Angle chasing: Finding the $\angle{NRQ}$ in a quarter circle 
Let $MON$ be a quarter circle. $P$ is the midpoint of $OM$ and $PQ$ is the angle bisector of $\angle{OPN}$. If $QR\parallel{OM}$, find $\angle{QRN}$.


Here is an angle-chasing problem which I am trying to solve. I couldn't make a good progress solving the problem. Though here is my approach:

If we assume $\angle{OPQ}=\theta$ and $NP$ intersects $QR$ at $A$, then we have: $$\angle{OPQ}=\angle{NPQ}=\theta,$$ $$\angle{ONP}=\frac{\pi}{2}-\theta,$$ $$\angle{NAQ}=2\theta,$$ $$\angle{PQR}=\theta.$$So, $\Delta{PAQ}$ is isosceles.

This doesn't help very much. How to solve the problem?

WLOG, we can consider the circle as the unit circle.
Preliminary result: $PN=\dfrac{\sqrt{5}}{2}$ by Pythagoras in $\Delta OPN$.
Using the bisector angle theorem, we have, for a certain $k$:
$$\dfrac{QN}{QO}=\dfrac{PN}{PO}=\sqrt{5} \implies OQ=\dfrac{QN}{\sqrt{5}}$$
As $OQ+QN=1$ we must have
$$OQ=\dfrac{1}{2 \Phi}$$
where $\Phi=\dfrac{1+\sqrt{5}}{2}$ is the Golden Ratio.
As a consequence, using Pythagoras in triangle $\Delta OQR$:
$$OQ^2+QR^2=OR^2 \ \implies QR = \sqrt{1-\dfrac{1}{4 \Phi^2}}=\dfrac{1}{2 \Phi}a$$
where $a:=\sqrt{4 \Phi^2-1}=\sqrt{2 \sqrt{5}+5}=\dfrac{\sqrt{5}}{\sqrt{-2 \sqrt{5}+5}}$
$$\tan \angle QRN =\dfrac{QN}{QR}=\dfrac{\sqrt{5}}{2 \Phi \dfrac{1}{2 \Phi}a}=\dfrac{\sqrt{5}}{a}=\sqrt{5-2 \sqrt{5}}=\tan \dfrac{\pi}{5}$$
(see here); therefore:
$$\color{red}{\angle QRN = \dfrac{\pi}{5}}$$
Remark: I wouldn't be surprized that a simpler solution is found for example through a connection with the pentagram.

Edit: In fact (see my remark) my intuition is verified. Thanks to @user3236 this issue is like proving the validity of Richmond's construction. Here is another site with complementary informations: https://www.cut-the-knot.org/pythagoras/RichmondPentagon.shtml).
