Interesting generalisation about a parallel ray between 2 mirrors I have been self studying from home nowadays and came across a result in a video lecture on 'Ray Optics' 

If an incident ray is parallel to one of the 2 mirrors kept an angle $\theta$ = $\frac{\fracπ2}n$ where $n \in N$, then the ray will fall normally on one of the mirrors and retrace its path. In the end, there will be $2n - 1$ total reflections.

The teacher gives examples by taking $\theta$ = 2 and 3, and supports the statement. 
Statement and example diagrams
I ask for a general proof of the statement. I feel like there must be a geometrical insight that leads our incident ray to be normal to one of the mirrors. I have some experience with olympiads and watch a lot of 3Blue1Brown videos, so I feel like there must be something I am missing.
I tried by constructing a quadrant of a circle on the $x-y$ plane and imagining one of my mirrors to be the $x$ - axis and the other mirror is made when a light beam $\parallel$ to the $x$ - axis intersects the circle. Then, I tried using parametric coordinates to lead me to the final point where the ray becomes normal to one of the mirrors. But, I was not able to think any further and also not able to define a point on the circumference where $\theta$ = $\frac{\fracπ2}n$.
 A: For a simple 'geometrical' answer the following sketch may suffice. Red lines represent the first mirror, and its images. The blue lines represent the second mirror and its images.
I think the diagram is self-explanatory to confirm that there are $2n-1$ reflections, and that the 'middle' one is perpendicular to the mirror. But for a wordy explanation see below the diagram.

Because the initial angle is $\frac{\pi}{2n}$, there will be $4n$ mirrors (and mirror images) in the circle (one at each of $0$,     $\frac{\pi}{2n}$, $\frac{2\pi}{2n}$, ... ,$\frac{4n-1\pi}{2n}$ or $2n$ in the half-circle. So any input beam parallel to one of the mirrors will intersect with precisely $2n-1$ of the mirrors.
Note that there will be a mirror image at $\frac{n\pi}{2n} = \frac{\pi}{2}$, which is perpendicular to the first mirror and also perpendicular to the light beam. That is the image point (or reflection) where the beam will strike normal to the mirror (the $n^{\text{th}}$ reflection). From the symmetry of the diagram it is easy to see that after that halfway point, the beam/mirror intersections as the beam exits are at the same angles as those that occurred on the way to the $n^{\text{th}}$ intersection.     
The same diagram can also be used to show that a light beam that is not parallel to one of the mirrors has a strict maximum of $2n$ reflections before exiting the 'mirror-maze'.
A: One useful way to study this is to look at what happens to the angles of the wavevectors (directions of travel). Consider this setup and look at what happens to the wavevectors:

If you look carefully you see that after the first reflection the incoming wavevector is reflected along the red line. The second reflection reflects the wavevector along the blue line.
Let $\theta$ denote the angle of the incoming wavevector and $\theta', \theta''$ the angle after the first and second reflection. Like in this diagram:

Let $\theta_1$ and $\theta_2$ denote the angle of the red line and blue line respectively. For reflections we have the following formula
$$\theta'=2\theta_1-\theta\\\theta''=2\theta_2-\theta'$$
Substituting this we get 
$$\theta''=\theta+2(\theta_2-\theta_1)$$
and we see that this is a pure rotation. We have two cases: an even number of reflections ($n=2m$) or an odd number of reflections ($n=2m+1$). You can show that
\begin{align}\theta^{(2m)}&=\ \ \ \theta+2m(\theta_2-\theta_1)\\
\theta^{(2m+1)}&=-\theta-2m(\theta_2-\theta_1)+2\theta_1\end{align}
Let's focus on an even number for now. We want the starting angle to be $\theta_2$ (so it is parallel to the second mirror) and after an even number of reflections it has to be equal to $\theta_2\pm\pi$ so it is normal to the second mirror.  This gives the following equation
$$\theta_2+2m(\theta_2-\theta_1)=\theta_2\pm\pi/2$$
and solving it gives the requested condition $\theta_2-\theta_1=\frac{\pi}{2(2m)}=\frac{\pi}{2n}$. I'll leave the case of odd reflections for you.
