Here is a possible proof "without inversion". (Note: Generally i am against settings of problems, which discriminate some structural part of mathematics, and ask for a solution "without" some ingredient, which would make the solution straightforward, simple, and easy to remember. But in this case i will accept the challenge. There will be some similitude argument instead... well, same two circles as in the solution using inversion are the main actors.)
Note that even the main wiki page for the inversion uses this problem to illustrate the use and usefulness of the inversion:
I will slightly change the notation to fit the order in the alphabet. So $M$ is not on $AB$, but on the side opposite to the vertex $A$, and similarly for $N,P$. (Else i would commit errors.) So let us state explicitly...
Let $\Delta ABC$ be a triangle. Let $I$ be the incenter, the intersection of the bisectors of the angles of the triangle. The incircle $(I)$ touches the sides of $\Delta ABC$ in the points $M\in BC$, $N\in CA$, and $P\in AB$.
Let $S,T,U$ be the intersections of the the angle bisectors $AI$, $BI$, $CI$ with the segments respectively perpendicular on them, $NP$, $PM$, $MN$.
Let $O$ be the circumcenter of $\Delta ABC$, the center of the circle $(ABC)$.
Let $o$ be the circumcenter of $\Delta MNP$, the center of the circle $(MNP)$. Of course, $o=I$, but it may be useful to use the notation $o$ when addressing this point for its quality of being the center of $(MNP)$. We use in the notation a small letter as possible, and with the same convention let $h$, $g$, $9$ be respectively for $\Delta MNP$ the orthocenter, the centroid, and the center of the nine point (Euler) circle $(STU)$. Here, $g=MS\cap NT\cap PU$, and $9$ is the mid point of $oh$.
We denote by $A'$ the intersection of the bisector $AI$ with the circle $(O)=(ABC)$. Construct $B',C'$ similarly.
Let $M'$ be the mid point of the segment $Mh$, so $M'\in(STU)$.
The we have:
(1) The Euler line (e) of $\Delta MNP$ is passing through $o=I$, $9$, $g$, $h$, in particular these points are colinear.
(2) $OA'\| oM\|S9M'$, and similarly $OB'\| oN\|T9N'$, $OC'\| oP\|U9P'$.
(3) The triangles $\Delta STU$ and $\Delta A'B'C'$ are perspective / similar / homothetic, the center of the perspectivity being $o=I$, the intersection of the bisectors $ASIA'$, $BTIB'$, $CUIC'$.
(4) The circles $(STU)$ and $(A'B'C')=(ABC)$ respect the same homothety. In particular, the center of homothety $o$ is on the line their centers $9$ and $O$.
(5) The points $o=I$, $9$, $g$, $h$, $O$ are on the Euler line $(e)$ of $\Delta MNP$.
Best Answer
Triangle ABC
Draw perpendiculars from vertex $B$ & $C$ on side $CA$ & $AB$ meeting them at points $E$ & $F$ respectively. They intersect at $H$. Draw $BO$,$OH$,$OC$ and $BI$. Extend $BI$ to meet $OH$ at $L$. Also, drop a perpendicular from the circumcentre on side $BC$ meeting it at point $M$.
Observe that, in $\triangle BOM$ & $\triangle BHF$,
$\left(i\right)$ $\angle BMO=\angle BFH=90^\text{o}$
$\left(ii\right)$ $\angle OBM=90-\frac{\angle BOC}{2}=90-\angle BAC=\angle ABE=\angle HBF$
$\left(iii\right)$ Since $\triangle BFC$ is a $30-60-90$ triangle, $BF=\frac{1}{2}BC=BM$
Hence, $\triangle BHF\cong \triangle BOM$ by $A-S-A$ criterion of congruence.
Thus, $BO=BH$.
Since $\angle HBF=\angle OBM$ and $BI$ bisects $\angle ABC$, $BI$ must also bisect $\angle OBH$.
In $\triangle BOH$, $BI$ bisects $\angle OBH$ and $BO=BH$; Hence, $BI$ is the perpendicular bisector of $OH$. Since $I$ lies on this perpendicular bisector, $\boxed {OI=IH}$.