Let $ AB\leq AC \leq BC$. Denote the incenter by $ I$. Let $ BC=a$, $ AC=b$, $ AB=c$, $ I_aT=IT=IB=IC=i$, $ AT=t$ and $ XT=x$. $ XI_a^2=XA.XT \Longrightarrow (i-x)^2=x(x+t) \Longrightarrow x=\frac{i^2}{2i+t}$. $ (1)$ Let lines passing from $ I$ and $ T$ and perpendicular to $ BC$, intersect $ BC$ in points $ D$ and $ M$, respectively. Let line passing from $ I$ and parallel to $ BC$, intersect $ TM$ and $ XA'$ in points $ F$ and $ G$, respectively. We have $ TF \parallel XG$, $ IF=DM$ and $ FG=MA'$. $ TF \parallel XG$ $ \Longrightarrow \frac{IF}{FG}=\frac{IT}{TX}\Longrightarrow \frac{DM}{MA'}=\frac{i}{x}$. Using $ (1)$ we have $ \frac{DM}{MA'}=\frac{2i+t}{i}=2+\frac{t}{i}$. $ (2)$ $ ABTC$ is a cyclic quadrilateral. By using Ptolemy's theorem, we have $ TB.AC+TC.AB=BC.AT \Longrightarrow i(b+c)=at \Longrightarrow \frac{t}{i}=\frac{b+c}{a}$. $ (3)$ $ AB\leq AC \Longrightarrow DM=BM-BD=\frac{a}{2}-\frac{a+c-b}{2}=\frac{b-c}{2}$. $ (4)$ From $ (2)$, $ (3)$ and $ (4)$, it is concluded that $ MA'=\frac{a(b-c)}{2(2a+b+c)}$. $ (5)$ We have $ \frac{BA'}{A'C}=\frac{BM+MA'}{CM-MA'}$. $ (6)$ From $ (5)$ and $ (6)$, it is concluded that $ \frac{BA'}{A'C}=\frac{\frac{a}{2}+(\frac{a}{2})(\frac{b-c}{2a+b+c})}{\frac{a}{2}-(\frac{a}{2})(\frac{b-c}{2a+b+c})}=\frac{(\frac{a}{2})(1+\frac{b-c}{2a+b+c})}{(\frac{a}{2})(1-\frac{b-c}{2a+b+c})}=\frac{a+b}{a+c}$. Similarly, we can find that $ \frac{CB'}{B'A}=\frac{b+c}{b+a}$ and $ \frac{AC'}{C'B}=\frac{c+a}{c+b}$. $ (\frac{BA'}{A'C})(\frac{CB'}{B'A})(\frac{AC'}{C'B})=(\frac{a+b}{a+c})(\frac{b+c}{b+a})(\frac{c+a}{c+b})=1$. By using Ceva theorem, it is concluded that $ AA'$, $ BB'$ and $ CC'$ are concurrent.

$ D, E, F$ are feet of the internal bisectors of the angles $ \angle A, \angle B, \angle C.$ $ EF$ cuts $ AI$ and $ X'$ and $ BC$ at $ D'.$ $ BC, AI, EF,$ are diagonals of the complete quadrilateral $ (AE, AF, IE, IF)$ $ \Longrightarrow$ the cross ratios $ (D', D, B, C) = - 1,$ $ (A, I, X', D) = - 1$ are harmonic. From the 1st one, $ D'$ is foot of the external bisector of the angle $ \angle A$ and $ D'A \perp AI_a.$ From the 2nd one, $ \frac {\overline{DX'}}{\overline{DI}} = \frac {2\ \overline {DA}}{\overline{DA} + \overline{DI}} = \frac {2(a + b + c)}{2a + b + c}.$ $ (P)$ is circumcircle of the $ \triangle II_bI_c,$ $ (O)$ is its 9-point circle, $ I_a$ its orthocenter, $ EF$ is the radical axis of $ (P), (O),$ $ D'X' \equiv EF \perp OI_a.$ $ U$ is tangency point of the incircle $ (I)$ with $ BC$ and $ A'' \in BC$ foot of perpendicular from $ X'$ to $ BC.$ $ \frac {\overline{DA''}}{\overline{DU}} = \frac {\overline{DX'}}{\overline{DI}}$ $ M \in (O)$ is the midpoint of $ II_a,$ $ (M)$ is the circumcircle of $ BICI_a.$ $ \overline XA \cdot {XT} = XI_a^2$ $ \Longrightarrow$ $ X$ is on radical axis $ s \perp OI_a$ of $ (O)$ and the point $ I_a.$ Tangent $ t \perp MI_a \equiv AI_a$ to $ (M)$ at $ I_a$ is radical axis of $ (M), I_a$ and $ BC$ is radical axis of $ (O), (M).$ They meet at the radical center $ R_a \in BC$ of $ (O), (M), I_a.$ $ R_aI_a \parallel D'A$ (both $ \perp AI_a$) and $ R_aX \parallel D'X'$ (both $ \perp OI_a$). Consequently, $ \frac {\overline{DA'}}{\overline{DA''}} = \frac {\overline{DX}}{\overline{DX'}} = \frac {\overline{DI_a}}{\overline{DA}}$ Substituting $ \frac {\overline{DA}}{\overline{DI_a}} = 1 + \frac {\overline{I_aA}}{\overline{DI_a}} = 1 - \frac {b + c}{a}$ $ \overline{DU} = \overline{DC} - \overline{UC} = \frac {ab}{c + b} - \frac {a + b - c}{2} = - \frac {(b - c)(b + c - a)}{2(b + c)}$ $ \overline{DA'} = - \frac {a}{b + c - a}\ \overline{DA''} = - \frac {2a (a + b + c)}{(b + c - a)(2a + b + c)}\ \overline{DU} = \frac {a(b - c)(a + b + c)}{(b + c)(2a + b + c)}$ $ \overline{A'B} = \overline{DB} - \overline{DA'} = - \frac {ca}{b + c} - \frac {a(b - c)(a + b + c)}{(b + c)(2a + b + c)} = - \frac {a(b - c)(a + b)}{2a + b + c}$ $ \overline{A'C} = \overline{DC} - \overline{DA'} = + \frac {ab}{b + c} - \frac {a(b - c)(a + b + c)}{(b + c)(2a + b + c)} = + \frac {a(b - c)(c + a)}{2a + b + c}$ $ \frac {\overline{A'B}}{\overline{A'C}} = - \frac {a + b}{c + a}$ By cyclic exchange, $ \frac {\overline{A'B}}{\overline{A'C}} \cdot \frac {\overline{B'C}}{\overline{B'A}} \cdot \frac {\overline{C'A}}{\overline{C'B}} = - \frac {a + b}{c + a} \cdot \frac {b + c}{a + b} \cdot \frac {c + a}{b + c} = - 1$ and $ AA', BB', CC'$ are concurrent by Ceva theorem. _________________________________________ Remark: $ X' \equiv AI \cap EF$ and $ A'' \in BC$ was defined as foot of perpendicular from $ X'$ to $ BC.$ If $ Y' \equiv BI \cap FD, Z' \equiv CI \cap DE$ and $ B'', C''$ are feet of perpendiculars from $ Y', Z'$ to $ CA, AB,$ then $ AA'', BB'', CC''$ are also concurrent.

Let $D,E,F$ be the midpoints of $BC,CA,AB.$ $I$ and $S$ are the incenters of $\triangle ABC$ and $\triangle DEF$ (Spieker point) and $S_a$ is the reflection of $A$ on $S,$ which is incenter of $\triangle UBC,$ being $U$ the reflection of $A$ on $D.$ If the A-excircle $(I_a)$ touches $BC$ at $P,$ then by the symmetry WRT $D,$ we deduce that $P$ is the tangency point of the incircle of $\triangle UBC$ with $BC,$ i.e. $S_a \in I_aP.$ Redefine $A' \equiv AS \cap BC$ and let $V$ be the reflection of $P$ on $A'.$ Parallel from $I$ to $BC$ cuts $AP,AA',AV$ at $Q,R,L,$ resp. Since $D$ is clearly midpoint of $IS_a,$ then $A'$ is midpoint of $RS_a$ $\Longrightarrow$ $(PS_a \parallel RV) \perp BC$ and since $DI \parallel AP$ (well-known), $PQDS_a$ is a parallelogram $\Longrightarrow$ $(DQ \parallel PS_a) \perp BC.$ As a result $Q \in DT$ and $VR$ is perpendicular bisector of $LQ$ $\Longrightarrow$ $V(R,Q,L,P)=-1.$ If $RV$ cuts $AI_a,AP$ at $X_a,P_a,$ we get then $V(R,Q,L,P)=(P_a,Q,A,P)=(X_a,T,A,I_a)=-1$ $\Longrightarrow$ $X$ is midpoint of $I_aX_a$ $\Longrightarrow$ $(XA' \parallel I_aP \parallel X_aV) \perp BC$ and similarly for $B'$ and $C'.$ So we conclude that $AA',BC',CC'$ concur at the Spieker point $S$ of $\triangle ABC.$

Let $ G, $ $ I, $ $ N_a $ be the centroid, incenter, Nagel point of $ \triangle ABC $, respectively. Let $ H_a, $ $ M_a, $ $ T_a $ be the projection of $ A, $ $ T, $ $ I_a $ on $ BC $, respectively. Obviously, $ M_a $ is the midpoint of $ BC $, so from $ XT $ $ \cdot $ $ XA $ $ = $ $ {XI_a}^2 $ $ \Longrightarrow $ $ A'M_a $ $ \cdot $ $ A'H_a $ $ = $ $ {A'T_a}^2 $ we get the circle with center $ A' $ passing through $ T_a $ is orthogonal to $ \odot (I_a) $ and the 9-point circle $ \odot (N) $ of $ \triangle ABC $, hence $ A' $ lies on the radical axis of $ \odot (N) $ and $ \odot (I_a) $ $ \Longrightarrow $ $ A'$ lies on the common tangent of $ \odot (N), $ $ \odot (I_a) $. On the other hand, if $ AI $ cuts $ BC $ at $ J_a $ and the tangent from $ J_a $ to $ \odot (I_a) $ ($ \neq $ $ BC $) touches $ \odot (I_a) $ at $ V_a $, then $ M_aV_a $ passes through the A-Ex-Feuerbach point $ F_a $ of $ \triangle ABC $ (well-known), so from $ (A',J_a;M_a,T_a) $ $ = $ $ F_a(F_a,J_a;V_a,T_a) $ $ = $ $ -1 $ $ \Longrightarrow $ $ A(A', I; G, N_a) $ $ = $ $ A(A',J_a; M_a,T_a) $ $ = $ $ -1 $, hence we get $ AA' $ passes through the Spieker point $ S_p $ of $ \triangle ABC $. Similarly, we can prove $ S_p $ lies on $ BB' $ and $ CC' $, so we conclude that $ AA', $ $ BB', $ $ CC' $ are concurrent at $ S_p $.

I think there is a projective solution for parallel model Let $ABC$ be a triangle and $DEF$ is cevian triangle of $P$. $K,L,M$ are midpoint of $PA,PB,PC$. Let $X,Y,Z$ lie on $PA,PB,PC$ such that $XA^2=XK.XD,YB^2=YL.YE,ZC^2=ZM.ZF$. $G$ is centroid of triangle $ABC$ and $Q$ is symmetric of midpoint $PG$ through $G$. $U,V,W$ lie on $EF,FD,DE$ such that $XU\parallel QA,YV\parallel QB,ZW\parallel QC$. Prove that $DU,EV,FW$ are concurrent. When $P$ is orthocenter, we get original problem, but using parallel projection from original problem, we get this problem, also.