This problem is too simple ,as this structure has appeared before.

Here is Chinese translation of geometry problems in 2017 ARMO:

Aha! Nice problem! Let $L$ be a point on $BC$ such that $\angle OIL=90^{\circ}$. It's clear that $IL$ is tangent to $(BIB')$ so it suffices to check $LI=LB'$. To see this actually holds, let $M$ be the point on the external bisector of angle $ABC$ such that $M, I, L$ are collinear. Claim: $IM=2IL$. (Proof) Let $A', C', L'$ be points symmetric to $I$ in $A, C, L$ respectively; $I_A, I_B, I_C$ be the excenters opposite to $A, B, C$ in $\triangle ABC$. Let $IL$ meet the circumcircle of triangle $I_AI_BI_C$ at $X, Y$. Note that $I$ is the midpoint of $XY$ and $A', C'$ lie on this circle so by Butterfly's Theorem we get $IM=IL'$ as claimed. $\square$ Evidently, $IM=2IL$ so the midpoint of $IM$ is equidistant from $B$ and $I$. Reflecting in $OI$ yields the conclusion.

How to use butterfly? I don't understand.

Well... This configuration is so popular that this is a bad problem for olympiad.

Lsway wrote: This problem is too simple ,as this structure has appeared before. Let $\{ D,A\}=AI \cap \odot O,\{ E,C\}=AI \cap \odot O.$ The line through $I$ perpendicular to $OI$ intersects $AC,DE$ at $F,G.$ Obviously $IG=IB$.From Butterfly Theorem we can know $IG=IF$.Hence $GBB^\prime F$ is a isosceles trapezoid $\Longrightarrow IF=B^\prime F$ $OI$ passes through the circumcenter of $\triangle BB^\prime I$$\Longrightarrow$$IF$ is tangent to $\odot (BB^\prime I).$ $\therefore F$ is the pole of $B^\prime I$ WRT $\odot (BB^\prime I)$ There's a typo in "$IG=IB$", which should be $IG=BG$.

A synthetic solution: First, we note that $B'$ lies on $(ABC)$, and the tangent line from $I$ to $(BIB')$ is perpendicular to $OI$. Let this line intersect line $AC$ at point $D$. Let $(B'ID)$ intersect $(ABC)$ again at $E$ and let $BI$ intersect $(ABC)$ at $F$ Claim:$E,D$ and $F$ are collinear. Proof: From the cyclic quadrilaterals, we have $\angle(B'EF)$$=$$\angle(B'BI)$$=$$\angle(B'ID)$$=$$\angle(B'ED)$ (the angles are directed) which means that $E,D$ and $F$ are on the same line. On the other hand, it is well known that $FI^2$$=$$FA^2$$=$$FD.FE$ (just consider that the triangles $FEI$ and $FID$ are similar from some angle chasing), which gives us that $\angle(DIF)$$=$$\angle(DEI)$$=$$\angle(DB'I)$. As the line $IO$ is perpendicular to the line $BB'$ and $ID$, we get that $BB' \parallel IO$ $\Rightarrow$ $\angle(B'BI)$$=$$\angle(DIF)$ $\Rightarrow$ $\angle(DB'I)$$=$$\angle(B'BI)$, which gives us the desired result.

Let the tangent of $(BIB')$ at $B$ meet $AC$ at $N$. Let $I'$ be the $B$-excentre of $\triangle ABC$ Claim. $I'N$ is parallel to $BB'$ Proof. Consider the $\sqrt{ac}$ inversion at $B$ together with a reflection w.r.t. the angle bisector of $\angle ABC$. Then $I$ and $I'$ are interchanged by a well-known fact, and since $$\angle NBI=\angle BB'I=\angle IBB'$$$N,B'$ are interchanged. Therefore, let $B"$ be the reflection of $B'$ w.r.t. $BI'$, then $$BN\times BB"=BI\times BI'$$so$$\angle NI'I=\angle NB"I=\angle BB'I=\angle IBB'$$as desired. Let $AC$ and $NI$ meet $BB'$ at $Y$ and $Z$, then notice that $$(Z,\infty_{BB'};B,Y)\overset{N}{=}(I,I';B,X)=-1$$hence $Z$ is the midpoint of $BY$. Let the tangent at $I$ to $(BB'I)$ meet $AC$ at $K$. By Menelaus Theorem, $$\frac{XN}{NY}=\frac{XI}{IB}=\frac{XK}{KY}$$Hence $(N,K;X,Y)=-1$, let $BK$ meet $(BB'I)$ at $M$, then $$(B,M;I,B')\overset{B}{=}(N,K;X,Y)=-1$$so $BM$ is the symmedian, and hence $B'K$ is also tangent to $(BB'I)$ as desired.

Let $J = (B'ID) \cap (ABC)$, Let $E$ be the arc midpoint; Let $D$ be on $AC$ such that $\measuredangle OID = 90^\circ$. Claim: $E-J-D$

By Shooting lemma and Incenter-excenter lemma, $MI^2=MA^2=MJ \cdot MD$; $\measuredangle DB'I = \measuredangle DJI = \measuredangle EID$ which is equal to $\measuredangle DIB'$ due to reflection, Hence we're done.

Since $OB=OB'$ we have $B' \in (ABC)$. Let $K$ be a point of intersection tangents to $(BB'I)$ in points $B'$ and $I$. We want to prove that $K \in AC$. Note that since $BI=B'I$, we have $\Delta BB'I \sim \Delta IB'K$. Now let's use complex coordinates. Let $(ABC)$ be a unit circle, $A$ has coordinate $a^2$, $C$ has coordinate $\frac{1}{a^2}$ and $B$ has coordinate $b^2$ so that $I$ has coordinate $-(1+ab+\frac{b}{a})=j$. Also let $B'$ has coordinate $x \not = b^2$ and $K$ has coordinate $k$. We have $$(j-b^2)\overline{(j-b^2)}=(j-x)\overline{(j-x)} \Rightarrow \frac{j}{b^2}+b^2\overline{j}=\frac{j}{x}+x\overline{j} \Rightarrow j\frac{x-b^2}{b^2x}=\overline{j}(x-b^2) \Rightarrow j=b^2x\overline{j}$$$$\Rightarrow x=\frac{j}{b^2\overline{j}}=\frac{-(1+ab+b/a)}{-b^2(1+\frac{1}{ab}+a/b)}=\frac{a+a^2b+b}{ab^2+a^2b+b}.$$Now since $\Delta BB'I \sim \Delta IB'K$, we get $\frac{b^2-x}{j-x}=\frac{j-x}{x-k}$ $$\Rightarrow k=\frac{j^2-2jx+b^2x}{b^2-x}=\frac{(1+ab+b/a)^2+2(1+ab+b/a)\frac{a+a^2b+b}{ab^2+a^2b+b}+b^2\frac{a+a^2b+b}{ab^2+a^2b+b}}{b^2-\frac{a+a^2b+b}{ab^2+a^2b+b}}=\frac{(a^2b+a+b)(a^2b+2a+b)}{a^2(b-1)(b+1)}.$$And we want to check that $k+\overline{k}=a^2+1/a^2$. Note that $$\overline{k}=\frac{-(a^2+ab+1)(a^2+2ab+1)}{a^2(b-1)(b+1)}$$so $$k+\overline{k}=\frac{(a^2b+a+b)(a^2b+2a+b)-(a^2+ab+1)(a^2+2ab+1)}{a^2(b-1)(b+1)}=\frac{a^4b^2+b^2-a^4-1}{a^2(b^2-1)}=\frac{(b^2-1)(a^4+1)}{a^2(b^2-1)}=\frac{a^4+1}{a^2}=a^2+1/a^2$$and it is end of solution!