Equivalence of different ways of geometrical multiplication There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).
Let $A,B$ be two integer points on the line $O1$:

Method 1


*

*Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).





*Draw a circle with radius $|OB|$ around $B$. 

*Let $C$ be the (other) intersection point of this circle with the line $O1$.



*Draw a circle with radius $|OB|$ around $C$.

*Do this $a-1$ times.

*The last intersection point $C$ is the product $A \times B$.

Method 2


*

*Construct a rectangle with side lengths $|OA|$, $|OB|$.





*Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).





*Draw a circle with radius $|O1|$ around $0$. 

*Let $C$ be the intersection point of this circle with the line $O1$.



*Draw a circle with radius $|O1|$ around $C$.

*Do this $c$ times.

*The last intersection point $C$ is the product $A \times B$.

Method 3


*

*Construct the line perpendicular to $O1$ through $O$.

*Construct the points $1'$ and $B'$.



*Draw the line $1'A$.

*Construct the parallel to $1'A$ through $B'$.

*The intersection point of this parallel with the line $O1$ is the product $A \times B$.

Method 4


*

*Construct the perpendicular line to $O1$ through $O$.

*Construct the point $1'$.

*Construct the circle through $1'$, $A$ and $B$.

*The intersection point of this circle with the line $O1'$ is the product $A \times B$.

Method 5
This method makes use of the parabola, i.e. goes beyond compass-ruler constructions. 


*

*Construct the unit parabola $(x,y)$ with $y = x^2$.

*Construct $B'$.

*Construct the line perpendicular to $O1$ through $A$.

*Construct the line perpendicular to $O1$ through $B'$.

*Draw the line through the intersection points of these two lines with the parabola.

*The intersection point of this line with the line $O1'$ is the product $A \times B$.


For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point) $A \times B$.

Note that the different methods take different amounts $\sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):


*

*Method 1: $\sigma \sim ab^2$

*Method 2: $\sigma \sim  ab$

*Method 3: $\sigma \sim  ab^2$

*Method 4: $\sigma \sim  a^2b^2$

*Method 5: $\sigma \sim  a^3b$
This is space complexity. Compare this to time complexity, i.e. the number $\tau$ of essential construction steps that are needed:


*

*Method 1: $\tau \sim a$

*Method 2: $\tau \sim  ab$

*Method 3: $\tau \sim  1$

*Method 4: $\tau \sim 1$

*Method 5: $\tau \sim  1$
From this point of view method 3 would be the most efficient.

Once again: 

I'm looking for other geometrical methods to multiply two numbers
  given as points on the number line $O1$ (is there one using the
  hyperbola?) and trying to understand better the "deeper" reasons why
  they all yield the same result (i.e. point).


Those answers I managed to visualize I will add here:
Method 6 (due to Cia Pan)

Method 7 (due to celtschk)

Method 8 (due to Accumulation)

 A: *

*Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.

*Construct the perpendicular at $O$.

*Construct the semicircle on the diameter $A'B$.

*Find $H$ at the intersection of the semicircle and the perpendicular.
$(OH)^2 = OA'\cdot OB = OA\cdot OB$.

*Draw line $1H$ and construct a perpendicular to it through $H$.

*Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1\cdot OK,$ hence $OK = OA\cdot OB.$
A: The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).


*

*Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).

*Select on $g$ an arbitrary point $P$ other than the origin.

*Draw a line through $1$ and $P$.

*Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.

*Draw a line through $P$ and $B$.

*Draw a parallel to that line through $Q$. The intersection with the number line is then $A\times B$.


Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).
With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.
A: If you construct two similar triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ such that $X_1Y_1=1$, $Y_1Z_1 = A$, and $X_2Y_2 = B$, then $Y_2Z_2=A*B$. 
Also, if you take any angle, mark $1$ and $A$ on one side, mark $B$ on another, draw a line from the $A$ point to the $B$ point, then construct a line parallel through that line through the $1$ point, it will intersect the other side a distance $\frac B A$ from the vertex. And $A*B$ is of course equal to $A/(1/B)$.
