Dear Mathlinkers, some ideas... 1. T1T2T3 is the Nagel triangle or the extouch triangle of ABC 2. T1T2T2 and ABC are orthologic and homothetic 3. T1T2T3 and H1H2H3 are orthologic... How to continue? Sincerely Jean-Louis

Let $I$ be the incenter of $\triangle A_1A_2A_3$,$I_3$ be its $A_3$-excenter and $\omega$ be its circumcircle.Denote by $S_1$ the midpoint of $\widehat{A_2A_1A_3}$ of $\omega$ and $S_2$ that of $\widehat{A_1A_2A_3}$,then $A_1,S_1,I_3$ are collinear and $A_2,S_2,I_3$ are collinear.Denote by $T'_1,T'_3$ the pedals of $I$ on $A_2A_3,A_1A_2$ respectively.Since $A_1T_3=A_2T'_3=A_2T'_1=A_3T_1$ and $A_1S_2=A_3S_2,\angle S_2A_1T_3=\angle S_2A_3T_1$,we obtain that $\triangle A_1S_2T_3\cong \triangle A_3S_2T_1$,which follows that $\angle S_2T_3A_2=\angle S_2T_1A_2$,then $S_2,T_3,T_1,A_2$ are concyclic.Similarly,$S_1,T_3,T_2,A_1$ are concyclic.Let the line $I_3T_3$ intersect the circumcircle of $\triangle A_2T_1T_3$ again at $U$,then $I_3U\cdot I_3T_3=I_3S_2\cdot I_3A_2=I_3S_1\cdot I_3A_1$,so $U$ lies on the circumcircle of $\triangle A_1T_2T_3$.Consequently,we get that $H_2T_1=UT_3=H_1T_2$.Moreover,$H_2T_1\parallel H_1T_2$ implies that $H_1T_1,H_2T_2$ have common midpoint,which,by the same reasoning,also lies on $H_3T_3$,so $H_1T_1,H_2T_2,H_3T_3$ are cocurrent,as desired. $Q.E.D.$

Easily generalisation: For a point $P$ in $ABC$, $A'B'C'$ be the pedal triangle, let $H_A$ be the orthocentre $AB'C'$ and define others similarly. Prove $A'B'C'$ and $H_AH_BH_C$ as perspective. To prove it, note parallelograms so $H_A$ is the reflection of $P$ in the midpoint of $B'C'$. Clearly, this forms a triangle both homothetic and congruent to $A'B'C'$, so the concurrence point can be defined as the midpoint of the segment adjoining $A'$ and the reflection of $P$ through the midpoint of $B'C'$. For the problem at hand, consider $P$ to be the Bevan point of $ABC$.