Angle trisection of $90^o$ Read on page #7 of article here, that angle of $90^o$ can be trisected. I went through this youtube video here, and here; and denote these two videos (denoting two separate methods) by (a), (b) respectively. These use elementary methods used in school days, but were never explained for the reason.
Anyway, I want to understand the logic behind both (a), (b) as it is a valid trisection.
In (a) there is an arc equal to the length of the radius, let $|r|$. Now, I hope that it is an approximation as there should be (as per me) no way to get the $\frac{2}{3}: \frac{1}{3}$ on the $90^o$ arc. It is only the chord joining the two ends (let, $yx$ with $y$ the coordinate on the $y$-axis, and $x$ the coordinate on the $x$-axis) that can be hoped to be approximately divided by the arc from point $y$ or $x$. Either end's arc will cut the hypotenuse $yx$ in two parts with ratios $\frac{1}{\sqrt{2}}= 0.707, 1-0.707=0.293$ in opposite sides from point $y,x$ respectively. So, if I am not wrong, it is an approximate approach and hence not mathematically or even geometrically valid.
Approach shown in (b) is even difficult to understand, and would request some help, as it is using a chord's division (named $DE$) in two parts (at midpoint $F$) by the angle bisector of $90^o$ to derive further angle bisections on both sides of the midpoint $F$. I am unable to comprehend even the mechanism in this approach, as the average of $45^o $ and $90^o$ is $67.5^o$, while the average of $0^o$ and $45^o$ is $22.5^o$.
Also, both of the above methods are faulty, then please give some link for correct way.

Edit Unable to understand, and hope that the approach (a) is valid, as a similar approach is used here, in Fig. 3.13. But, am unable to understand that too, but being a credible reference source, can infer (a) is at least a valid one.

Edit 2 I am posting an image of the reference in the 'Edit' made earlier, for a different, but linked question. It is given on the earlier page (#53) that on bisecting the smaller angle $\widehat {AON}$ obtained by bisecting, get the trisection. By this logic, the approach (b) is also valid. But, how can it be understood, is not clear, as double bisection to lead to $\frac{1}{4}$th of the original angle.


Edit 3 Regarding bisection leading to trisection, have found a suitable basic geometrical technique at: http://www.lamath.org/journal/Vol1/trisection_using_bisection.pdf, on pg. #3,4, by using the two unequal circles' intersection formed from the bisection of the original interval.
The article titled: 'Trisection using Bisection' describes also the nth-section of any interval using unmarked ruler and collapsing compass, in the last section.
 A: 
The line $AB$ is perpendicular to $BC$. The circles have, respectively, centers $B$ and $C$ and radius $BC$. The segments $BD$, $BC$ and $CD$ are equal by construction. Can you prove that the angle $\widehat{CBD}$ is twice the angle $\widehat{DBA}$?
A: First on method (a):

When you draw the first arc, we have $BD=BQ$ and then using the same radius, we have $DQ=BD=BQ$, hence $\triangle BQD$ is equilateral. Hence $\angle DBQ = 60^\circ$ and $\angle QBE = 90^\circ - \angle DBQ = 30^\circ$. By symmetry, $\angle DBP = 30^\circ$ too and $\angle PBQ = 30^\circ$.
For method $(b)$:
If I understand it correctly, it doesn't work.

Suppose all radius chosen is $1$, then $D$ and $E$ lies on the unit circle. By construction of $F$, $D, E, F$ lies on the line $x+y=1$. Then an arc is drawn from $F$ and another is drawn from $E$. These arcs has equation $(x-0.5)^2+(y-0.5)^2=1$  and $(x-1)^2+y^2=1$. If the claim is correct, let the intersection of the arc be $A$ then $\angle ABC$ should form $30^\circ$, however, if this is the case, point $A$ should lies on the straight line $y= \tan\left( \frac{\pi}6 \right)x$. So go to any graphic software and plot out those curve and we can find that it is not the case that the straight line (black line in my plot) passes through point $A$ ( The intersection of purple and orange arc).

A comment about the title of method $(b)$, if I am not mistaken, it has been proven previously that we can't  trisect an arbitrary angle. 
