Acute triangle $ABC$ is inscribed in circle $\omega$. Let $H$ and $O$ denote its orthocenter and circumcenter, respectively. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Rays $MH$ and $NH$ meet $\omega$ at $P$ and $Q$, respectively. Lines $MN$ and $PQ$ meet at $R$. Prove that $OA\perp RA$.
I will be very grateful if someone can verify this proof . I am very new to radical axis.
Also, Please post your solutions too. We learn a lot from other's solutions too. Thanks in advance.
My Proof: Before proceeding further, I would like to state a lemma.
Lemma: Let $ABC$ be a triangle with orthocenter $H$, and suppose that $E$ and $F$ are the feet of the $B$ and $C$-altitudes. Suppose that the circumcircle of triangle $AEF$ meets the circumcircle of triangle $ABC$ again at $K$. Let $M$ be the midpoint of $BC$. Then we have $K, H,$ and $M$ are collinear.
Proof of the Lemma: Sine $HF\perp AB$ and $HE\perp AC$, we note that $H\in (AEF)$. So $\angle AKH= \angle AFH = 90^{\circ}$
Let $KH\cap(ABC)=X$. Note that since, $\angle AKH=90^{\circ}$, we have $X=$ diametrically opposite point of $A$.
But by a known lemma , we know that $H,M,X$ are collinear . So we have $K$,$ H,$M$ are collinear.
Now, using this Lemma, we claim that $MNPQ$ is cyclic
Claim: $MNPQ$ is cyclic
Proof: By the above Lemma, we get $H'MHP$ and $QHN{H'}_1$ are collinear, where $H"M=HM$ and $H'$ is the antipode of $P$ wrt $(ABC)$ and $N{H'}_1=HN$ and ${H'}_1$ is the antipode of $Q$ wrt $(ABC)$.
Hence by $POP$, $\Bbb P(H, (ABC))=HH'\cdot HP=QH\cdot H{H'}_1$.
But $HM=\frac {1}{2} HH'$ and $HN=\frac {1}{2}H{H'}_1= HN \implies HM\cdot HP=QH\cdot HN$.
Hence by converse of $POP$ , we have $MNPQ$ cyclic .
Claim: $AMON$ is cyclic with diametre $AO$ .
Proof of the Claim: Just note that $AM \perp OM$ and $AN \perp ON$ .
Main Proof : Now, using the fact that pairwise radical axis of 3 circles concur,
we get that for circles $(MNPQ),(ABC),(AMON)$ ; the pairwise radical axis concur at $PQ\cap MN=R$.
But note that the radical axis of $(ABC)$ and $(AMON)$ is nothing but the line tangent to $(AMON)$ at $A$.
Since $AO$ is the diameter of $(AMON)$, hence $OA\perp RA$