Find the length of the segment $QA$ in acute-angled $\triangle ABC$ Problem

Let $ABC$ be an acute-angled triangle. The external bisector of $\angle BAC$ meets the line $BC$ at point $N$. Let $M$ be the midpoint of $BC$. $P$ and $Q$ are two points on line $AN$ such that, $\angle PMN=\angle MQN=90^{\circ}$. If $PN=5$ and $BC=3$, find the length of $QA$. (BdMO 2021 Higher Secondary P6)

My attempt
Some angle chasing leads to $$\triangle PMN \sim \triangle QMN \sim \triangle PQM.$$ Then, $$\frac{PN}{MN}=\frac{MN}{QN} \implies MN^2=5QN \tag {1}$$ and $$\frac{PN}{PM}=\frac{PM}{PQ} \implies PM^2=5PQ \tag {2}$$ I couldn't achieve something useful from them.
I also tried adding the internal bisector of $\angle BAC$, then if the bisector is $AD$ we have $AD \parallel QM$ and $\triangle ADN \sim \triangle QMN$. But these are also not useful.(At least I can't proceed further)
Edit: After Intelligenti pauca's comment, I tried the problem in Geogebra and found that $A,B,C,P$ are concyclic. But I am unable to prove this. Here is the figure. Can someone prove this?

Also, Dr. Mathva provided a solution using projective geometry. But as I don't know projective geometry, I need a solution with Euclidean Geometry.

So, how to solve this problem?
Thanks in advance.
 A: Define $D$ to be the foot of the internal $A$-angle-bisector on $BC$. In the following, I will use basic concepts and notions of projective geometry; I will later try to come up with a solution employing more basic geometry :)

Notice that $\angle NAD=\frac{180^\circ-A}2+\frac{A}2=90^\circ$, which due to $\angle BAD=\angle DAC$ forces immediately $(N,D;B,C)=-1$. It is now well-known that this implies $MD\cdot MN=\left(\frac12 BC\right)^2$. As a sketch of the proof, consider the circle $\Gamma$ with diameter $BC$; the tangent through $N$ touches $\Gamma$ at $X$. Now just use similarity and/or Power of a Point. Alternatively, observe that $(N,D;B,C)=-1$ implies that $N,D$ are inversive conjugates with respect to $\Gamma$.
Observe, furthermore, that due to $\triangle AND\sim \triangle QNM\sim MNP$ we have $$\frac{DM}{AQ}=\frac{NM}{NQ}=\frac{NP}{NM}\implies DM\cdot NM=AQ\cdot NP\implies AQ\cdot NP=\left(\frac12 BC\right)^2$$
Plug in the values to obtain $$AQ=\frac{\left(\frac12 BC\right)^2}{NP}=\frac{\frac94}{5}=\fbox{$\displaystyle \frac9{20}$}$$
A: Claim: $P$ lies on $(ABC)$.
Proof: Let $P'$ be the intersection of the perpendicular bisector of $BC$ and circumcircle of $\triangle ABC$. Then, $\angle BP'C=\angle A$ and $\angle P'BC=\angle P'CB=90^ \circ -\frac{\angle A}{2}$. Then $\angle BAP'=90^ \circ +\frac{\angle A}{2}$. But note that $\angle BAP=\angle A +90^ \circ -\frac{\angle A}{2}=90^ \circ +\frac{\angle A}{2}$. So, $P$ and $P'$ are the same point, which implies $P$ lies on $(ABC)$. QED
Now using PoP, $$AN \cdot PN=BN \cdot CN=(MN-BM)(MN+CM)=MN^2-BM^2$$ $$\implies AN \cdot PN=MN^2-\frac{9}{4} \tag{1}$$
And $\triangle QMN \sim \triangle PMN$, then $$QN \cdot PN=MN^2 \tag{2}$$
Subtracting (1) from (2), $$QA \cdot PN=\frac{9}{4}$$ $$\implies QA= \boxed {\frac{9}{20}}$$
