Back to Plato’s Line

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.

The diagrams I ended up with outlined a process that works, but I was never entirely happy with the last steps. They depended on using a compass to repeat a length as well as on two points lining up — concrete requirements that depend on drawing accuracy.

Last week I had a lightbulb moment and realized I didn’t need them. Lurking right in front of my eyes is a solid proof that’s simple, clear, and fully abstract.

And while all three of those are desirable traits in almost anything, that last one was my potion’s fly. A geometric proof shouldn’t depend on duplicating a length with a compass. If anything, a crudely drawn geometry should suffice to illustrate the purely abstract geometrical reasoning behind the proof. The accuracy of the drawing shouldn’t matter at all. The proof should be entirely Platonic, so to speak.

It turns out that, after completing only the first four steps detailed in the February post, the proof is sitting right there. It’s almost embarrassing I didn’t see it. (Though I have found that sometimes it takes a while for the simple to reveal itself.)

I’ll recap the four steps again here, but I’ll assume you’ve read the earlier post (and also Tina’s, Plato’s Divided Line and Cave Allegory, which goes more into what the Divided Line means; I’m just about the geometry). The end point turns out to be roughly the same, but the new way of showing it is much shorter, clearer, and simpler.

§ §

Figure 1.

Step 1: We start with a vertical line that we’ll divide (and sub-divide again) per Plato’s instruction, “Let us represent them as a divided line, partitioned into two unequal segments…”

The “Divided Line” extends from point a to point b. We divide it by some arbitrary ratio at point c and use the perpendicular line technique to draw a horizontal line extending to the right to some comfortable length.

We label the lower segment (the line ac) as X and the upper segment (the line cb) as Y. This gives us the ratio Y:X (or Y/X) that we’ll use to further sub-divide the segments.

We can put point c anywhere along the line ab, the geometry works regardless. Plato specified unequal segments, so point c should be off-center, and most authors put the smaller segment below and the larger one above.

Keep your eye on that Y:X ratio. Whatever its value happens to be, as a ratio, it’s a significant part of the process and the proof here. What’s important about ratios is that you can have something like:


Where the variables can have various values, but each fraction evaluates to the same value (the ratio). For example:


In which case the ratio is 0.5.


Figure 2.

Step 2: Draw horizontal lines to the right from points a and b to match the line from point c. Then draw a vertical line on the right to create a rectangular box.

Note that the box can be taller than wide, or wider than tall, or exactly square with all sides equal. It makes no difference to our geometry.

We do assume the vertical lines are perfectly vertical and the horizontal lines perfectly horizontal. Or, actually, more critically, that the lines are straight and meet at 90° angles.

There is, in fact, no requirement the diagram be aligned at any orientation, just that the lines be true — straight and perpendicular where they meet. (We don’t care if the picture hangs crooked on the wall, but we do care that the lines in the picture are right.)

The compass-and-straight-edge technique illustrated in the earlier post shows how to create perpendicular and parallel lines, but we could draw these diagrams freehand because what’s important is the geometric logic involved.

We need only imagine a perfectly drawn box for that logic to be apparent. None of it depends on the accuracy of the drawing.


Figure 3.

Step 3: Draw a diagonal line (red line in Figure 3) from point b to point d. This line intersects the line ce at point g. Draw a vertical line perpendicular to ce at that intersection.

We can think of the red line as a “mirror” that reflects the left “wall” down onto the “floor”. In particular, we’re reflecting the line cg down as the line gf.

This new line divides the floor by the same ratio (Y:X) as the wall. We’ll label the floor segments U and V. They are “reflections” of the wall segments X and Y, respectively.

Since the floor is (in this case) shorter than the wall, U and V, are both shorter than their respective X and Y segments. This is an artifact of how wide the box is. If the box were wider than tall, the U and V segments would be longer than their respective X and Y segments. This box is taller than wide, so floor segments are compressed. (If the box was square, they’d be equal lengths.)

All that matters is that we’ve divided the floor, V:U, by the same Y:X ratio as the wall:


The equality of ratios allows an important identity:


This equality, YU=XV, will be crucial in the proof!

As an aside, when the box height and width aren’t the same, physical mirrors don’t work. Flat mirrors don’t compress or expand lengths. An actual mirror works here only when the box is square, which puts the mirror at a 45° angle. What we’re doing here isn’t actually reflection but geometric projection. But the mirror intuition remains apt.


Figure 4.

Step 4: Draw two diagonal lines (red lines in Figure 4), one from point b to point e and one from point c to point d. These intersect the middle vertical line at points k and h, respectively.

From points k and h, draw perpendicular (horizontal) lines to sub-divide the upper and lower segments. We label the four segments, from bottom to top, A, B, C, and D.

This completes Plato’s instructions. We’ve divided the line by some arbitrary ratio and sub-divided the parts using the same ratio.

Altogether, we now have:


Which all express the same ratio (whatever that happens to be).

This step sub-divides the segments because the red lines represent (projective) mirrors that reflect the entire “floor” (which is divided by the ratio) onto their respective segments of the “wall”. Since each mirror reflects the entire floor onto the segment, that segment is divided by the ratio. (Which makes it a good illustration of how projection expands or compresses lengths but maintains relative sizes within the projection. Here the entire floor is projected onto two different lengths.)

As an aside, the projection in Step 3 does reverse the “image” onto the floor but reflecting it back to the wall in this step reverses it again. The double-reverse preserves the ratio’s orientation on the wall (in this case, larger above smaller), where it matters.

So far this is all the same as last time. Now we enter new territory.


Figure 5.

Step 5: No more drawing is required, although you can fill in the triangles as I’ve done in Figure 5. What’s important is focusing on the two triangles formed in Step 4 when we added the two “mirrors”.

We’ll call the larger upper triangle (points cbe) Tri-Y and the smaller lower triangle (points acd) Tri-X.

Tri-Y and Tri-X share two important properties: Firstly, both have the same width, U+V. Secondly, both are right triangles (their lower-left corners are 90° angles).

We can note that, since each occupies one-half of their respective Y and X segments, their areas must have the same Y:X ratio (which is true and easy to demonstrate).

More importantly, the slopes of their hypotenuses also have the Y:X ratio to each other. The formula for slope is just:


For our two triangles, Tri-Y and Tri-X, we have:


We can factor out the common 1/U+V, which leaves us with just the Y and X, which tells us their slopes do have the Y:X ratio to each other.

So, now we have:



Figure 6.

Step 6: Now focus on the two smaller triangles shown in Figure 6. We’ll call the upper-right one (points gke) Tri-C and the lower one (points mch) Tri-B. The big triangles in Step 5 were associated with the Y and X line sections, these triangles are associated with the C and B line sections.

Both triangles are right triangles (they have 90° angles in their lower-left corners). More importantly, each hypotenuse lies on the hypotenuse of one of the big triangles from Step 5.

That means we know the slopes of line ke and line ch — they are the slopes for Tri-Y and Tri-X, respectively.

We also know that the widths for Tri-C and Tri-B are U and V (respectively). All of these things are fixed by the construction. They’re consequences of how we’ve defined the various lines. It would be true even if we never made a diagram (it has reality in the Platonic realm).

Now we can ask ourselves about the B and C segment lengths. We see that B is the height of Tri-B and that C is the height of Tri-C. We can calculate this height by multiplying the slope times the width.

This is the same as calculating how far we travel by multiplying our (constant) speed by the time we spend. If we go 3 MPH for 5 hours, we go 15 miles. We can think of the line slope here as speed and the horizontal axis as time (in this case, going right-to-left). The vertical axis is distance traveled.

To calculate the heights of Tri-C and Tri-B:


We can again factor out the common 1/U+V. This leaves us with YU and XV as the respective proportional heights of Tri-C and Tri-B. (Proportional because we factored out that common value.)

YU and XV, therefore, are also the proportional heights of the C and B segments.


We know from above that YU=XV (because Y/X=V/U), so, by construction and geometry, Tri-C and Tri-B must have the same heights. Therefore, segments C and B must have the same lengths. (If proportional heights are the same, actual heights are the same.)

The equality is inherent in the geometry. There is no way for the middle segments to not be equal. (Of course, we knew that all along from the algebra.)

This proof makes physical sense. We can go 5 MPH for 3 hours or we can go 3 MPH for 5 hours. Either way, we go 15 miles. Imagine we have: Y=5 MPH, X=3 MPH; V=5 hours, U=3 hours. Then going 5 MPH for 3 hours is YU and going 3 MPH for 5 hours is XV. We know we get 15 miles in either case, so, indeed, YU=XV.

§ §

This all works because the segment lengths have a relationship that is locked by the single ratio. The Y:X segment divides into D:C and B:A sub-segments. The ratio-locking also creates an inverted X:Y ratio between Tri-C and Tri-B. Note how Tri-C straddles the smaller U and C segments, whereas Tri-B straddles the larger V and B segments.

The inverted X:Y ratio cancels out the Y:X ratio, leaving a 1:1 ratio, so B=C. That’s what YU=XV really says. It’s bigger1*smaller2=smaller1*bigger2.

This almost amounts to a statement about the paradox of the Divided Line, the perception that the four segments, A, B, C, D, should each increase in relative size versus the algebraic and geometric reality that B=C. The relationship between Tri-B and Tri-C crosses the main segment line and pulls on the respective B:A and D:C sub-segment pairs. The opposing tensions force the B and C segments into equality.

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

About Wyrd Smythe

The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe

3 responses to “Back to Plato’s Line

  • Wyrd Smythe

    From the post: “We can note that, since each [of Tri-Y and Tri-X] occupies one-half of their respective Y and X segments, their areas must have the same Y:X ratio (which is true and easy to demonstrate).”

    Here is that demonstration…

    The area of a triangle is:


    So, for our two triangles here, we have:

    \displaystyle\textrm{Area}(\textrm{Tri-Y})=\frac{1}{2}\left[({U\!\!+\!\!V})\times{Y}\right]= \frac{U\!\!+\!\!V}{2}{Y}\\[0.95em]\textrm{Area}(\textrm{Tri-X})=\frac{1}{2}\left[({U\!\!+\!\!V})\times{X}\right]=\frac{U\!\!+\!\!V}{2}{X}

    We can factor out the U+V/2, which is common to both, and see that their areas have the same Y:X ratio as used to divide the line.


  • Wyrd Smythe

    I’ve been trying to think of a better way to summarize this:

    Figure 5 highlights two right triangles whose areas and slopes we know have the Y:X ratio to each other. The widths are equal (U+V), so the heights have the Y:X ratio to each other.

    Figure 6 highlights two smaller right triangles with the same respective slopes as the bigger two. Since the widths of these have a V:U ratio (which is the same as a Y:X ratio), then the heights must be equal.

  • Plato’s Divided Line | Logos con carne

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