Let $\triangle ABC$ be an acute triangle with incenter $I$ and circumcenter $O$. The incircle touches sides $BC,CA,$ and $AB$ at $D,E,$ and $F$ respectively, and $A'$ is the reflection of $A$ over $O$. The circumcircles of $ABC$ and $A'EF$ meet at $G$, and the circumcircles of $AMG$ and $A'EF$ meet at a point $H\neq G$, where $M$ is the midpoint of $EF$. Prove that if $GH$ and $EF$ meet at $T$, then $DT\perp EF$.
My Progress: After seeing this problem, the first thing that struck my mind was sharky devil lemma (not a very known lemma )
Here's the lemma: In triangle $ABC$, let $DEF$ be the contact triangle, and let $(M)$ be the midpoint of the arc $(BC)$ not containing $(A)$ in $(ABC)$. Suppose ray $MD$ meets $(ABC)$ again at $R$. If $I$ is the incenter of $(ABC)$ and ray $RI$ intersects $(ABC)$ again at $A'$, then $A'$ is the antipode of $A$. If $P=RA'\cap EF$, then $DP\perp EF$.
Anyways, here's the problem's diagram:
Here $J$ is defined as $(ABC)\cap (AEF) .$
Now, if I am able to show that $JITA'$ are collinear, then I am done.
Moreover, I got that $T$ is the radical centre of $(AEF),(GHA')$ and $(AHG)$.
Here, I defined $K$ as $AT\cap (AEF)$.
Now,I thought of using Phantom points . So I defined $T'= \overline{JIA'}\cap EF$ .
We want to show that $T'=T $ . To show that $T'=T $ , we can also show $G,T',H$.
Now, note that $AM\perp EF$.
Let $AJ\cap EF=L$.
So, by radical axis lemma on $(AEF),(ABC) ,(GH'EF)$ , we get $AJ,EF,GA'$ concur at $L$ .
Also we have $T'KMI$ and $AJT'M$ cyclic .
Again by radical axis lemma on $(AEF),(AJT'M),(T'MKI)$ , we get $AJ,TM(EF),KI$ concur at $L$.
Note that $\angle AGA'=90=\angle AMF$ . Since $LFE$ and $LGA$ are collinear , we get $(AHMGLK)$ concyclic.
Also note that $T'$ is the orthocentre of $\Delta ALI$.
This is what I got till now. Now after showing that J,I,T are collinear , by applying sharky devil lemma , we will be done . I know that this problem has a 1 para solution( by @Anand ), but can someone provide a non-projective solution ?
Thanks in Advance!