Let $O$ be circumcenter,$(I)$ tangent to sides at $D,E,F$. Reflects $I$ over $O$ to get $B_e$. Since $O$ midpoint $IB_e$, $D,E,F$ are projections of $I$ on $BC,CA,AB$ so clearly $A_1,B_1,C_1$ are projections of $B_e$ on $BC,CA,AB$. We can easily see $O_B,O_C$ are midpoints of $LB,LC$ so $O_BO_C//BC$. Similarly $O_AO_BO_C$ and $ABC$ are hom mothetic so clearly the lines through $O_{A}, O_{B}, O_{C}$ respectively parallel to the internal angle bisectors of $\angle A,\angle B, \angle C$ are concurrent. (In this problem,we can ignore $I$ to generalize it)

Since $AB_1 = BA_1$, we have $\triangle AB_1T \cong \triangle BA_1T$ where $T$ is the midpoint of arc $AB$, containing $C$, so $A_1B_1CT$ is cyclic, so $OO_C$ is the perpendicular bisector of $CT$ (which is $C$-external bisector) and hence parallel to the internal bisector of $C$, done! (Alternatively, suspect that $O$ is the desired point from an accurate diagram and show $OO_C$ has angle $\frac{\angle ACB}{2}$ with $AC$ by projecting $O$ and $O_C$ to $M$ and $M_C$ and computing $\frac{OM - O_CM_C}{MM_C}$.)

Let $I_A$, $I_B$, $I_C$ be the excenters of triangle $ABC$, and let $V$ be the Bevan point of triangle $ABC$. [asy][asy] import geometry; unitsize(1 cm); pair[] A, B, C, I, O; pair X; A[0] = (1,3); B[0] = (0,0); C[0] = (4,0); I[1] = excenter(B[0],C[0],A[0]); I[2] = excenter(C[0],A[0],B[0]); I[3] = excenter(A[0],B[0],C[0]); A[1] = (I[1] + reflect(B[0],C[0])*(I[1]))/2; B[1] = (I[2] + reflect(C[0],A[0])*(I[2]))/2; C[1] = (I[3] + reflect(A[0],B[0])*(I[3]))/2; O[1] = circumcenter(A[0],B[1],C[1]); O[2] = circumcenter(B[0],C[1],A[1]); O[3] = circumcenter(C[0],A[1],B[1]); X = reflect(O[2],O[3])*(A[1]); draw(X--A[0],green); draw(X--B[0],green); draw(X--C[0],green); draw(I[1]--X, gray(0.7)); draw(I[2]--X, gray(0.7)); draw(I[3]--X, gray(0.7)); draw(circumcircle(A[0],B[1],C[1]),red); draw(circumcircle(B[0],C[1],A[1]),red); draw(circumcircle(C[0],A[1],B[1]),red); draw(A[0]--B[0]--C[0]--cycle); draw(I[1]--I[2]--I[3]--cycle); dot("$A$", A[0], N); dot("$B$", B[0], SW); dot("$C$", C[0], E); dot("$A_1$", A[1], NE); dot("$B_1$", B[1], NE); dot("$C_1$", C[1], W); dot("$I_A$", I[1], S); dot("$I_B$", I[2], NE); dot("$I_C$", I[3], W); dot("$O_A$", O[1], NE); dot("$O_B$", O[2], NW); dot("$O_C$", O[3], NE); dot("$V$", X, W); [/asy][/asy] Then $A_1$ lies on $VI_A$, $B_1$ lies on $VI_B$, and $C_1$ lies on $VI_C$. Since $\angle AC_1 V = \angle AB_1 V = 90^\circ$, the circumcircle of triangle $AB_1 C_1$ is also the circle with diameter $AV$. In other words, $O_A$ is the midpoint of $AV$. Similarly, $O_B$ is the midpoint of $BV$, and $O_C$ is the midpoint of $CV$. That means triangles $ABC$ and $O_A O_B O_C$ are homothetic. Therefore, the lines through $O_A$, $O_B$, $O_C$ that are parallel to the angle bisectors of $\angle A$, $\angle B$, $\angle C$ concur at the incenter of triangle $O_A O_B O_C$.