Let $O_1$ and $O_2$ denote the centers of $\odot(AII_1)$ and $\odot(AII_2).$ Then $\angle AI_1I=\angle ADI_2=90^{\circ}-\tfrac{1}{2}\angle ADB$ and similarly $\angle AI_2I=\angle ADI_1.$ Hence $\angle IAO_1=90^{\circ}-\angle AI_1I=90^{\circ}-\angle ADI_2=\angle AI_2I$ $\Longrightarrow$ $AO_1$ is tangent to $\odot(AII_2)$ $\Longrightarrow$ $\odot(AII_1)$ and $\odot(AII_2)$ are orthogonal. Now according to the problem Exsimilar center, all lines $MN$ go through the exsimilicenter of the circumcircle $(O)$ and incircle $(I)$ of $\triangle ABC,$ obviously fixed.

My solution: Let $ I, O $ be the incenter, circumcenter of $ \triangle ABC $, respectively . From incenter of triangle we get $ M, N $ are the tangent point of the Thebault circles of the cevian $ AD $ , so from Mabey Thebault have ovelooked we get $ MN $ pass through the exsimilicenter $ S $ of $ (I) \sim (O) $ . ie. $ MN $ pass through a fixed point $ S $ Q.E.D

My solution: Note that $\angle AI_2I+\angle AI_1I=(\frac{\angle B}{2}+\frac{\angle BAD}{2})+ (\frac{\angle C}{2}+\frac{\angle CAD}{2})=\frac{A+B+C}{2}=90$ so $\odot (\triangle AI_1I),\odot (\triangle AI_2I)$ are orthogonal now apply an inversion with center $A$ and radius $\sqrt{AB.AC}$ and then reflection throw the bisector of $\angle A$ note that $I\longleftrightarrow I_A$,$B\longleftrightarrow C$ under the inversion and $\odot (\triangle AI_1I),\odot (\triangle AI_2I)$ are taken to two perpendicular lines throw $I_A$ and $M,N$ are taken to intersections of these two perpendicular lines with $BC$(call them $M',N'$) let $X$ projection of $I_A$ on $BC$ and note that $XM'.XN'=I_AX^2$ is fixed this means that power of $X$ WRT $\odot (\triangle AM'N')$ is fixed so all of the circles $\odot (AM'N')$ are coaxal so $MN$ passes throw the fixed point DONE

Notice that $$\angle AMI+\angle ANI=\angle AI_I+\angle AI_2I=\angle BIC-\angle I_1AI_2=90^{\circ}.$$Apply $\sqrt{bc}$ inversion and denote by $X'$ the image of the point $X$. Then $I'$ is the $A$ excenter and $M',N'$ are points on $BC$ such that $\angle M'I'N'=90^{\circ}$. It suffices to show that all circles $(AMN)$ have two common points. Let $E$ be the touch point of the $A$ excircle with side $BC$. Note that $$EM'\cdot EN'=EI'^2=\text{constant}.$$Therefore, $AE$ is a common radical axis of all the circles $(AMN)$. The result follows.

Let $\overline{IM},\overline{IN}$ meet $\odot(ABC)$ again at points $M',N'$, respectively. An easy angle chase shows $\overline{M'N'}$ is a diameter of $\odot(ABC)$, thus $O \in \overline{M'N'}$ (where $O$ is the circumcenter of $\triangle ABC$). Now we prove a more general Lemma: (which clearly finishes the problem) Lemma: Given a conic $\mathcal C$ and any two points $O,I$ in the plane. Consider any pair $M',N'$ of points lying of $\mathcal C$ such that $O \in \overline{M'N'}$. Let lines $IM',IN'$ meet $\mathcal C$ again at $M,N$, respectively. Then all such lines $MN$ pass through a fixed point. Proof: We have that there is a involution swapping all such pairs $\{M',N'\}$, by definition. Then due to projection through $I$ we obtain: a involution swaps all such pairs $\{M,N\}$. Hence all such lines $MN$ pass through a fixed point. $\square$