a) Relabel $E$ be $S$ and $F$ be $R$. Let $E, F$ be contact points of $(I)$ with $CA, AB;$ $B_e$ be Bevan point of $\triangle ABC$. We have $\dfrac{\overline{LS}}{\overline{LR}} = \dfrac{\overline{LD}}{\overline{LI_a}} = \dfrac{\overline{LI}}{\overline{LB_e}}$, so $R$ $\in$ $(B_e, B_eI_a)$. Hence if we let $I_b, I_c$ be $B$ - excenter, $C$ - excenter of $\triangle ABC$ then $I$ is orthocenter of $\triangle I_aI_bI_c,$ so $AI = AR$ b) First, we will prove that $N$ be second intersection of $(I_aBC)$ and $(I_aI_bI_c)$. We change the original problem into the following configuration Configuration. Given $\triangle ABC$ with altitude $AD, BE, CF$ and orthocenter $H$. Let $D'$ be orthogonal projection of $H$ on $EF,$ $S$ be a point on $AEF$ such as $AS$ $\parallel$ $DD'$. $SD'$ intersects $(AEF)$ again at $R$. Prove that $R$ $\in$ $(ABC)$ Proof. Let $E', F'$ be orthogonal projection of $H$ on $FD, DE$ then it's easy to see that $\triangle ABC$ and $\triangle D'E'F'$ are homothetic. So $AS$ is $A$ - symmedian of $\triangle ABC,$ which means $AS$ passes through midpoint $M$ of $EF$. Redefine $R$ is second intersection of $(ABC)$ and $(AEF)$. Since $$(EF, HR) = A(EF, HR) = A(CB, DR) = - 1$$we have $HERF$ is harmonic quadrilateral. Hence if we let $Q$ be second intersection of $HM$ with $(AEF)$ then $RQ$ $\parallel$ $EF$. So $$(SD', SR) \equiv (SD', SH) + (SH, SR) \equiv (MD', MH) + (QH, QR) \equiv 0 \pmod \pi$$or $S, D', R$ are collinear In the main problem, let $U$ be midpoint of arc $BC$ not containing $A,$ $W$ be intersection of $MN$ with $UO$. We will prove that $W$ is fixed, hence $T$ lies on $\bigodot(UW)$ which is fixed. We consider the following configuration Configuration. Given $\triangle ABC$ inscribed in $(O)$ with $B, C$ are fixed, $A$ moves on $(O)$. Let $H$ be orthocenter of $\triangle ABC$. $\bigodot(AH)$ intersects $(O)$ again at $R;$ $D$ $\equiv$ $AH$ $\cap$ $BC$. $RD$ intersects the line through $O$ and perpendicular to $BC$ at $W$. Prove that $W$ is fixed Proof. Let $M$ be midpoint of $BC,$ tangents at $B, C$ of $(O)$ intersect at $S,$ $RD$ intersects $(O)$ again at $V,$ $J$ be $A$ - Humpty point of $\triangle ABC,$ $JV, AS$ intersect $BC$ at $Y, X$. We have $\dfrac{VB}{VC} = \dfrac{DB}{DC} \cdot \dfrac{RC}{RB} = \dfrac{AB}{AC},$ then $V$ $\in$ $AS$. We have $$V(MX, YD) = (MX, YD) = \dfrac{\overline{YM}}{\overline{YX}} : \dfrac{\overline{DM}}{\overline{DX}} = \dfrac{\overline{YM}}{\overline{DM}} \cdot \dfrac{\overline{DX}}{\overline{YX}} = \dfrac{\overline{YJ}}{\overline{DA}} \cdot \dfrac{\overline{DA}}{\overline{YV}} = - 1 = V(MS, JW)$$Then $W$ is midpoint of $MS$. But $(O), B, C$ are fixed then $M, S$ are fixed, hence $W$ is fixed

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