Recently my friend Tina, who writes the blog **Diotima’s Ladder**, asked me if I could help her with a diagram for her novel. (Apparently all the math posts I’ve written gave her ideas about my math and geometry skills!)

What she was looking for involved **Plato’s Divided Line**, an analogy from his runaway bestseller, the *Republic* (see her post *Plato’s Divided Line and Cave Allegory* for an explanation; I’m not going to go into it much here). The goal is a geometric diagram proving that the middle two segments (of four) must be equal in length.

This post explores and explains what I came up with.

[Update 5/24/2022] See my updated post, *Back to Plato’s Line* for a better proof!

Our starting point is what **Plato**, in the *Republic*, says to **Glaucon** about the line. He has introduced the idea of splitting the world into two classes of experience, the visible and the intelligible (also called opinion and knowledge). Plato then says:

Let us represent them as a divided line, partitioned into two unequal segments, one to denote the visual and the other the intelligible order. Then, using the same ratio as before, subdivide each of the segments. Let the relative length of these subdivisions serve as indicators of the relative clarity of perception all along the line.

A question arises from that last sentence. It can be easily demonstrated that the middle two segments must have the same length, but Plato’s further explanation of all four segments seems to imply an inequality, that each higher segment, representing greater clarity of thought, should be *larger* than the one below it.

A good starting point for something like this is throwing some numbers at the problem and seeing what happens. I started by trying some simple fractions:

In all cases, the left column is the initial division, as Plato said, *“into two unequal segments.”* I used one-third, one-fourth, and one-fifth, respectively, for the lower segment. The upper segment, then, is whatever remains (one minus the lower segment).

The middle column represents the subdivision *“using the same ratio as before”* applied to both the upper and lower segments. To get the actual values for the sub-segments we multiply, giving us the right-hand column. For example, one-third of one-third is one-ninth. Note that the fractions in the right column sum to one (as do the fractions in the first column).

What the examples show is that the middle segments are equal in all three cases.

Of course, these numbers might just happen to work out that way, so the next step is abstracting this to a general case. In the examples above, note that we start by picking some fraction, **1/ d**, for the lower segment (1/3, 1/4, 1/5). Then the upper segment is:

This might seem to exclude a fraction such as two-fifths (or three-eighths), but in fact:

So, our abstraction works for any fraction because, if necessary, we can normalize it to **1/ d**. Our abstraction is then:

And, again, the middle two segments are shown to be equal.

**§**

That seems conclusive, but if we’re skeptical, we can simplify by assigning length ** a** to the lower segment. Then the upper segment has length

**1-a**, and therefore:

And, once again, the middle segments are equal.

We can simplify it even more by using **1- a=b**. Then, using just

**and**

*a***, we have:**

*b*Which is about as simple as it gets. Bottom line, if a line is divided according to some ratio (** b/a**), and the divisions are sub-divided by that

*same*ratio, the middle two segments must always be equal in length.

**§**

Which is well and good, but totally algebraic. **Algebra**, as we learned it in school, doesn’t show up until about the 9th century, long after Plato. The Greeks did have some geometric algebra as well as facilities with numbers, so it’s not hard to believe the math above may have been perfectly well known to Plato.

But at that time, the Greeks were more about geometry, and, in any event, what’s needed for Tina’s book is a *diagram* — something geometric and visually appealing. Something sufficiently interesting for Plato scholars and not pure gibberish to dismay the mathematically (and especially geometrically) inclined.

The math above provides confidence that a geometry-based exploration must agree, if not hopefully somehow *prove*, that the middle segments must be equal. It turns out to be slightly tricker than it is with algebra. Similar to the example fractions above, we can show that any geometric example we make has equal middle segments but finding a proof they *must* be equal using only geometry is a bit harder.

**§ §**

A key limit in how we approach the geometry is that our only tools are the ability to draw a straight line (using a straight edge) and the ability to draw a circle of any radius (using a compass). There are two important tricks I want to cover before I proceed.

**The first trick:** Given any straight line, draw a *perpendicular* line (a line at right angles) intersecting the first line at some point, **P**.

*Figure 1* illustrates this trick.

The black horizonal line is our initial straight line. We wish to create a perpendicular line (blue) that intersects the first line at point **P**.

**Step 1:** Use a compass to draw a circle of some radius, ** r1**, with point

**P**as the center point of the circle. This creates two new points,

**A**and

**B**, where the circle intersects the horizontal line.

**Step 2:** Use a compass to draw two new circles, one with point **A** as its center, the other with point **B**. Both circles must have the *same* radius, ** r2**. The circles intersect at two points,

**C**and

**D**.

**Step 3:** Use a straight edge to draw a line from point **C** to point **D**. That line passes through point **P** and is perpendicular to the horizontal line.

**Note:** The radius of the circles isn’t important, but larger circles make for more accuracy. It is necessary that ** r1<r2**, usually by a fair amount (in

*Figure 1*

**is a bit more than twice**

*r2***). The crucial thing is that the two circles drawn from points**

*r1***A**and

**B**

*must have the same radius*.

**§**

**The second trick:** Given two lines, prove they are (or are *not*) parallel. Alternately, given some straight line, construct a second line parallel to it at some given distance.

*Figure 2* illustrates.

The two black lines slanting down to the right may, or may not, be parallel. We want to find out which.

**Step 1:** Using a straight edge, draw a line (blue) that intersects both black lines. It intersects them at point **P** and point **A**.

**Step 2:** Using a compass and point **P** as the center, draw an arc that intersects both black lines. Set the radius, ** r**, of the compass such that it is the distance from point

**P**to point

**A**along the blue line. The arc intersects the lower black line at point

**B**.

**Step 3:** Using the compass set at the same radius, ** r**, draw two new arcs using point

**A**for one and point

**B**for the other. The arcs will intersect at some point

**C**. If point

**C**lies on the upper black line, then the black lines are parallel. If it lies off the line, they are not.

**Note:** What we’re doing is creating a *parallelogram*. If we succeed, the original lines are parallel (and so are both blue lines). We can use the same technique to create the upper parallel line (an exercise left to the reader).

**§**

This is the part of geometry that concerns itself with what can be done using only those two tools, the straight edge and the compass. Note that a straight edge is not a ruler — this type of geometry doesn’t allow measuring lines or angles other than proportionally.

It’s a fun kind of geometry, often taught early in school, and it’s surprising how much is possible using only those two simple tools.

**§ §**

Knowing we’re on firm ground mathematically, let’s see what we can do geometrically.

**Step 1:** We again start as simple as possible with a straight vertical line *“partitioned into two unequal segments”* as shown in *Figure 3*.

We label the bottom point ** a**, the top point

**. We pick an arbitrary point**

*b***to divide the line unequally. (In fact, everything we do here works fine in the degenerate case of equal segments.)**

*c*We label the *lengths* of the bottom segment ** X** and the top segment

**. This gives us a ratio**

*Y***for the two segments. The total length of the line is**

*Y:X***.**

*X*+*Y*Note that a ratio, **Y:X**, is the same thing as the fraction, **Y/X**. Here we always put the larger upper length above (or first, in a ratio) and the smaller lower length below (or second) to reflect the idea behind Plato’s vertical line.

**Step 2:** Using the trick described above for creating perpendicular lines, draw three straight horizontal lines perpendicular to points ** a**,

**, and**

*b***, as shown in**

*c**Figure 4*.

We also pick an *arbitrary* point ** d** on the lower line and draw a straight vertical line perpendicular to it. The new vertical line intersects the middle horizontal line at point

**.**

*e*(We care only about the part of those lines to the right (so that’s all I’ve shown). The perpendicular trick obviously results in some extension to the left and, in the case of the new vertical line, below. For clarity, I’m not showing the arcs of the trick or the extra bits beyond the parts we need.)

Note that the distance ** ad** (that is, from point

**to point**

*a***) can be anything, but for precision should be as long as convenient and comfortable.**

*d***Step 3:** Draw a straight (slanted) line connecting point ** b** to point

**, as shown in red in**

*d**Figure 5*. This line intersects the horizontal line

**at point**

*ce***.**

*g*Draw a new vertical straight line perpendicular to line ** ce** such that it intersects point

**. The new vertical line intersects the lower horizontal line**

*g***at point**

*ad***.**

*f*This subdivides the horizontal by same ratio (but not necessarily the same *lengths*) as the vertical. This gives us the following equal ratios:

The horizontal ratio matches the vertical one because point ** g** projects down to point

**. Think of the slanted red line**

*f***as a mirror reflecting the vertical line**

*bd*

*ab**proportionally*onto the horizontal line

**. (Because the “mirror” is straight, the reflection can shrink or expand but**

*ad**must remain proportional*.)

(If we happened to use the same length ** ad** as we did for

**(by using a compass to make them equal), then**

*ab***and**

*ac*=*df***. The diagram would be square, and the “mirror” would have a 45° angle, making the reflection necessarily the same size.)**

*bc*=*af***Step 4:** Now we subdivide the vertical line segments and complete Plato’s Divided Line.

Draw a straight slanted line from point ** b** to point

**and another from point**

*e***to point**

*c***(shown in red in**

*d**Figure 6*). The upper one intersects the middle vertical line at point

**, the lower one at point**

*k***.**

*h*Draw two new straight horizontal lines perpendicular to the vertical lines, one to intersect point ** k** and the other to intersect point

**. These intersect the original vertical line (**

*h***) at points**

*ab***and**

*m***.**

*n*The new diagonals, line ** be** and line

**, create two new “mirrors” that each reflect the entire horizonal line**

*cd***onto, respectively, the upper and lower segments of line**

*ad***. Since we know that line**

*ab***has the same ratio**

*ad***as the original vertical line (**

*af:df***), we also know the reflections must be proportional.**

*bc:ac*Therefore, we have subdivided the segments according to the original ratio as specified by Plato. We label the segments, bottom to top, as ** A**,

**,**

*B***, and**

*C***.**

*D*Now all we have to do is prove that ** B=C**.

**§ §**

My first attempt to find a proof didn’t give me the strong result I wanted. It’s not difficult to *show* **B=C** in drawings with different sizes and ratios but *proving* it requires an abstraction as it did for the math above. I’ll get to that proof, but first a small detour…

Below is the *second* candidate diagram I came up for Tina’s novel (see below for the first candidate):

The goal is showing that the triangles **P**, **Q**, and **R**, are equal to each other. In particular, to show that **P=R**, because that means ** t=u** (in this early version I labeled the four segments

**,**

*s***,**

*t***, and**

*u***, rather than**

*v***,**

*A***,**

*B***, and**

*C***).**

*D*It’s easy to use a compass to show that the longer lines (the horizonal baselines or the diagonal hypotenuses) are the same length, but we could have done that with the **B** and **C** (or ** t** and

**) segments. There is some value in having longer lines to minimize error, but nothing here**

*u**requires*the equality.

And I know from drawing this diagram many, many times that precision is a problem. If I wasn’t very careful — *especially* when not using graph paper — I got a diagram in which they are *not* equal.

There is some logic we can apply, though. *Figure 7* continues the diagram we began above (and may make the construction of the triangles more apparent than in the prototype above).

**Step 5:** Extend the horizontal line ** mh** to the right such that

**=**

*hp***. A compass set to the distance**

*mh***allows determining the distance**

*mh***.**

*hp***Step 6:** Draw a (new) straight line from point ** n** through point

**to the extended line**

*g***. We expect it to intersect at point**

*mhp***. (In any carefully drawn diagram, it will.)**

*p*Note that the line ** ngp** forms another “mirror” that reflects the horizontal line

**onto the vertical line**

*mhp***. Such a mirror always reflects the same proportion (although it may shrink or expand the whole), so**

*mcn***=**

*mh:hp***.**

*mc:cn*Since we explicitly set ** mh**=

**, therefore**

*hp***=**

*mc***(or**

*cn***B**=

**C**).

On the other hand, if our attempt to construct the triangles fails, if the line from point ** p** to point

**does not intersect point**

*m***, or if a line from point**

*g***through point**

*m***does not intersect at point**

*g***, then**

*p***B≠C**.

We also note that only in the case the segments are equal are the diagonal lines parallel to each other. This gives us one more way to demonstrate **B**=**C**. If we create the three triangles regardless of the lengths (that is, we create the lines ** mhp** and

**regardless of anything else — possibly ending up with the left or right example in**

*ngp**Figure 8*), if we can show the diagonals to be parallel, then

**P**=

**Q**=

**R**and

**B**=

**C**.

Every carefully drawn diagram will demonstrate this just as every fraction we chose at our starting point does. We have a proof by *example*, but this is akin to always finding white swans. We can’t be sure that the next diagram (or fraction) won’t be a black swan.

We require a geometric abstraction just as we required an algebraic one.

**§ §**

Back from our detour, *Figure 9* picks up where *Figure 6* (from **Step 4**) above left off.

Note that *Figure 9* shows the **Y** and **X** lengths and assigns lengths **U** and **V** to the subdivided horizonal line ** ad**. We know from

**Step 3**above that

**Y:X**=

**V:U**(or, as fractions, that

**Y/X**=

**V/U**).

We start by focusing on the larger upper triangle formed by points * c*,

*, and*

**b***, and the smaller lower triangle formed by points*

**e****,**

*a***, and**

*c***.**

*d*Since they share the same base length (**U**+**V**) and have heights according to our ratio **Y:X** (literally, the height of the upper triangle is **Y** and of the lower one **X**), the area of the triangles has the same ratio:

We can factor out the (**U**+**V**)/**2**, which leaves **Y** and **X**, our ratio.

Secondly, more importantly, note that the (red) diagonal lines (** be** and

**) have a**

*cd**slope*, which is their vertical distance divided by their horizontal distance (or height divided by their width). We call these

**S**and

*y***S**, respectively:

*x*We can factor out the **1**/(**U**+**V**), again leaving **Y** and **X**, so these slopes also have the same ratio, **S y:Sx**, as our line segments (

**Y:X**).

Now we’re ready for…

**New Step 5:** Reflect the diagonal line ** cd** by drawing a straight line from point

**to a new point**

*c***that we create using a compass set to the distance**

*r***(so that**

*ed***). Because this line has the same slope as**

*ed*=*er***, but in the opposite direction, we label the slope of the new line**

*cd***-S**. (As shown in

*x**Figure 9*.)

Note that this line intersects point ** k**, and we have another demonstration that

**B=C**, because, if

**=**

*ed***, then**

*er***=**

*gh***, because the triangle**

*gk***is an**

*crd**isosceles*triangle.

What’s more important is the new triangle formed by ** cre**, which by construction is identical to the original triangle

**. We know the triangles**

*acd***and**

*cbe***have the**

*acd***Y:X**ratio to each other, so the triangles

**and**

*cbe***must also have that ratio.**

*cre*Our advantage now is that their bases overlap and their heights (which have the ratio **Y:X**) start at the same baseline and extend upwards. The proof we seek is now in our hands.

**§**

*Figure 10* focuses on these two triangles, ** cbe** and

**. Crucially, we have the respective slopes**

*cre***S**and

*y***S**

*x**and these intersect each other*.

**New Step 6:** The slopes intersect at point ** k**, which obviously means they have equal heights above the baseline at that point. The “mirrors” formed by

**S**and

*y***S**project point

*x***k**down onto point

**g**, which we previously gave the lengths

**V**and

**U**.

We can think of line slope as akin to (constant) velocity. If the “slope” is 60 MPH, it means we’d go 60 miles in *one* hour, but it also means we’d go 30 miles in *one-half* hour (or 15 miles in *one-quarter* hour). We just multiply the “slope” by the time interval.

Likewise, we can determine the *height* of any point along the line slope by multiplying the slope by the length (along the baseline) at that point. Specifically, we can determine the height of point ** k** on the slope

**S**by:

*x*And the height of point ** k** on slope

**S**(starting from point

*y***and heading**

*e**left*because we must start from zero) by:

As we did with the slopes above, we can factor out the **1**/(**U**+**V**), leaving us with:

We know from above that **V/U**=**Y/X** and that leads to the following identity:

Because we multiply both **Y/X** and **V/U** by **X×U**, which cancels out the lower part of the fraction leaving us with the crucial identity **Y×U=X×V**.

Crucial because we just established that the height of point * k* is, in one case

**X×V**and in the other

**Y×V**. Since those equal each other, we know that point

*is a*

**k***necessary height equality*between the two slopes where they intersect.

And since triangle ** cre** is a mirror version of triangle

**, that means the line segments**

*acd***and**

*gh*

*gk**must*be equal. Or in other very important words:

**B=C**.

Geometrically speaking.

**§**

This makes perfect sense. Given the proportional ratio between the slopes, it seems almost obvious that the rise of slope **X** over distance **V** *must be the same* as the rise of slope **Y** over distance **U**.

Put in terms of velocity, if we go half as fast for twice the time, we cover the same distance as if we went twice as fast for half the time. Concretely, if we go 30 MPH for one hour, we go 30 miles, and if we go 60 MPH for one-half hour, we also go 30 miles.

We no longer depend on drawing precise diagrams, numeric examples, or even abstract algebra, but on simple logic. In Plato’s Divided Line, the middle segments, **B** and **C**, have to be equal.

**§ §**

I set a limit of 4000 words for this post, and since I’m under 3500 at this point, I’ll include a few extras. I mentioned above there was an earlier candidate diagram I came up with for Tina’s novel. Here is one version of it:

It illustrates a given ratio (in this case **Y:X=2:1**) and includes some bits of math illustrating the necessary algebraic equality of the middle segments. It doesn’t prove anything, and I couldn’t come up with a good way to show a proof.

The bold vertical lines are intended as segments (**B:A** or **D:C**) and combining appropriate pairs (indicated by the blue rectangles joining pairs) would give the entire Divided Line. The red shading demonstrates that the **B** segment (upper part of a left member of the pair) equals the **C** segment (lower part of the right member). Since the slopes of the connecting lines are all parallel, we see that **B=C** in all cases.

There might be a way to demonstrate geometric necessity, but if so, it eluded me. In any event, I sense that any proof would require a fair bit of algebra, and what I came up with above seems much simpler and clearer.

(I think Fred Brooks, in *The Mythical Man-Month* (1975), was mostly right when he wrote that designers should be prepared to throw away their first design because it’s likely to be wrong anyway. I’ve found over and over that one needs to chew on a problem for a while to really see the shape of it.)

**§**

For fun, here’s a version of the proof done using only a compass and straight edge (no graph paper). This is how Plato or Glaucon might have done it.

As you see, the compass work makes it a little messy. (What you *don’t* see is all the attempts that weren’t precise enough that lines didn’t intersect correctly. I generated *a lot* of scrap paper on this project!)

**§**

Lastly, we look at the diagram as a sort of mechanical framework we can move around for different subdividing ratios (*see Figure 11*).

Imagine that the lines can telescope to fit appropriately as we move the mechanism around. Also imagine that the green circles are joints that bind the lines but allow them to slide through them. The two green squares (upper left and lower right) are fixed points that only allow pivoting.

Note the shaded red “kite” shape in the lower half. It appears to be symmetrical about its horizontal axis (and, as we’ve shown above, it truly is).

We can consider what happens when we resize the mechanism or move the middle horizontal line up or down to change the ratio between the top and bottom.

The green circle on the blue diagonal line fixes the intersection with the middle vertical and horizontal lines, so moving the horizontal line down moves the vertical one right and vice versa. Because all our lines are straight, the movement is linear.

As those two lines move, the red diagonal lines also move in sync. Here’s what different configurations of the mechanism look like:

The symmetry remains constant, and this gives us a mechanical sense of how and why **B=C**.

We can take it a step further. The top row of *Figure 12* shows the degenerate case where the top and bottom segments are (contrary to Plato) equal in length. In this case it’s obvious that **B=C** because **A=B=C=D** *by construction*.

Not shown (because it would be boring) is the other degenerate case where the lower segment collapses to zero length, giving us **A=B=C=0** (and **D**=*the whole line*). Again, by construction, **B=C**.

Since the mechanism moves in linear fashion (no curves), and the degenerate end cases have **B=C**, we are further assured that **B=C** for all intermediate cases.

**§ §**

And on that note, we’re done. *Quod Erat Demonstrandum*!

Stay geometric, my friends! Go forth and spread beauty and light.

∇

February 28th, 2022 at 9:13 am

Appendix IAs a reference point for what Plato’s Divided Line is all about, here is a table I made that tried to summarize the content I’d cribbed from other examples:

On the edges are a bit of example math (left) and the abstract algebra (right) that, respectively, demonstrate and prove the equality.

February 28th, 2022 at 9:13 am

Appendix IIIf you want to explore the mathematical side, here is how to derive the necessary numeric values for each of the four segments:

Given any desired ratio:

Calculate:

And that provides the values for any

b/a. (The middle two, of course, are equal.)For example, given

5/3…:QED!😉February 28th, 2022 at 9:35 am

Note that the final fractions sum to one (

64/64).February 28th, 2022 at 9:13 am

Appendix IIIThe debate over Plato’s Divided Line is from the perceived tension between the easy-to-show equality of the middle segments and Plato’s discussion of what each segment means. (As the table in Appendix I above shows, as Plato explores them, these segments are seen from multiple perspectives and are rich in meaning.) The question arises: Did Plato know the middle segments were equal? If so, why does he imply each level is greater than the one below it? Was he in error? Was he kidding? Was he creating a deliberate paradox to foment thought?

Note, however, that Plato never explicitly says each segment is larger than the one below. That (as I understand it) is only

impliedby his statements (although, again as I understand it, the implication is quite strong, hence the debate).My take, for whatever it’s worth, is that, of course Plato knew the segments were equal. I don’t see how he would have gotten that wrong, the Greeks were plenty facile with numbers and geometry. Therefore, I think he was either teasing or, more likely, creating a deliberate paradox for followers to chew on.

My (very uninformed) guess is that crossing the major line, the first division of unequal parts, from Opinion to Knowledge, causes a bit of a “reset” that puts one in new and unfamiliar territory — the realm of nontangible thought and knowledge. And while abstractions, understanding, and critical thinking, may morally and intellectually be closer to “the Good”, larger in spirit, that they are a new beginning reduces them in some sense creating the equality between our experience of the tangible world and this new experience of abstraction.

I tried to echo this in the post, first moving from tangible example fractions to an abstraction that collects their meaning, and then moving from example geometry to an abstraction that doesn’t require examples.

February 28th, 2022 at 10:07 am

Up until last night, all the examples of the Divided Line I’d seen made the upper segments larger than the lower with the implication that moving towards the Good was greater. Tina sent me a link to a video where the lecturer had the

lowersegments larger (the middle ones still being equal, of course). He claimed doing it the other way was an error. For him, segment D, at the top, was the smallest. (Perhaps because it’s the rarest?)(I’ll have to watch that bit again to see exactly why. It was something about the proportions. At the time I was more interested in his geometry, which presented a different approach to subdividing the major segments. I plan to show that in an Appendix below once I create and scan a diagram.)

February 28th, 2022 at 9:14 am

Appendix IVI mentioned above that Tina sent me a link to a presentation that divides the line using a different, and rather interesting, technique.

Step 1:Begin, as before, with a vertical line intended to divide and sub-divide. In this case, however, use a compass to draw a large circle (with center point) that covers most of the line.pThe two points where the circle intersects the line become points

anda, the bottom and top, respectively.bStep 2:Pick an arbitrary pointto divide the line abc“into two unequal segments.”Draw a horizontal line perpendicular to pointthat intersects the circle (on the left side) at pointc.dStep 3:Draw straight lines from pointsandato pointb. This creates triangledinscribed inside the semi-circle.adbNote that, because we inscribed the triangle inside a half-circle, by Thales’s theorem, this is necessarily a right-triangle. The angle at point

is 90°.dStep 4:Draw two lines (red in figure below) from pointto each side of the triangle such that they are perpendicular to the side they intersect.cThe intersection points are

(lower) ande(upper).fStep 5:Draw two horizontal lines, one from pointand one from pointe, such that they are perpendicular to the vertical line.fThese lines intersect the vertical line at points

andg, respectively.hAnd that’s it. We’ve divided and sub-divided Plato’s Line per his instructions.

Why does this work?We have a large trianglethat we know to be a right-triangle. Inside the large triangle, we created two smaller trianglesadbandcfb. By construction, these are also right-triangles,aecand each shares an angle with the larger triangle. They also both contain a 90° angle, which means they necessarily share the third angle. Thus, they are equivalent triangles — the same triangle but scaled.We also know the hypotenuse (long side) of the main triangle is

. Our ratio isac+cb, so the upper (red) triangle has the same ratio to the large triangle as the upper segmentac/cbhas to the whole line. Likewise, the lower (red) triangle has the same ratio to the larger triangle as does the lower segment to the whole line.cbIf point

divides the line by some ratiod, then pointsac/cbandenecessarily also divide their respective segments by that ratio.fIn this case, rather than reflecting in a “mirror” we’ve scaled a triangle proportionally, but it’s the same basic idea. Start with the main division and then find a way to apply that to each segment formed by that division.

It’s a clever trick. It does rely on Thales’s theorem, and that theorem does predate Plato as a theorem. Euclid proved it (in

Elements) around 300 B.C. (after Plato), but the theorem, and perhaps even a proof, would have been known to him.What this technique doesn’t appear to do is prove the necessity of the equality

B=C. It’s possible one exists, but nothing immediately springs out at me. The speaker in the video used algebra to prove the equality and used a compass todemonstratethe equality in the diagram but didn’t present a geometric proof.February 28th, 2022 at 6:53 pm

The technique above requires drawing a line from some point

such that it is perpendicular to a given line. The trick for doing that is similar to drawing a perpendicular line to some pointpona given line (described in the post).Step 1:Using the pointas the center point, draw an arc that intersects the given line at pointspanda.bStep 2:Using pointsandaas center points, draw two arcs that intersect on the opposite side of the line from pointb.pStep 3:Draw a line from pointto the intersection of the arcs. The new (blue) line will be perpendicular to the original (black) line.pFebruary 28th, 2022 at 9:14 am

Appendix VHere is an algebraic proof of

B=C.Step 1:Start with the four sub-segments (bottom to top),A,B,C&D. We also have the two original segments,X=A+BandY=C+D, and that is our ratioY:X(also known as the fractionY/X).Thus, we have as axioms:

Step 2:From the above, we can create two important equalities. The first is:Because we multiply both sides by

Cand then byA, which cancels out the lower part of the fractions leaving the equality.The second equality is similar:

Because we multiply both sides by

A+Band then byC, which again cancels out the lower part of the fractions leaving the equality.Step 3:The proof begins with the second equality above:Step 4:A bit of math gives us:Because, on the left, we multiply

Dby(A+B).Step 5:Using the first of the two equalities in Step 2, we substitute D×A for C×B:Step 6:A bit of math gives us:Because, on the left, we factor out the common

B.Step 7:Finally, we see that:Because we can divide both sides by

1/(C+D), which leaves justBandC.QEDFebruary 28th, 2022 at 12:14 pm

Ugh. Looks like you had fun, in a masochistic kinda way.

I kept reading those =! as Equal-NOT.

February 28th, 2022 at 12:19 pm

Ha, yeah, or worse, the “factorial of equals” (whatever TF

thatis)!For me, exploring new territory like that (math, geometry, coding, a new language) is just fun. Effortful fun, to be sure, but it’s like runners who love running or those who love working out or those who love the effort of building something. There’s a joyful aspect to the progress and growth that more than compensates for the effort.

March 1st, 2022 at 4:46 pm

I’m lucky to have found someone who finds this sort of thing rewarding, and you’ve done a fantastic job!

March 1st, 2022 at 7:12 pm

Well, thank you, I’m glad you approve! I had fun; the old dog enjoys chewing a new bone once in a while! 🐶🦴😋

March 1st, 2022 at 8:23 pm

Reblogged this on Diotima's Ladder and commented:

Fantastic work on the mathematical features of Plato’s Divided line over at Logos Con Carne——using ancient methods!

March 1st, 2022 at 8:23 pm

I hope you don’t mind that I’ve reblogged this chez moi. I want as many people as possible to see it! 🙂

March 1st, 2022 at 8:50 pm

No problem! Spread the “wealth”! 😁

March 12th, 2022 at 12:31 am

I keep saying synchronicity rules my life. Here it is again. In this post, I casually mention

The Mythical Man Monthby Fred Brooks, and the next video byComputerphileis:I rest my case.

April 19th, 2022 at 7:09 am

[…] Last February I posted about how my friend Tina, who writes the Diotima’s Ladder blog, asked for some help with a set of diagrams for her novel. The intent was to illustrate an aspect of Plato’s Divided Line — an analogy about knowledge from his worldwide hit, the Republic. Specifically, to demonstrate that the middle two (of four) segments always have equal lengths. […]

August 7th, 2022 at 9:41 pm

[…] Here’s a geometric proof of the equality of the two middle segments of the line. And here’s a video with a different explanation of the same thing. If you’re into that. […]