Dear Mathlinkers, according to the cevian nests theorem, they are concurrent...for the first step. Sincerely Jean-Louis

Dear Mathlinkers, more simple, the orthic triangle of DEF is homothetic to ABC, then they are concurrent...for the first step. Sincerely Jean-Louis

jayme wrote: according to the cevian nests theorem, they are concurrent...for the first step. What is cevian nests theorem?

see for example http://perso.orange.fr/jl.ayme vol. 3 the cevian nests theorem or on search on mathlinks Sincerely Jean-Louis

For the second step, prove that the concurency point of $AX,\ BY,\ CZ,$ lies on the segment line $OI,$ where $O,\ I$ are the circumcenter and the incenter of $\vartriangle ABC$ respectively. Kostas Vittas.

Dear Mathlinkers, let P be the homothetic center... Consider the circumcircle of ABC and XYZ... and we are done with the idea of Kostas... Sincerely Jean-Louis

Thank you both jayme and vittasko. I've edited the problem's statement.

Let O and H be the circumcentre and orthocentre of DEF respectively. Then XYZ and ABC are homothetic, and also OAC and HXZ are homothetic so we are done

we know that the triangles$ABC$ and $DEF$ are Homogeneouses .now That's enough to proof the line $OI$ is the Euler line of the triangle $DEF$. we prove that $OI$ pass trough to $K$ (K is orthocenter of the triangle DEF). We separated $AK$ and $BL$ on the segments $AC$ and $BC$ respectively.($AK=BL=c$) and we separated $AM$ and $CN$ on the segment $AB,CB$ respectively.($AM=CN=b$).we know that $OI$ is perpendicular to $LK,MN$ so $LK//MN$ .$<KLC=<BMN=q$ we see $LC/KC=a-c/b-c=sin(q+C)/sin(q)$ and we see $BN/BM=a-b/b-c=sin(B-q)/sin(q)$ .now in the triangle $DEF$ the intersection of altitude from $D$ by $OI$ is $Q$. distance from $I$ to $EF$ is $ID.cosD=rsin(A/2)$ so $A1Q=2rsin(A/2)$ and $<QID=<180-q$ and $<QDI=(B-C)/2$$ so 2sin(A/2)=sin(q)/[sin(q-(B-c)/2)]$ this means $sin(q+c)-sin(B-q)=sin(q)$ .so $D=Q$

Let $H$ be the orthocenter of $\triangle DEF$, $M$ be the common point of $AI$ and $EF$. Let $AX$ intersects $IH$ at $P$. Because $HX \parallel AI$, we have $\frac{PH}{PI}=\frac{HX}{AI}=\frac{HX.IM}{r^2}=\frac{HX.HD}{2r^2}$. Because $HX.HD=HE.HY=HF.HZ=k$, we deduce that $P$ is a fixed point on the Euler line of $\triangle DEF$, so $AX,BY,CZ$ concurs at $P$.

My solution: Let $ H $ be the orthocenter of $ \triangle DEF $ . Since $ \triangle ABC $ and $ \triangle XYZ $ are homothetic , so $ AX, BY, CZ $ are concurrent at their homothety center $ T $ . Since $ \triangle XYZ \cap H \sim \triangle ABC \cap I $ , so $ T $ lie on $ HI $ which is the Euler line of $ \triangle DEF $. Q.E.D Remark: $ T $ is the isogonal conjugate of isotomic conjugate of $ H $ WRT $ \triangle DEF $

Let $H$ be the orthocenter of $\triangle DEF$. We have $\frac{\sin \angle XAB}{\sin \angle XAC}= \frac{XF}{XE}$. Similarly, we get \[\frac{\sin \angle XAB}{\sin \angle XAC} \cdot \frac{\sin \angle ZCA}{\sin \angle ZCB} \cdot \frac{\sin \angle YBC}{\sin \angle YBA}= \frac{XE}{XF} \cdot \frac{YF}{YD} \cdot \frac{ZD}{ZF}=1.\] Hence, $AX,BY,CZ$ are concurrent at $T$. On the other hand, since $ID \perp BC,ID \perp YZ$ so $YZ \parallel BC$. We also get $XY \parallel AB, XZ \parallel AC$. Therefore $\frac{TY}{TB}= \frac{TX}{TA}= \frac{TZ}{TC}$. We also have $YH \parallel BI, XH \parallel AI, ZH \parallel CI$. $YH \cap TI=H'$. We have $\frac{TH'}{TI}= \frac{TY}{TB}= \frac{TX}{TA}$. Hence, $H'X \parallel AI$. Thus, $H \equiv H'$ or $H,T,I$ are collinear or $H$ lies on Euler line of $\triangle DEF$.

My solution: Let $I_A, I_B, I_C$ the excenters with relation to $A, B, C$, respectively. Notice that $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic. Now, let $V$ be the reflection of $I$ through $O$ (indeed, $V$ is the Bevan point of $\triangle ABC$). Since $I$ is the orthocenter of $\triangle I_AI_BI_C$ and $O$ is the center of the nine points circle of $\triangle I_AI_BI_C$, follows that $V$ is the circuncenter of $\triangle I_AI_BI_C$, since $O$ is the midpoint of $IV$. In this way, follows that $\triangle DEF \cup I$ and $\triangle I_AI_BI_C \cup V$ are homothetic. Since $\triangle XYZ$ and $\triangle ABC$ are the orthic triangles of $\triangle DEF$ and $\triangle I_AI_BI_C$, respectively, finally follows that $\triangle XYZ \cup I$ and $\triangle ABC \cup V$ are homothetic, in other words, $AX, BY, CZ, IV$ are concurrent, but $IV$ and $IO$ are the same straight line, thus $AX, BY, CZ$ concur on $IO$.

It's surprisingly simple by Monge d'Alembert (sorry if I misquoted the theorem I am about to use) Let $H$ be the orthocenter of triangle $DEF$ and $N$ be the isogonal conjugate of the Nagel point of $ABC$. Then the circumcircle of $ABC$ and the nine point circle of $DEF$ are homothetic w.r.t some point $T$ because $XYZ$ and $ABC$ are similar and since the corresponding sides are parallel we have $AX,BY,CZ$ concurrent at $T$. Now, the circumcircle of $DEF$ and its nine point circle are homothetic about $H$. The circumcircle of $DEF$ and $ABC$ are homothetic about $N$. So, by Monge's theorem, $H,N,T$ are col-linear.