$(A_{4},A,B,C)=-1$ $\Rightarrow$ $ AA_4 , BB_4 , CC_4 $ are concurrent in lemone poin of $ABC$

nima1376, you are wrong, the quadrilateral $ABA_4C$ is not harmonic in general. The concurrency point is not the Lemoine point of $\triangle ABC$ but its center $X_{57},$ i.e. the perspector of $\triangle ABC$ and the orthic triangle of $\triangle A_1B_1C_1.$ Let $A_0,B_0,C_0$ be the projections of $A_1,B_1,C_1$ on $B_1C_1,$ $C_1A_1,$ $A_1B_1.$ $D$ and $M$ denote the midpoints of $\overline{BC}$ and the arc $BAC$ of the circumcircle $(O).$ From $(B,C,A_1,A_3)=-1,$ we get $A_3D \cdot A_3A_1=A_3B \cdot A_3C=A_3A_4 \cdot A_3A_2$ $\Longrightarrow$ $DA_1A_4A_2$ is cyclic $\Longrightarrow$ $\angle A_2A_4A_1=\angle A_2DA_1=90^{\circ}$ $\Longrightarrow$ $M \in A_1A_4.$ Furthermore, $A_1A_0 \perp A_3A_0$ implies that $B_1C_1$ bisects $\angle BA_0C$ externally or $\angle BA_0C_1=\angle CA_0B_1.$ Since $\angle BC_1A_0=\angle CB_1A_0,$ then $\triangle BA_0C_1 \sim \triangle CA_0B_1$ $\Longrightarrow$ $C_1A_0:B_1A_0=BC_1:CB_1=BA_1:CA_1,$ therefore isosceles $\triangle AB_1C_1 \sim \triangle MCB$ are similar with corresponding cevians $AA_0$ and $MA_1$ $\Longrightarrow$ $\angle MA_1C=\angle AA_0B_1.$ Hence since $A_0A_1A_4A_3$ is cyclic on account of the right angles at $A_0,A_4,$ we deduce that $A,A_0,A_4$ are collinear and analogously $B_0 \in BB_4$ and $C_0 \in CC_4.$ By Cevian Nest Theorem the conclusion follows.

Luis González wrote: nima1376, you are wrong, the quadrilateral $ABA_4C$ is not harmonic in general. The concurrency point is not the Lemoine point of $\triangle ABC$ but its center $X_{57},$ i.e. the perspector of $\triangle ABC$ and the orthic triangle of $\triangle A_1B_1C_1.$ sorry for my bad mistake another solution let $O$ is circumcircle of triangle $ABC$ $A_{2}A_{1}\cap O=K$ . $(B,C,A_{1},A_{4})=-1\Rightarrow (K,A_{4},B,C)=-1\Rightarrow \frac{BA_{4}}{A_{4}C}=\frac{KB}{KC}$ but $K$ is center of spiral similar goes $BC_{1}$ to $CB_{1}$ $\Rightarrow \frac{BK}{CK}=\frac{BC_{1}}{CB_{1}}$ so we are done with ceva...

North Korea is NOT Korea! "Korea" means South Korea. Why is this in Korea category?

First we claim that $\dfrac{A_{4}B}{A_{4}C}=\dfrac{A_{3}B}{A_{3}C}$. For the sake of argument, let us assume that $A_3$ is on the ray $BC$. Now $\angle BA_{4}A_{3}=180^{\circ}-\dfrac{\angle A}{2}$. So, $\angle A_{4}A_{3}C=\angle A_{4}A_{3}B=180^{\circ}-\angle BA_{4}A_{3}-\angle A_{4}BA_{3}$ $=\dfrac{\angle A}{2}-\angle A_{4}BA_{3}=\dfrac{\angle A}{2}-\angle A_{4}BC=\angle A_{2}AC-\angle A_{4}AC$ $=\angle A_{2}AA_{4}=\angle A_{2}BA_{4}=\angle A_{2}BA_{4}$. This means $A_{2}B$ is tangent to the circle $BA_{3}A_{4}$ and $A_{2}C$ is tangent to the circle $CA_{3}A_{4}$. From these information, we have that $A_{4}B=\dfrac{A_{2}B\times A_{3}B}{A_{2}A_{3}}$ and $A_{4}C=\dfrac{A_{2}C\times A_{3}C}{A_{2}A_{3}}$. Since $A_{2}B=A_{2}C$ we have $\dfrac{A_{4}B}{A_{4}C}=\dfrac{A_{3}B}{A_{3}C}$ as we claimed. By similar arguments, we can show that, $\dfrac{B_{4}C}{B_{4}A}=\dfrac{B_{3}C}{B_{3}A}$ and $\dfrac{C_{4}A}{C_{4}B}=\dfrac{C_{3}A}{C_{3}B}$. Now apply Menelaus's theorem for the lines $A_{3}B_{1}, B_{3}C_{1}, C_{3}A_{1}$. Then we have three equations: 1) $\dfrac{BA_{3}}{A_{3}C}\cdot \dfrac{CB_{1}}{B_{1}A}\cdot \dfrac{AC_{1}}{C_{1}B}=1$ 2) $\dfrac{CB_{3}}{B_{3}A}\cdot \dfrac{AC_{1}}{C_{1}B}\cdot \dfrac{BA_{1}}{A_{1}C}=1$ 3) $\dfrac{AC_{3}}{C_{3}B}\cdot \dfrac{BA_{1}}{A_{1}C}\cdot \dfrac{CB_{1}}{B_{1}A}=1$ Multiply (1),(2),(3) and use the fact $AC_1=AB_1, BC_1=BA_1, CA_1=CB_1$ to get that, $\dfrac{A_{3}B}{A_{3}C}\cdot \dfrac{B_{3}C}{B_{3}A}\cdot \dfrac{C_{3}A}{C_{3}B}=1$ Thus by our previous argument, 4) $\dfrac{A_{4}B}{A_{4}C}\cdot \dfrac{B_{4}C}{B_{4}A}\cdot \dfrac{C_{4}A}{C_{4}B}=1$ Now,by sine law, we have that $\dfrac{A_{4}B}{A_{4}C}=\dfrac{\sin \angle BAA_4}{\sin \angle CAA_4}$. Deriving similar expressions for $\dfrac{B_{4}C}{B_{4}A},\dfrac{C_{4}A}{C_{4}B},$ and using them in (4), we get that , $\dfrac{\sin \angle BAA_4}{\sin \angle CAA_4}\cdot \dfrac{\sin \angle CBB_4}{\sin \angle ABB_4}\cdot \dfrac{\sin \angle ACC_4}{\sin \angle BCC_4}=1$. This the trigonometric form of Ceva's theorem. So $AA_4, BB_4, CC_4$ are concurrent.

Dear Mathlinkers, the problem have something to do with the A-mixtilinear incircle of ABC... see http://perso.orange.fr/jl.ayme , A new mixtilinear incircle adventure I, G.G.G. vol. 4, p. 20-21. Sincerely Jean-Louis

is this too hard for a problem 1 on an olympiad?

To Konigsberg; depends To jayme, yes indeed, my solution shows this directly. Let the perpendicular to $I$ through $AI$ meet $BC$ at $A_5$, then note $A(A_5, A_4; B, C) = -1$. We know that $A_5B_5C_5$ are collinear and perpendicular to $HI$, so $AA_4, BB_4, CC_4$ are concurrent.

My solution (for the original problem) : Since $ AA_1, BB_1, CC_1 $ are concurrent at $ X_{7} $ of $ \triangle ABC $ , so from Desargue theorem we get $ A_3 , B_3 , C_3 $ are collinear . ... $ (\star ) $ Since $ A_4A_2, B_4B_2, C_4C_2 $ is the external bisector of $ \angle CA_4B, \angle AB_4C, \angle BC_4A $, respectively , so combine with $ ( \star ) $ we get $ \frac{CA_4}{A_4B} \cdot \frac{AB_4}{B_4C} \cdot \frac{BC_4}{C_4A}=\frac{CA_3}{A_3B} \cdot \frac{AB_3}{B_3C} \cdot \frac{BC_3}{C_3A}=1 $ . ie. $ AA_4, BB_4, CC_4 $ are concurrent Q.E.D __________________________________________________ My solution ( for $ AA_4 \cap BB_4 \cap CC_4 \equiv X_{57} $ ) : Let $ D $ be the projection of $ A_1 $ on $ B_1C_1 $ and $ X $ be the midpoint of arc $ BAC $ . Since $ (B, C; A_1, A_3)=-1 $ , so $ \frac{CA_4}{A_4B}=\frac{CA_3}{A_3B}=\frac{CA_1}{A_1B} $ , hence we get $ A_4A_1 $ is the bisector of $ \angle CA_4B $ and $ X \in A_4A_1 $ . Since $ D(B, C; A_1, A_3)=-1 $ , so $ DA_1 $ is the bisector of $ \angle BDC $ , hence we get $ \triangle DC_1B \sim \triangle CB_1D $ . Since $ \frac{B_1D}{DC_1}=\frac{CD}{DB}=\frac{CA_1}{A_1B} $ , so we get $ \triangle AC_1B_1 \cap D \sim \triangle XBC \cap A_1 $ , hence from $ \angle BAD=\angle BXA_1=\angle BAA_4 $ we get $ D \in AA_4 $ . ie. $ X_{57} $ of $ \triangle ABC $ lie on $ AA_4 $ Similarly, we can prove $ X_{57} \in BB_4 $ and $ X_{57} \in CC_4 $ , so we get $ AA_4, BB_4, CC_4 $ are concurrent at the $ X_{57} $ of $ \triangle ABC $ . Q.E.D

To prove $AA_4, BB_4, CC_4$ are concurrent at $X_{57}$ from my post, we can do this: Let $\ell$ be the radical axis of $\odot ABC, \odot A_1B_1C_1$. Let the medial triangle of $A_1B_1C_1$ be $A_6B_6C_6$ and note that $\ell$ is the radical axis of $\odot A_6B_6C_6$ by harmonic conjugates (fact 1). So, note $AA_5 \cap B_1C_1 \in \ell$ for obvious reasons, and because of the fact 1 we conclude $AA_5$ meets $B_1C_1$ at the same point where the orthic axis of $A_1B_1C_1$ meets $B_1C_1$, so $AA_4$ contains the foot of the $A$ altitude of $A_1B_1C_1$.

This problem is really weak in the sense that we don't need $A_1,B_1,C_1$ to be the tangency points of the incircle with the sides of $\triangle{ABC}$; it is enough to have $AA_1,BB_1,CC_1$ concurrent. Let $\{X_A\}=A_1A_2\cap (ABC)$. As $(A_3,A_1,B,C)=-1$, by perspectivity from $A_2$ we get that $X_ABA_4C$ is harmonic. This yields $\dfrac{A_4B}{A_4C}=\dfrac{X_AB}{X_AC}$, but $X_AA_2$ is the bisector of $\widehat{BX_AC}$, whence $\dfrac{A_4B}{A_4C}=\dfrac{A_1B}{A_1C}$. Writing the analogous relations and multiplying them we get that $$\dfrac{A_4B}{A_4C}\cdot \dfrac{B_4C}{B_4A}\cdot \dfrac{C_4A}{C_4B}= \dfrac{A_1B}{A_1C}\cdot \dfrac{B_1C}{B_1A}\cdot \dfrac{C_1A}{C_1B}=1$$which is equivalent to the fact that $AA_4,BB_4,CC_4$ are concurrent.

Let $(J_a)$ be a circle tangent to $BC$ at $A_1$ and tangent to $(ABC)$ at some point, say $T_a$. Define $J_b,J_c;T_b,T_c$ similarly. We claim that $A_4\equiv T_a$ etc. Let $X_a$ denote the midpoint of arc $BAC$ of $(ABC)$. If $\mathcal{H}$ denotes the homothety centered at $T_a$ that takes $(J_a)$ to $(ABC)$, then it takes $BC$ to the line $\ell$ parallel to $BC$ and tangent to $(ABC)$, clearly at $X_a$. Then $\mathcal{H}$ takes $A_1$ to $X_a$; so $T_a,A_1,X_a$ are collinear. Since $X_aA_2$ is a diameter of $(ABC)$, we have $X_aT_a\perp T_aA_2\implies A_1T_a\perp T_aA_2 \;(\star )$. Next, observe that $(B,C,A_1,A_3)=-1\implies T_a(B,C,A_1,A_3)$ is a harmonic pencil. But $T_aX_a$ and hence $T_aA_1$ is the bisector of $\angle BT_aC$, so we have $A_1T_a\perp T_aA_3$. Combining this with $(\star )$ gives $A_3,T_a,A_2$ are collinear, so $T_a\equiv A_4$. But from Concurrency on OI we have $AT_a$ etc. are concurrent at the Isogonal Mittenpunkt, as desired. $\blacksquare$

laFiesta wrote: North Korea is NOT Korea! "Korea" means South Korea. Why is this in Korea category? I guess they is not good geography

$\textbf{Proof :}$ Points $A_3, B_3, C_3$ lie on same line $l$. Consider projective transformation which sends circle $(ABC)$ into circle and line $l$ to infinity line. Name $A', B',\ldots, A_2', \ldots , A_4'$ images of points $A, B,\ldots, A_2, \ldots , A_4$ wrt this projective transformation. So we have that $A_2'A_4'|| B'C'$, $B_2'B_4'||A'C'$, $C_2'C_4'||A'B'$. Let lines $A'A_2'$, $B'B_2'$, $C'C_2'$ concurrent at point $X$. Then lines $A'A_4'$, $B'B_4'$, $C'C_4'$ concurrent at point which is isogonal to $X$ wrt $A'B'C'$. $\Box$

Let $A_2A_1\cap \odot ABC=\{A_5\}$ $(A_5,A_4;B,C)=-1$ $\frac{A_5B}{A_5C}=\frac{CA_1}{BA_1}=\frac{BA_4}{CA_4}$ and hence $BA_4\cdot CB_4\cdot AC_4=A_4C\cdot AB_4 \cdot BC_4$ so we're done.

Nice Problem!!! North Korean TST 2013 P1 wrote: The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $ BC, CA, AB$ at $ A_1 , B_1 , C_1 $ respectively. The line $AI$ meets the circumcircle of $ABC$ at $A_2 $. The line $B_1 C_1 $ meets the line $BC$ at $A_3 $ and the line $A_2 A_3 $ meets the circumcircle of $ABC$ at $A_4 (\ne A_2 ) $. Define $B_4 , C_4 $ similarly. Prove that the lines $ AA_4 , BB_4 , CC_4 $ are concurrent. Clearly $\{A_2,B_2,C_2\}$ are the midpoints of $\widehat{BC},\widehat{AC},\widehat{AB}$ respectively and let $\{X,Y,Z\}$ be the midpoints of $\widehat{BAC},\widehat{CBA},\widehat{ACB}$ respectively. Then $(A_3,A_1;B,C)\overset{A_4}{=}(A_2,A_1A_4\cap\odot(ABC);B.C)$. So, $A_4A_1\cap\odot(ABC)= X$. Now if $P$ is the foot of perpendicular from $A_1$ to $B_1C_1$ then $AP,XA_1$ concur at $A_4$ from this problem $\longrightarrow$Two Lines meet on a circle. Analogously we get that if $\triangle PQR$ is the orthic triangle of $\triangle A_1B_1C_1$, then $\{\overline{A-P-A_4}\},\{\overline{B-Q-B_4}\},\{\overline{C-R-C_4}\}$. Now as $\{AA_1,BB_1,CC_1\}$ and $\{A_1P.B_1Q,C_1R\}$. Hence by Cevian Nest Theorem we conclude that $AA_4,BB_4,CC_4$ are concurrent. $\blacksquare$

Suppose that $A_2A_1$ intersects the circle again at $A'$, and $A_1A_4$ intersects at $A_5$. We have $$(A_3A_1;BC)\stackrel{A_2}{=}(A_4A';BC)\stackrel{A_1}{=}(A_5A_2;BC)=-1$$Thus, $A_4A_5$ is an angle bisector, and $\frac{\sin\angle BAA_4}{\sin\angle CAA_4}=\frac{BA_4}{CA_4}=\frac{BA_1}{A_1C}$. Multiplying similar expressions for $B_4,C_4$, we are done by Trig Ceva and Ceva on the Gergonne point.

[asy][asy] import graph; size(15cm); real a=105; real b=205; real c=335; real r=10; path omega=circle(origin, r); pair A=r*dir(a); pair B=r*dir(b); pair C=r*dir(c); pair I=incenter(A,B,C); pair A1=foot(I,B,C); pair B1=foot(I,C,A); pair C1=foot(I,B,A); dot(A1); dot(B1); dot(C1); draw(A--B--C--cycle); draw(omega,blue); pair X=10*C1-9*B1; pair P=10*B-9*C; pair A3=intersectionpoints(B1--X,B--P)[0]; pair A2=r*dir((b+c)/2); pair B2=r*dir((a+c)/2+180); pair C2=r*dir((a+b)/2); dot(A2); dot(B2); dot(C2); pair A4=intersectionpoints(A2--A3,omega)[1]; draw(A2--A3); dot(A4); draw(B1--A3); draw(A3--A2, magenta); pair D=(B+C)/2; dot(D); draw(B--A3); draw(A1--A4--A2--D--cycle,red+ linewidth(1)); draw(A--A4,green); label("$A$",A,dir(90)); label("$B$",B,dir(-135)); label("$C$",C,dir(-45)); label("$D$",D,dir(-45)); label("$A_1$",A1,dir(90)); label("$A_2$",A2,dir(-90)); label("$A_3$",A3,dir(-90)); label("$A_4$",A4,dir(-90)); label("$B_1$",B1,dir(45)); label("$C_1$",C1,dir(-45)); draw(B--A4--C,dotted); label("$B_2$",B2,dir(45)); label("$C_2$",C2,dir(135)); [/asy][/asy] As mentioned earlier, $A_1,B_1,C_1$ may be any points on their respective sides, not necessarily the touch points, as long as $AA_1,BB_1,CC_1$ are concurrent. First, note that it is well-known that $(A_3A_1;BC)=-1$, and let $D$ be the midpoint of $\overline{BC}$. By EGMO lemma $9.17$(midpoints of harmonic bundles) combined with PoP, $A_3A_4\cdot A_3A_2=A_3B\cdot A_3C=A_3A_1\cdot A_3D$, thus $A_1A_4A_2D$ is cyclic. Because $\angle A_1DA_2=\pi/2$, this implies $\angle A_1A_4A_2=\pi/2$. By another well-known property of harmonic bundles, $\overline{A_4A_2}$ bisects $\angle BA_4C$. By angle bisector theorem and law of sines, we have $\prod_{\text{cyc}}^{}\frac{\sin\angle BAA_4}{\sin\angle A_4AC}=\prod_{\text{cyc}}^{}\frac{2R\sin\angle BAA_4}{2R\sin\angle A_4AC}=\prod_{\text{cyc}}^{}\frac{BA_4}{A_4C}=\prod_{\text{cyc}}^{}\frac{BA_1}{A_1C}$(angles and lengths directed, $R$ is the radius of $(ABC)$) Because $AA_1,BB_1,CC_1$ concur(at the Gergonne point for the specific case of the problem), the last cyclic product is $1$ by Ceva's theorem. Because the first cyclic product is 1, by trig Ceva, the lines $AA_4,BB_4,CC_4$ are concurrent, as desired.

laFiesta wrote: North Korea is NOT Korea! "Korea" means South Korea. Why is this in Korea category? Why? North Korea is still Korea. It encompasses the entire peninsula, should it not?

Let $D_A = AA_4 \cap BC$ , $S = (AC_1IB_1) \cap (ABC)$ and $E = AS\cap BC$. It is well-known that $S,A_1,A_2$ are collinear. Now we have $-1 =(A_3,A_1;B,C) \stackrel{A_2} = (A_4,S;B,C) \stackrel{A} = (D_A,E;B,C)$. By radical axis theorem on $(ABC) , (BIC) , (ASC_1IB_1)$ we get that $SI$ is tangent to both $(BIC)$ and $(ASC_1IB_1)$. Hence, we have $\triangle EIB \sim \triangle ECI$. So, $\frac{EB}{EI} = \frac{EI}{EC} = \frac{IB}{IC} \implies \frac{EB}{EC} = \frac{IB^2}{IC^2}$. So, $-1 = (D_A,E;B,C) \implies \frac{D_AB}{D_AC} = \frac{EB}{EC} = \frac{IB^2}{IC^2}$. Writing up similar expressions for the others, we get that $AA_4 , BB_4 , CC_4$ are concurrent by the converse of Ceva's Theorem. $\square$

By Ceva-Menelaus, we have $-1 = (A_3, A_1; B, C)$. Now, let $A_5$ denote the midpoint of arc $BAC$. Because $$\angle A_3A_4A_5 = 180^{\circ} - \angle A_2A_4A_5 = 90^{\circ}$$and $A_4A_5$ bisects $\angle BA_4C$, the Right Angle and Bisectors Lemma implies $$- 1 = (A_3, A_4A_5 \cap BC ; B, C)$$so $A_1, A_4, A_5$ are collinear. Let $P, Q, R$ be the feet of the $A_1$-altitude, $B_1$-altitude, $C_1$-altitude respectively of $A_1B_1C_1$. A well-known lemma implies $P \in AA_4, Q \in BB_4, R \in CC_4$. Now, let $H$ denote the orthocenter of $A_1B_1C_1$. Applying the Cevian Nest Lemma on the Gergonne Point of $ABC$ and $H$ implies $AP, BQ, CR$ are concurrent, which finishes. $\blacksquare$ Remark: The well-known lemma I cited is proven in many solutions to TSTST 2020/2.

Let $X=(AI)\cap (ABC)$. So by spiral similarity we have $\triangle XBC_1\sim\triangle XCB_1$. So we get $XB/XC=BA_1/CA_1$. Hence $A_1$ lies on the bisector of $\angle BXC$, and that gives $X$, $A_1$, $A_2$ are collinear. So \[ -1=(BC;A_3A_1)\stackrel{A_2}{=}(BC;A_4X)\implies \frac{BX}{CX}=\frac{BA_4}{CA_4} \]Now we have \[ \frac{\sin \angle BAA_4}{\sin \angle CAA_4}=\frac{\sin \angle BCA_4}{\sin \angle CBA_4}=\frac{BA_4}{CA_4}=\frac{BX}{CX}=\frac{BA_1}{CA_1} \]Now by sine ceva we are done.

Excellent Problem! Kim Jong Un’s math problems never disappoints me. Step 1 Redefine $A_4,B_4,C_4$ Note that $AA_1,BB_1,CC_1$ are concurrent $\implies$ $(A_3,A_1,B,C)=-1$ Let $M$ be a midpoint of segment $BC$; $A_3A_1\cdot A_3M=A_3B\cdot A_3C=A_3A_4\cdot A_3A_2$. We conclude that $A_1,M,A_2,A_4$ are concyclic which means $\angle A_1A_4A_2= 90^\circ$. So, $A_4,A_1,A’$ are collinear where $A’$ is a midpoint of arc $BC$ containing $A$. Similarly, $B_4,B_1,B’$ and $C_1,C_4,C’$ are collinear where $B’$ is a midpoint of arc $CA$ containing $B$ and $C’$ is a midpoint of arc $AB$ containing $C$. Step 2 Finish by simple trig-ceva calculation Observe that $A_1A_4$ bisects $\angle BA_4C$ $\implies \frac{BA_4}{A_4C}=\frac{BA_1}{A_1C}$ Apply law of sines yields $\frac{sin BAA_4}{sin A_4AC}=\frac{BA_4}{A_4C}=\frac{BA_1}{A_1C}$ Apply trigonometric ceva’s theorem to $\triangle ABC$ yields $AA_4, BB_4$ and $CC_4$ are concurrent, as desired.

bored In accordance with the absolutely outrageous point naming, let $A_5$ be the $A$-sharkydevil point, and define $B_5$, $C_5$ similarly. Now $A_2$, $A_1$, $A_5$ collinear, so \[ (A_3, A_1; B, C) \overset{A_2}{=} (A_4, A_5; B, C). \]Thus, if $A_6$ is the foot of the altitude from $A_1$ to $\overline{B_1C_1}$, then $\overline{AA_6A_4}$ collinear (this is well-known), and Cevian Nest finishes.

laFiesta wrote: North Korea is NOT Korea! "Korea" means South Korea. Why is this in Korea category? DPRK is the only Korea. What do you mean South Korea ????

Let $A_5$ be the $A$-Sharkydevil point, or $(AB_1C_1) \cap (ABC)$. Inverting around the incircle yields that $A$ goes to the midpoint of $B_1C_1$ and cyclic variants, implying that $(ABC)$ goes to the nine point circle of $A_1B_1C_1$, since the image of $A_5$ lies on $B_1C_1$ and is not $A$, we know that it is the foot of the altitude from $A_1$ to $B_1C_1$ (calling it $A_6$), giving the result that $A_5, A_6,I$ are collinear. By spiral similarity, we know that $A_5I$ maps to $A_5A_2$, and $B_1C_1$ goes to $BC$, and the $A_6$ goes to $A_1$, since $A_6A_1 \parallel IA_2$ and $A_1$ lies on $B_1C_1$, so $A_5, A_1, A_2$ are collinear. Since $AA_1, BB_1, CC_1$ are concurrenct by Ceva, harmonic lemmas give $(A_3, A_1; B,C)$ harmonic, projecting over $A_2$ gives $(A_4, A_5; B,C)$ harmonic, projecting through $A$ gives $(A_5A \cap B_1C_1, A_4A \cap B_1C_1;B_1,C_1)$ harmonic, projecting through $A_5$ gives $(A, (A_4A \cap B_1C_1)A_5 \cap (AI); B_1, C_1)$ harmonic, so $(A_4A \cap B_1C_1), A_5$ are collinear with $I$, so $A_4A \cap B_1C_1 = A_6$. By Cevian Nest, it is obvious that $AA_6, BB_6, CC_6$, concur, so we are done.