TL;DR

Here is an exercise for fun. It is a variation of a problem first worked out by Descartes. It is to show the length of the vertical segment is the square root of the product of the lengths of the horizontal segments.

Right triangles ADC and CDB are similar. Therefore
##\dfrac{b}{h}=\dfrac{h}{a}\implies h^2=ab\implies h=\sqrt{ab}.##
I am pretty sure it was proven this way in my geometry class.

On edit.
Segment ##h## is the geometric mean of the segments into which it splits the diameter. Now I remember the bottom line of my geometry lesson!

My question is, how did Descartes do it and what exactly did he do?

##r = 2R \cos \theta = 2\frac{b+a}{2} \cos \theta = (b+a) \cos \theta ##

the distance from the origin is ##\sqrt{h^2 + b^2}## and ##\cos \theta = \frac{b}{\sqrt{h^2 + b^2}}##

##\sqrt{h^2 + b^2} = \frac{(b+a)b}{\sqrt{h^2 + b^2}}##

##h^2 + b^2 = b^2 + ab##

##h^2 = ab##

##h = \sqrt{ab}##

Statement of the question
You are given an actual drawn segment that we will call ##b##. How would you draw another actual segment ##h## that is the square root of the given segment ##b##?

Preliminary analysis
The statement of the question is insufficient to answer the question and more information is needed. A person trained in physics will realize soon that drawing two line segments next to each other, one of which is purportedly the square root of the other, is like comparing apples and oranges - both are fruit and that's that. Even though both are line segments, their units are [L] and [L^1/2].

However, note that any two segments can be in a square root relationship. All one has to do is find and draw a unit length segment ##a## that validates the relationship. Validation requires that one of the segments be the geometric mean of the unit segment ##a## and the other segment. Trivially, when the two segments are equal the unit length and the two segments are all the same. When the two segments are unequal, it doesn't matter which of the given unequal segments one calls the "geometric mean" ##h## and which the "other segment" ##b.## The key expression to be satisfied is $$\left(\frac{b}{h}\right)\times\left(\frac{a}{h}\right)=1.$$ Construction

Proof
The proof, using similar triangles, is shown in post #6 and will not be repeated here.

Illustrative example
Given two straight line segments, ##b = 1.96~##pu^* and ##h=0.815~##pu, draw the unit length segment ##a## in pu such that one segment is the square root of the other.

If you choose to use a ruler with units to draw ##b## (as opposed to a straight edge without units), you might as well draw the square root segment directly in those units.

Once you draw the figure in post #1, you have defined the units ##a## in which ##h## is the square root of ##b##. Furthermore, the figure can be drawn if you are given (on a piece of paper) any two of the three segments involved and all you have is a compass and straightedge with no markings.

The bottom line is that there three segments representing (1) a number, (2) the square root of this number and (3) a unit segment. If you are given segments representing any two of the three, you can construct the third. I note that (3), the unit segment, is necessary to make the conversion from "segment representation" to numerical representation.

This point is made clear if similar triangles are used in the Gemini problem. We have the basic relation $$\frac{AB}{DB}=\frac{DB}{BC}\implies AB\times BC=(DB)^2. $$ With ##AB=5BC##, $$5 BC^2=(DB)^2\implies \frac{DB}{BC}=\sqrt{5}.$$ This last expression says that "segment ##DB## expressed in terms of unit segment ##BC## is the dimensionless quantity ##\sqrt{5}.##"

Considering the obfuscatory numero-geometric solution proposed by Google Gemini, I would also add (sorry for gloating) that it also illustrates the superiority of human intelligence over the artificial variety.

I don't think so. In what units would you draw "1" in this construction? Angstroms, centimeters, inches, feet, furlongs, light years, smoots? If you draw it as a fraction of ##b##, how would you choose that fraction?

If you choose to use a ruler with units to draw ##b## (as opposed to a straight edge without units), you might as well draw the square root segment directly in those units.

Once you draw the figure in post #1, you have defined the units ##a## in which ##h## is the square root of ##b##. Furthermore, the figure can be drawn if you are given (on a piece of paper) any two of the three segments involved and all you have is a compass and straightedge with no markings.

The goal is to construct the square root geometrically, not calculate it or measure it.

EDIT: But I agree if one just draws an arbitrary segment ##b## one has a problem and must define what 1 is. We can always draw ##a## and define it as 1 or not with respect to ##b## and the relationship holds.

It looks like you missed the importance of defining the unit segment in post #1 and treated ##a## as if it were a number,

Note that Descartes is careful to define the unit segment before he proceeds to do anything else.
He begins the paragraph you marked with "For example, let AB be taken as unity ..." Two paragraphs down, he says "If the square root GH is desired, I add, along the same straight line, FG equal to unity ..." (emphasis mine.)

It looks like you missed the importance of defining the unit segment in post #1 and treated ##a## as if it were a number,

Remember that ##a##, as a line segment, is always unity, as per Descartes's construction, no matter how long you draw it in the diagram relative to ##b##. The special case is all three segments ##a##, ##b## and ##h## are equal.

Can you post what John Wallis did in this context? I am not familiar with that.

Then you say to yourself, "Wait a minute, I know that 2.54 cm is 1 inch. What if I did the construction in inches? You add another inch to ##b## and get a circle of diameter 2 in. which is the construction in red. Note that the square root of ##b## labeled "h (in)" is (surprise, surprise!) equal to 1 in.

So you see, being given just a segment on a piece of paper is not enough to produce a second segment that is equal to the square root of the first because you get a different construction depending on the units you choose. Descartes teaches us that the length of the segment ##a## that we must add to the given segment ##b## doesn't matter as long as we understand that it is a unit segment.

In other words, Descartes says:
Given ##b##, add a segment having length ##a## of your choice to it, and proceed with the construction to find line segment ##h.## If you fashion a ruler with spacings equal to your chosen ##a## and measure ##b## and ##h## with it, you will find that ##h=\sqrt{b}.##

$$(BP)^2+(AB)^2=(AP)^2$$ $$(BP)^2+(BC)^2=(PC)^2$$ $$(AP)^2+(PC)^2=(AC)^2$$

Note that triangle APC is always a right triangle.

In Wallis's expression, let ##BP=h~;~~AB=b~;~~BC=a~## to obtain ##h=\sqrt{ab}~## which has been shown in many different ways here.

Thanks, but how is that different from what we have been talking about?

In Wallis's expression, let ##BP=h~;~~AB=b~;~~BC=a~## to obtain ##h=\sqrt{ab}~## which has been shown in many different ways here.