Our next task is to demonstrate the sizes of the individual arcs cut off between the equator and the ecliptic along a great circle through the poles of the equator. As a preliminary we shall set out some short and useful theorems which will enable us to carry out most demonstrations involving spherical theorems in the simplest and most methodical way possible.
In opening the next chapter in Book 1, Ptolemy again gives us a goal to work towards, namely, the length of the chord shown in solid red below1.
But before we do that, we’re going to have to lay out some lemma2 to get us there. There’s going to be several, but for this post, I’m only going to address the first two which come from triangles, whereas the remaining involve circles and a bit of new notation that I’ll want to introduce before getting into them.
Although we’re working towards spherical geometry, there’s still plenty of triangles and circles we’ll still be able to draw within the sphere. After all, any plane through a sphere cuts a circle. And if you have a chord across that circle and connect it to a side or center, then you have a normal Cartesian triangle.
So it’s not surprising that Ptolemy starts with some triangles:
The only thing to note in this diagram is that $\overline{DG} \parallel \overline{EH}$3.
According to Ptolemy, because these two lines are parallel,
$$\frac{\overline{AG}}{\overline{AE}} = \frac{\overline{DG}}{\overline{EH}}$$
Ptolemy didn’t give a proof for this, but it comes from some similar triangles. In particular $\triangle{ADG}$ is similar to $\triangle{AHE}$. So the ratio of their sides will be the same as the ratio of their bases.
Next Ptolemy does something a little funny and tosses a $\overline{DZ}$ into the mix.
$$\frac{\overline{AG}}{\overline{AE}} = \frac{\overline{DG}}{\overline{DZ}} \cdot \frac{\overline{DZ}}{\overline{EH}}$$
Obviously the $\overline{DZ}$’s will cancel so we haven’t really changed anything here, simply altered the look of it so we can do a substitution here in a moment. But before that, we’ll apply the same logic as previously to note that $\triangle{BDZ}$ and $\triangle{BHE}$ are similar which means that, again, the ratio of their bases and sides will be similar. This allows us to state:
$$\frac{\overline{DZ}}{\overline{EH}} = \frac{\overline{BZ}}{\overline{BE}}$$
And wouldn’t you look at that, we have a $\frac{\overline{DZ}}{\overline{EH}}$ in common to both of our equations allowing for some substitution [13.1]4.
$$\frac{\overline{AG}}{\overline{AE}} = \frac{\overline{DG}}{\overline{DZ}} \cdot \frac{\overline{BZ}}{\overline{BE}}$$
This is the first lemma Ptolemy set out to prove.
For the next part, we begin with nearly the same figure as the previous one, but we’ll tack on a bit more:
Here, we’ve dropped the previous parallel line and extended $\overline{GD}$ out to point H such that $\overline{AH} \parallel \overline{BE}$.
This time, instead of looking at the ratio of a piece of a side to the full side, we’ll look at the pieces of the sides individually:
$$\frac{\overline{EG}}{\overline{AE}} = \frac{\overline{GZ}}{\overline{HZ}}$$
Again, this comes from similar triangles, but it’s also in Euclid’s Elements in Book VI, proposition 2.
Using $\overline{DZ}$ again just to break things up, we can say:
$$\frac{\overline{GZ}}{\overline{HZ}} = \frac{\overline{GZ}}{\overline{DZ}} \cdot \frac{\overline{DZ}}{\overline{HZ}}$$
This next step is a little tricky, but $\triangle{ADH}$ is similar to $\triangle{BDZ}$. The reason is that two of their sides are common ($\overline{HD}$ extends right into $\overline{DZ}$ as does $\overline{BD}$ into $\overline{DA}$). In addition, $\angle{DBZ} = \angle{HAD}$ because these are alternate interior angles. That’s enough to justify those two triangles being similar, and in fact, you can see it if you imagine flipping $\triangle{BDZ}$ up around point D so it rests inside $\triangle{ADH}$. Anyway, similar triangles again allows us to state:
$$\frac{\overline{DZ}}{\overline{BD}} = \frac{\overline{DH}}{\overline{AD}}$$
Doing a bit of algebraic rearranging we can state:
$$\frac{\overline{DH}}{\overline{DZ}} = \frac{\overline{AD}}{\overline{BD}}$$
Now we’ll return to the componendo theorem5 we saw previously since we have the handy $\frac{a}{b} = \frac{c}{d}$ form where $a = \overline{DH}$, $b = \overline{DZ}$, $c = \overline{AD}$, and $d = \overline{BD}$. Thus, applying the theorem we get
$$\frac{\overline{DH} + \overline{DZ}}{\overline{DZ}} = \frac{\overline{AD} + \overline{BD}}{\overline{BD}}$$
But if we look at what the numerator in both of these physically represent in the diagram, we see that $\overline{DH} + \overline{DZ} = \overline{HZ}$ and $\overline{AD} + \overline{BD} = \overline{AB}$. Substituting these in:
$$\frac{\overline{HZ}}{\overline{DZ}} = \frac{\overline{AB}}{\overline{BD}}$$
Let’s flip both of those fractions over:
$$\frac{\overline{DZ}}{\overline{HZ}} = \frac{\overline{BD}}{\overline{AB}}$$
Which is something we had before we went on this most recent detour of similar triangles and componendo. So we can substitute that in:
$$\frac{\overline{GZ}}{\overline{HZ}} = \frac{\overline{GZ}}{\overline{DZ}} \cdot \frac{\overline{BD}}{\overline{AB}}$$
One more substitution will finish this one off, so let’s notice that $\triangle{GZE}$ and $\triangle{GHA}$ are similar (or apply the theorem from Euclid again). As with above, that means the ratio of their similar sides will be equal as well, allowing us to state:
$$\frac{\overline{GZ}}{\overline{HZ}} = \frac{\overline{EG}}{\overline{AE}}$$
That substitutes right into the above giving us:
$$\frac{\overline{EG}}{\overline{AE}} = \frac{\overline{GZ}}{\overline{DZ}} \cdot \frac{\overline{BD}}{\overline{AB}}$$
And that’s the second lemma Ptolemy wants to introduce [13.2].
- I kinda hate having this picture here. Although it does accurately illustrate what Ptolemy opens the chapter with, we’ve got a few posts of proofs to get through before we’ll actually engage with this. Feels like I’m teasing you…
- A lemma is a proof done as part of another proof. A pre-proof, if you will.
- $\parallel$ is the symbol for parallel.
- As is often done in math, numbers to important theorems are given to make referencing them later easier. The 13 here comes from the fact that we’re currently in chapter 13 of Book 1. Since I’ve been breaking up these posts differently than Ptolemy has broken up his books, I’ve been omitting the chapters.
- Ptolemy said he used the dividendo theorem, but I’m not sure how, so I did it this way