1.Denote the reflection of X w.r.t YZ is E, we will prove the EA,OI,BC are concurrent. 2.Let I1I2I3 be the excentral triangle. The lines Y Z and I2I3 are parallel because both are perpendicular to AI. Similarly, ZX//I3I1 and XY//I1I2. Hence, the excentral triangle and the intouch triangle are homothetic and their Euler lines are parallel. Now, I and O are the orthocenter and nine-point center of the excentral triangle. On the other hand, I is the circumcenter of the intouch triangle. Therefore, the line OI is their common Euler line, contains the orthocenter H' of XY Z. 3.AI/IA1=BA/BA1 XH’=2r*cos(A/2) XE=2h EH’= XE- XH’ where h, r are the attitude from X to YZ and the incircle radii we can calculate AI/IA1= EH’/ XH’= (sinB + sinC)/sinA Therefore, we complete the proof.

Maxim Bogdan wrote: Let $ ABC$ be a non-isosceles triangle, in which $ X,Y,$ and $ Z$ are the tangency points of the incircle of center $ I$ with sides $ BC,CA$ and $ AB$ respectively. Denoting by $ O$ the circumcircle of $ \triangle{ABC}$, line $ OI$ meets $ BC$ at a point $ D.$ The perpendicular dropped from $ X$ to $ YZ$ intersects $ AD$ at $ E$. Prove that $ YZ$ is the perpendicular bisector of $ [EX]$. Take a look at http://forumgeom.fau.edu/FG2004volume4/FG200411index.html

This is what I wrote for the RMC (and in the contest). Practically, it is a compilation of the solution given in the note April cited above. There is also another synthetic solution by Dinu Serbanescu, but I shall write it when I have more time. Theorem (Emelyanov). Let $ ABC$ be a nonisosceles triangle, for which denote by $ X$, $ Y$, $ Z$ the tangency points of its incircle with the sides $ BC$, $ CA$ and $ AB$, respectively. Let $ D$ be the intersection of $ OI$ with the sideline $ BC$, where $ O$, $ I$ are the circumcenter and incenter, respectively. The line perpendicular to $ YZ$ through $ X$ cuts $ AD$ at $ E$. Prove that the line $ YZ$ is the perpendicular bisector of $ EX$. Solution. We begin with a preliminary result. Lemma. The orthocenter $ H'$ of the intouch triangle $ XYZ$ lies on the line $ OI$. We will give two different proofs for this useful result. First proof. Let $ I_{a}I_{b}I_{c}$ be the excentral triangle of $ ABC$. The lines $ YZ$ and $ I_{b}I_{c}$ are parallel because both are perpendicular to $ AI$. Similarly, $ ZX \| I_{c}I_{a}$ and $ XY \| I_{a}I_{b}$. Hence, the excentral triangle and the intouch triangle are homothetic and their Euler lines are parallel. Now, $ I$ and $ O$ are the orthocenter and the nine-point center of the excentral triangle. On the other hand, $ I$ is the circumcenter of the intouch triangle. Therefore, the line $ OI$ is their common Euler line, and thus it contains the orthocenter $ H'$ of $ XYZ$. $ \hfill \square$ Second proof. Let $ A'$, $ B'$, $ C'$ be the midpoints of $ YZ$, $ ZX$ and $ XY$, respectively. Note that \[ IA \cdot IA' = \left(\frac {r}{\sin{\frac {A}{2}}}\right) \cdot \left(r \sin{\frac {A}{2}}\right) = r^{2}.\] Similarly $ IB \cdot IB' = IC \cdot IC' = r^{2}$. Therefore, the inversion $ \Psi(I,r^{2})$ with respect to the incircle of $ ABC$ maps the vertices $ A$, $ B$, $ C$ to $ A'$, $ B'$, $ C'$, and moreover maps the circumcircle of $ ABC$ into the circumcircle of $ A'B'C'$. But if a circle with center $ P$ is inverted in a circle with center $ Q$, then its image is a circle whose center lies on the line $ PQ$. Thus, the circumcenter $ O'$ of $ A'B'C'$ lies on the line $ OI$. Since $ A'B'C'$ is the medial triangle of $ XYZ$, its circumcenter $ O'$ is the nine-point center of $ XYZ$. Following the fact that $ I$ is the circumcenter of $ XYZ$, it becomes clear now that $ OI$ is the Euler line of $ XYZ$. Obviously, the orthocenter $ H'$ lies on this line. $ \hfill \square$ Returning to our problem, we notice that the Lemma motivates us to rewrite the last part into the following equivalent version: Claim. Let $ E$ be the reflection of $ X$ into the line $ YZ$. Denote by $ D$ the intersection of $ AE$ with the sideline $ BC$. Prove that $ D$ lies on the line $ OI$. Proof. Denote by $ T$ the intersection of $ AI$ with $ BC$. Since $ XE$ and $ AT$ are both perpendicular to $ YZ$, they are parallel and therefore the collinearity is equivalent with $ EH'/H'X = AI / IT$. It is known that \[ \frac {AI}{IT} = \frac {CA + AB}{BC} = \frac {\sin{B} + \sin{C}}{\sin{A}}.\] For example, this follows from Van Aubel's theorem. For any acute triangle, $ AH = 2R \cos{A}$. The angles of the intouch triangle are \[ X = \frac {B + C}{2},\ \ Y = \frac {C + A}{2},\ \ Z = \frac {A + B}{2}.\] It is clear that $ XYZ$ is always acute, and \[ XH' = 2r\cos{X} = 2r \cos{\frac {B + C}{2}} = 2r \sin{\frac {A}{2}}.\] Thus, \begin{eqnarray*} \frac {EH'}{H'X} & = & \frac {EX - H'X}{H'X} = \frac {EX \cdot YZ}{H'X \cdot YZ} - 1 \\ & = & \frac {4[XYZ]}{H'X \cdot YZ} - 1\ \ \ ([\mathcal{P}] \text{denotes the area of the polygon}\ \mathcal{P}) \\ & = & \frac {2r^{2}(\sin{2X} + \sin{2Y} + \sin{2Z})}{2r\sin{X} \cdot 2r \cos{X}} - 1 \\ & = & \frac {\sin{2Y} + \sin{2Z}}{\sin{2Y}} = \frac {\sin{B} + \sin{C}}{\sin{A}}. \end{eqnarray*} This completes our proof. $ \hfill \blacksquare$ Remarks. We invite the reader to deduce the following (simple) consequences: Corollary 1. The line joining $ D$ to the projection of $ X$ on $ YZ$ passes through the midpoint of the bisector of angle $ A$. Corollary 2. The $ OI$-line is parallel to $ BC$ if and only if the projection of $ X$ on $ YZ$ lies on the line joining the midpoints of $ AB$ and $ AC$.

Since IO is the Euler line of XYZ, we can restate the enunciation as a more useful-general result. Proposition: Triangle $\triangle ABC$ is scalene with euler line $e.$ $\triangle A_0B_0C_0$ is the tangential triangle of $\triangle ABC,$ $A_0,B_0,C_0$ against $A,B,C.$ If $D$ is the reflection of $A$ across $BC,$ then lines $e,B_0C_0, DA_0$ concur. Using barycentric coordinates with respect to $\triangle ABC,$ we have $A_0 (-a^2:b^2:c^2) \ , \ B_0 (a^2:-b^2:c^2) \ , \ C_0 (a^2:b^2:-c^2) \ , \ D(-a^2:2S_C:2S_B)$ $e \equiv S_A(S_B-S_C)x+S_B(S_C-S_A)y+S_C(S_A-S_B)z=0$ $\Longrightarrow \ DA_0 \equiv 2(b^2S_B-c^2S_C)x+a^2(a^2-b^2)y-a^2(a^2-c^2)z=0$ $\Longrightarrow \ B_0C_0 \equiv c^2y +b^2z=0$ Then, lines $e,B_0C_0, DA_0$ are indeed concurrent at a point $U$ with coordinates $U \equiv (a^2b^2(a^2-b^2)+a^2c^2(a^2-c^2):-2b^2(b^2S_B-c^2S_C):2c^2(b^2S_B-c^2S_C))$

Dear Mathlinkers, for a synthetic proof you can see http://perso.orange.fr/jl.ayme vol. 4 Symérique de OI... p. 22. Sincerely Jean-Louis

I hope I didn't mislabel any points. Here's a rather complicated solution using homothety, I wonder if someone simplify this? Let $\triangle A'B'C'$ be the triangle such that $\triangle ABC$ is the orthic triangle of $\triangle A'B'C'$. It is well known that $I$ is the orthocenter of $A'B'C'$ and $A'$ is the excenter of $ABC$. Let $O'$ be the circumcenter of $\triangle A'B'C'$; then by Euler Line, $O', O, I$ are collinear. Let the $A$-excircle of $\triangle ABC$ touch $BC, CA, AB$ at $X', Y', Z'$, respectively. Let $AX$ intersect the $A$-excircle at $W$; reflect $W$ over $Z'Y'$ to get $W'$. Reflect $A'$ over $BC$ to get $A''$. Note that $\triangle Z'WY'$ is homothetic to $\triangle CAB$, so $A''O' \parallel W'A'$. Let $W'A$ meet $A''O'$ again at $P$; because of parallel lines, $AP = AW'$. Since $A'$ is also the midpoint of $X'W$, and $A'A \parallel WW'$, $X'P \parallel A'A$. Let $Q$ be the reflection of $X$ over $YZ$; then from a homothety centered at $A$ taking the incircle of $\triangle ABC$ to its $A$-excircle, $Q$ lies on $AW'$ and $XQ \parallel WW' \parallel X'P$. Since we also have $IQ \parallel W'A' \parallel PB$ and $IX \parallel A'O'$, $\triangle O'X'P$ is homothetic to $\triangle IXQ$, so $OI$, $BC$, and $AQ$ concur. Then $Q = E$, so $YZ$ bisects $EX$, as desired.

i hope this wasn't posted above next $H,R,H'$ are the orthocenter of $XYZ$ and second intersections of $XI,XH$ with the circle $XYZ$. Inversion in $I$ with radius $IY$ pictures $M,N,P$ to $A,B,C$($M,N,P$ are midpoints of $YZ,ZX,XY$) and therefore the Euler's circle of $XYZ$ into circle $ABC$. so $O$ is on the line connecting $I$ and the center of circle $MNP$(line $IH$) so $I,O,H$ are collinear $AI\cap BC=D$ then $\angle H'XR=\angle XID$ and $\angle XH'R=\angle IXD=90$ so $IXD\sim XH'R$ and $ID/XI=2*XI/XH'$ and $ID*XH'=2*XI^2$ since $\angle H'YZ=\angle HXZ=\angle HYZ$ and $HH'\perp YZ$ then $H,H'$ are symmetric wrt $YZ$. next since $XH*AI=2*MI*AI=2*IY^2=2*IX^2$ then using the previous identity we have $XH/XH'=ID/IA=XH/EH$ so $EH=XH'$ and now $E,X$ are symmetric wrt $YZ$. so $YZ$ is the perpendicular bisector of $EX$. Remark: proving that $H,I,O$ are collinear can be solved without inversion considering the feet of altitudes in $XYZ$(points $P,Q,S$) and the homothethy that pictures $PQS$ into $ABC$)

plane geometry wrote: 3.$XH'=2r*cos(A/2)$ $XE=2h$ $EH'= XE- XH'$ where $h, r$ are the attitude from $X$ to $YZ$ and the incircle radii we can calculate $AI/IA_{1}= EH'/ XH'= (sinB + sinC)/sinA$ Therefore, we complete the proof. there is typo mistake: $XH'=2rsin(\frac{A}{2})$ not cosine and the rest follows..