XC^2$.

As $XA = XC$, it follows that $XD' \cdot XE = XA^2$, so triangles $XD'A$ and $XAE$ are similar, so $\angle XAE = \angle AD'X = \angle AD'D = \angle ACD = \gamma$.
Finally, note that $\angle XAD = \angle XAC - \angle DAC = 90^\circ - \beta/2 - \alpha/2 = \gamma/2$, so $\angle IAE = \angle DAE = \angle XAE - \angle XAD = \gamma/2$. As $\angle AIB = 90^\circ + \gamma/2$, the claim follows.

Let $W$ be the mid-arc point of $CAB$ and let $I'$ be the reflection of $I$ across $O$. As $W, X, Y$ are the mid-arc points of $CAB, ABC, BCA$, respectively, their reflections across $O$ are the mid-arc points opposite. These latter form a triangle with orthocentre $I$, so $I'$ is the orthocentre of triangle $WXY$.
Reflection across $O$ maps lines $XV, WY$ and $WX$ to the perpendicular bisectors of $AI, BI$ and $CI$, respectively, so $XY \perp AI, WY \perp BI$ and $WX \perp CI$. By the claim, $AE \perp IF$ and $AF \perp IE$, so $I$ is the orthocentre of triangle $AEF$ and hence $EF \perp AI$ as well.
Triangles $AEF$ and $WXY$ have therefore corresponding parallel sides, so they are homothetic from some point $R$. This homothety maps $I$ to $I'$, as they are corresponding orthocentres. Hence the lines $AW, EY, FY$ and $II'$ are concurrent at $R$. As $I, O$ and $I'$ are collinear and $AW$ is the external bisector of $\angle BAC$, the conclusion follows.