Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear.
Define : $F,X,Y$ as midpoints of $BC,CA,AB$ .
$N_9$ as the nine point center
$O$ as the circumcentre
$H'$ as the orthocentre
$BX \cap (O_1) = L$
$CY \cap (O_2)= I $
Claim: $ABCL$ is a parallelogram
Proof: Since $AX=CX$ by midpoint condition and $BX=XL$ by POP ( Taking power of $X$ wrt both circles $XD.XL=CX^2= XD.XB$ )
Similarly $ABCI$ is a parallelogram
Claim: $ALI$ are collinear
Proof: That follows from BC parallel condition
We also know that $H, M,N_9,O$ are collinear , $M$ is the centroid
Now what we noticed was that $OO_2H'O_1$ are parallelogram with $N_9$ as the intersection of the diagonals .
What I think is that showing $OO_2H'O_1$ a parallelogram is enough , since we know that $N_9$ is the midpoint of OH'
There's also nice dilations happening , like $N_9$ dilating $O$ to $H'$ and $O_2$ to $O_1$ with scale $-1$ ( observation )
dilation centred at $X$ and $Y$ with scale factor -1 too.
Moreover, we also know that $OO_2 \perp BE$ , so it's enough to show that $O_1H' \perp BE$ .
Also I want to find a pure synthetic method ( not using tring, cord , Bary, vector , etc ) but can include inversion or projective .
Thanks in advance!
Here's the diagram link: https://www.geogebra.org/geometry/g3mbkhkp
EDIT: Since @Anand told me to define $IB\cap LC$ , I defined $IB\cap LC=J$
Since $IL || BC$ , $A$ is midpoint of $IA$ and $F$ is midpoint of $BC$ , we get that $JFA$ is collinear and $B,F,C$ are midpoints of $IJ,AJ,LC$
So $AJ,BL,CI$ concur at $K$ , $K$ is the centroid and we get that $K$ is the centre of dilation with scale factor $-2$ .
also $K$ dilates $O$ to $H'$ too .
That's how much I could proceed till now :(