Let $I$ be the incenter. Let $DX$ intersect the incircle at $J$, and the tangent to incircle at $J$ intersect $BC$ at $K$. Note that the circles $ATFIE$ and $BIE$ are tangent to each other at $I$ since their centers both lie on $AI$. Thus the common tangent to these two circles is parallel to $EF$. But $IK\perp JD$ and $JD\perp FE$. So $IK$ is this common tangent. So the radical axes of circles $AFIE, BIE, ABC$ concur, ie. $AT,BC,IK$ are concurrent. But $K\in BC$, so $K,T,A$ collinear. Now let $EF\cap JD=H$. With respect to the incircle, $EF,JD$ are polars of $A,K$ respectively. Thus the polar of $H$ is $AK$, thus $IH\perp AK$. If $IH\cap AK=T'$, then obviously $T'$ lies on circle $AFIE$. Thus $T'\equiv T$, and $T,H,I$ collinear. Now \[\angle FTH=\angle FTI=\angle FAI=\angle FXH\] Thus $T,F,H,X$ are concyclic, and $TX\perp TF$, as desired.

Let $M$ be the midpoint of the arc $BC$ on $(ABC)$ not containing $A$. Clearly $\angle AFT=\angle AIT=\angle AET\implies\angle BFT=\angle MIT=\angle CET$. Also $\angle AFT=\angle ABT=\angle ACT$. And so $\triangle TFB\sim\triangle TIM\sim\triangle TEC$. Now $BD=BF$ and $CD=CE$ imply that $\frac{BD}{DC}=\frac{BF}{CE}=\frac{BT}{TC}$ since $\triangle TFB\sim\triangle TEC$. And so by the converse of the angle bisector theorem, $TD$ is the angle bisector of $\angle BTC$ and so $T,D,M$ are collinear. Now let $G$ be the foot of the perpendicular from $D$ to $EF$. Angle chasing shows $\triangle GDE\sim\triangle DBI$ and $\triangle GDF\sim\triangle DCI$. Therefore $\frac{BD}{DI}=\frac{DG}{GE}$ and $\frac{CD}{DI}=\frac{DG}{GF}$ and thus $GE\cdot BD=DG\cdot DI=CD\cdot GF$ which implies $\frac{GF}{GE}=\frac{BD}{CD}=\frac{BF}{EC}$ since $BD=BF$ and $CD=CE$. But $\triangle TFB\sim\triangle TEC$ so $\frac{GF}{GE}=\frac{BF}{EC}=\frac{FT}{TE}$ and so by the converse of the angle bisector theorem, $TG$ bisects angle $FTE$. Then $\angle GTF=\frac{1}{2}\angle FTE=\frac{1}{2}A=\angle FAI=\angle FXG$ since of course $AI||XG$. Thus $CTGF$ is cyclic and so $\angle XTF=\angle XGF=90^{\circ}$.

Let $Y$ be the antipode of $A$ with respect to the circumcircle of $ABC$. Let $Z$ be the foot of the perpendicular from $D$ to $EF$. Note that $TFIEA$ is cyclic. Hence $\angle FTI = \frac{1}{2} \angle BAC$. $TX \perp TF \Leftrightarrow TFZX$ cyclic $\Leftrightarrow \angle FTZ = \angle FXZ = \frac{1}{2} \angle BAC \Leftrightarrow T, Z, I$ collinear. Note that $Y, I, T$ are collinear so we want to prove that $Y, I, Z$ are collinear. To do this I used complex numbers. Set the incircle to be the unit circle and represent the complex number of a point by its lower case affix. It is well known that $z=\frac{1}{2} (d+e+f-\frac{ef}{d}) \implies \frac{z}{\overline{z}} = \frac{def(d+e+f)-e^2f^2}{de+ef+fd-d^2}$. It is also a well known complex number result that $o=\frac{2def(d+e+f)}{(d+e)(e+f)(f+d)} \implies y= \frac{4def(d+e+f)-2ef(d+e)(f+d)}{(d+e)(e+f)(f+d)} \implies \frac{y}{\overline{y}} = \frac{ef(d^2+de+df-ef)}{de+ef+df-d^2}=\frac{z}{\overline{z}}$ so we are done.

Another solution: Let $AO$ cut the circumcircle of $ABC$ at $G$, as $\angle ATG = 90^{\circ}$ and $\angle ATI = 90^{\circ}, T, I, G$ are collinear. Let $DX$ cut $EF$ at $L$. Since $XL\parallel AI$, if we can show that $T, L, I$ are collinear, then it follows that $TXLF$ cyclic and $TX\perp TF$. To prove that $T, L, I$ are collinear is the same as to prove that $L, I, G$ are collinear. Let $AI$ cut the circumcirle of $ABC$ again at $K$, the midpoint of the lower arc $BC$. Denote $\angle A, \angle B, \angle C$ the angle of $\Delta ABC$, its incirle's radius $r$ and its circumcirle's radius $R$. Notice that $\triangle IDN\sim AKG$, therefore $\frac{DN}{KG} = \frac{r}{AK} = \frac{r}{2R.cos(A/2+C)}$, but $\frac{LH}{DN} = cos(A/2+C)$, hence $\frac{LH}{KG} = \frac{r}{2R}$. Also notice that $IK = BK = \frac{BC}{2.cos(A/2)} = \frac{2R.sinA}{2.cos(A/2)} = 2R.sin(A/2)$, therefore, $\frac{HI}{IK} = \frac{r.sin(A/2)}{2R.sin(A/2)} = \frac{r}{2R}$. So $\frac{LH}{KG} = \frac{HI}{IK}, L, I, G$ are collinear. q.e.d.

Another way to prove the collinearity of $L, I, G$: Notice that $KI = KB = KC$ Let $KO$ cut the circumcircle again at $J$. $\Delta EHI\sim \Delta JCK, \frac{HI}{IK} = \frac{HI}{KC} = \frac{IE}{KJ} = \frac{IE}{AG}$ If $Y$ is the projection of $D$ on $AK$ then $\Delta IDY\sim \Delta AKG, \frac{LH}{KG} = \frac{DY}{KG} = \frac{ID}{AG} = \frac{IE}{AG}$. Hence $L, I, G$ are collinear

goodar2006 wrote: The incircle of triangle $ABC$ for which $AB\neq AC$, is tangent to sides $BC,CA$ and $AB$ in points $D,E$ and $F$ respectively. Perpendicular from $D$ to $EF$ intersects side $AB$ at $X$, and the second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX\perp TF$. Proposed By Pedram Safaei Let $J$ be the intersection of $XD,EF$. According to topic http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=499885, we have $T,I,J$ are collinear. $\widehat{FTI}=\widehat{FAI}=\widehat{FXJ}$. Thus, $T,X,J,F$ are concyclic. Hence $\widehat{XTF}=90^0$. We are done.

My solution: Let $ H=DX \cap EF $ Since $ H(F, D; B, C)=-1 $ and $ \angle FHD=90 $ , so $ \angle BHD=\angle DHC $ . ie. $ \angle FHB=\angle CHE $ hence we get $ \triangle BHF $ ~ $ \triangle CHE $ and $ \frac {BD}{DC}=\frac {BH}{HC}=\frac {FH}{HE} $ . Since $ T $ is the Miquel point of complete quadrilateral $ \{ EF, FB, BC, CE \} $ , so $ T $ is also the Miquel point of complete quadrilateral $ \{ HF, FB, BD, DH \} $ (i.e. $ T, F, H, X $ are concyclic.) , hence we get $ \angle FTX=180-\angle XHF=180-90=90 $ . Q.E.D

$S$ is the foot of the altitude of $D$ on $EF$ it suffices to prove that $XTSF$ is cyclic.It is well know that t is the center of spiral similarity that takes $EC$ to $FB$ so we have $\triangle TEC\sim \triangle TFB$ so $\frac{TC}{TB}=\frac{EC}{FB}=\frac{CD}{DB}$ thus $TD$ is the angle bisector of the $\angle CTB$ but from the similarity $\angle ETC=\angle FTB$ so $TS$ is angle bisestor of the $\angle FTE$ thus $\angle FTS=\angle SXF=\frac{\alpha}{2}$ and we are done

Note that $(T,G)$ are swapped under an inversion about the incircle, as $\odot(ABC)$ is sent to the nine-point circle of $\triangle DEF$ and line $EF$ is sent to $\odot(AEF)$, hence $I, T,$ and $G$ are collinear. Then $AI\parallel GX$ since they are both perpendicular to $EF$, hence we have $$\angle GTF=\angle ITF=\angle IFG=\angle IAE=\angle FAI=\angle FXG$$so quadrilateral $TXGF$ is cyclic, giving $\angle XTF=\angle XGF=90^{\circ}$ as desired$.\:\blacksquare\:$

Let $U=DX \cap EF$ and let $V=EF \cap BC.$ Since $T$ is center of spiral similarity mapping $BF$ to $EC$ we have $\frac{TB}{TC}=\frac{BD}{DC}. (*)$ Claim: $UD$ bisects $\angle BUC$ Proof: $(B,C;D,V)=-1$ and $\angle DUV=90^{\circ}$ thus the claim. So, now we have $\frac{BU}{UC}=\frac{BD}{DC}(**).$ Now from $(*),(**)$ we have that $\odot(DTU)$ is Apollonius circle for $B,D.$ But since $\angle DUV=90^{\circ}$ we have that $VDUT$ is cyclic. Clearly, $T$ is Miquel point of complete quadrilateral $BCEFVA$ hence $TFBV$ is cyclic. But we proved that $VDUT$ is cyclic hence $T$ is Miquel point of $BDUFVX$ hence $TUFX$ is cyclic and we are done. $\blacksquare$

goodar2006 wrote: The incircle of triangle $ABC$ for which $AB\neq AC$, is tangent to sides $BC,CA$ and $AB$ in points $D,E$ and $F$ respectively. Perpendicular from $D$ to $EF$ intersects side $AB$ at $X$, and the second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX\perp TF$. Proposed By Pedram Safaei Let $DX\cap FE=K$. Claim:- $\overline{T-K-I}$ Invert around the Incircle. Let this map be $\Psi$. So, $$\Psi:K=EF\cap \odot(X_5 \text{ of } \triangle DEF)\leftrightarrow \odot(AEF)\cap\odot(ABC)\neq A=T\implies \Psi:K\leftrightarrow T\implies \overline{I-K-T}$$So, $DX\|AI\implies FTXK$ is a cyclic quadrilateral by Reim's Theorem. Hence, $TX\perp TF$ @below Oh! Actually I thought that to be well known that's why I didn't write it. . Edit:-Added it.

@above u need to prove $\{T,K,I\}$ collinear to apply reims and thats kindof the main part of the problem (see #5)

Dear Mathlinkers, https://artofproblemsolving.com/community/c6t48f6h1430728_problem_for_you Sincerely Jean-Louis

Let $H$ be the foot from $D$ to $EF$, and let $J$ be on $BC$ such that $IJ\parallel EF$. Notice that $J$ lies on the polar of $H$, so $AJ$ is the polar of $H$. Let $M$ be the midpoint of arc $BC$ not containing $A$. Since $MI\perp EF$, we get $JI^2=JB\cdot JC\implies J\in AT$. Since $\angle ITA=90^{\circ}$, we get $I,H,T$ collinear. Then $\angle HXF=\frac{1}{2}\angle A$ and $\angle HTF=\angle ITF = \angle IAF = \frac{1}{2}\angle A$, so $XHFT$ is cyclic. Thus, $TX\perp TF$.

Let $F'$ be the antipode of $F$ on $(AEF)$, $K$ be the foot of the altitude from $D$ to $EF$. By inversion centered at $I$ with radius $ID$, $(ABC)$ goes to the ninepoint circle of triangle $DEF$ and $(AEF)$ goes to $EF$ thus, $I,K,T$ are collinear. By Pascal at $ITF'EFA,$ we get that $K,TF'\cap AB,AI_{\infty}$ are collinear. This line is $\overline{DKX}$ hence $T,X,F'$ are collinear which gives the desired result.$\blacksquare$