Let $\omega$ be a circle with center $O$, and let $MI_B \cap \omega = D$. Also, draw diameters $BP$ and $DE$ in the circle $\omega$.

Lemma. Triangle IBD is isosceles.

Proof. Let $F$ be the intersection of lines $DM$ and $AC$, and let $G$ be the reflection of $M$ across $AC$. By the Incenter-Excenter Lemma, the center of circle ($AIC$) is $M$. Let $\Phi$ denote the inversion centered at $M$ with respect to the circle ($AIC$).

Under $\Phi$, point $I$ is fixed, and $D$ and $F$ are inverse images of each other. Also, it can be verified that $MI^2 = MO \cdot MG$, so $O$ and $G$ are also inverses.

Since $M, I_B$, and $F$ are collinear, so are $G, I$, and $F$. Denote this line by $\ell$. Since $\ell$ does not pass through the inversion center $M$, its image is a circle — namely, the circle ($DIO$). Therefore, quadrilateral $DIOM$ is cyclic, which implies:
$$\angle OIM = \angle ODM = \angle EDM = \angle EBM = \angle EBI.$$
Hence, $OI \parallel BE$, so $OI \perp BD$. Thus, $IB = ID$, and triangle $IBD$ is equilateral. $\square$

Let the line $PI$ intersect the circle $\omega$ at point $Q$. Then we aim to show that the points $M, T, Q$ are collinear. Since under the inversion $\Phi$, the circle $\omega$ and the line $AC$ are mapped to each other, the image of point $A$ under this inversion is the point $S = BM \cap AC$. Because point $I$ is fixed under $\Phi$, the circles with diameters $BI$ and $IS$ are mapped to each other under the inversion. Since $IT \perp TS$, point $T$ lies on the circle with diameter $IS$. Likewise, since $BQ \perp QP$, point $Q$ lies on the circle with diameter $BI$. Thus, under the inversion $\Phi$, the image of point $Q$ is point $T$. Therefore, the points $M, T, Q$ lie on a straight line.

In the hexagon $MDNHPQ$, by the converse of Pascal's Theorem, the points $D, I$, and $N$ are collinear. Considering Lemma 1 and the symmetry of point $I$, we have $IM = IN$, so triangle $\triangle IMN$ is an isosceles triangle.

Now, let us move on to the main solution. Under the inversion $\Phi$, the points $R$ and $N$ are images of each other, so $MI^2 = MR \cdot MN$. Hence, $\triangle IMR \sim \triangle NIM$. On the other hand, since we already proved that $IM = IN$, it follows that $IR = RM$, which means that triangle $\triangle IRM$ is isosceles.