We shall next show how the remaining individual chords can be derived from the above, first of all setting out a theorem which is extremely useful for the matter at hand.
Having derived a handful of special angle-chord relationships, Ptolemy next set out to derive a more general theorem to get the rest. So Ptolemy constructs a new diagram from which to start his calculations.
Let’s start by breaking this diagram down. At its core, it’s a quadrilateral (a figure with 4 sides) inscribed in a circle. That’s figure ABGD1.
From there, the diagonals are connected: $\overline{BD}$ and $\overline{AG}$.
The funny part here is $\overline{BE}$. It doesn’t follow naturally, but Ptolemy is defining it as being drawn in such that $\angle{ABE} = \angle{DBG}$ .
That’s all our givens, so time to see what we can do with them.
First off, since we just stated $\angle{ABE} = \angle{DBG}$, that means if we add the same angle to both ($\angle{DBE}$) then the resulting angles ($\angle{ABD}$ and $\angle{EBG}$) are also the same.
In addition $\angle{BDA}$ and $\angle{BGE}$ are equal because they both have vertices on the circumference and subtend the same chord ($\overline{AB}$).
Next off, we have some similar triangles. We just established that $\angle{BDA}$ and $\angle{BGE}$ are the same angle, and we previously established that $\angle{ABD}$ and $\angle{EBG}$ are equal. That’s two out of three of the angles in $\triangle{BGE}$ which, since angles in triangles add up to 180º, means that the third one must be equal too.
Thus, $\triangle{BGE}$ is a similar triangle to $\triangle{ABD}$, where similar means the angles are all the same even though the sides can be different (but proportional) lengths.
Next, we will take advantage of that property about the proportional lengths.
$$\frac{\overline{BG}}{\overline{EG}} = \frac{\overline{BD}}{\overline{AD}}$$
Next we repeat the last few steps for the other angles:
$\angle{BAE} = \angle{BDG}$ because they again subtend the same arc.
This gives us $\triangle{ABE}$ and $\triangle{BGD}$ being similar, so we can again start taking ratios of their parts:
$$\frac{\overline{AB}}{\overline{AE}} = \frac{\overline{BD}}{\overline{DG}}$$
These fractions are a bit unwieldy so to simplify, let’s cross multiply each in turn:
$$\overline{AD} \cdot \overline{BG} = \overline{BD} \cdot \overline{EG}$$
$$\overline{DG} \cdot \overline{AB} = \overline{BD} \cdot \overline{AE}$$
Now let’s add those together.
Some of you might be getting worried that we’re randomly adding things together. Isn’t that against the rules since adding things changes the things?
It’s ok in this case because what we’re adding is equal on both sides even if it looks different. It’s like adding 1 to one side of an equation and $\frac{6}{6}$ to the other side. They look different, but they’re the same so it balances out and while the result may look different, it’s ultimately the same so it’s all good.
Unfortunately, they’re not like terms, so they don’t go together too well:
$$\overline{AD} \cdot \overline{BG} + \overline{DG} \cdot \overline{AB} = \overline{BD} \cdot \overline{EG} + \overline{BD} \cdot \overline{AE}$$
Not pretty and there’s only one bit of simplification we can do:
$$\overline{AD} \cdot \overline{BG} + \overline{DG} \cdot \overline{AB} = \overline{BD} \cdot ( \overline{EG} + \overline{AE})$$
Fortunately, that gives us one more piece of simplification. If we look back at the diagram, we see that $\overline{AE}$ and $\overline{EG}$ are part of the same line, $\overline{AG}$. So $\overline{AE} + \overline{EG} = \overline{AG}$, which we can substitute into the part int he parentheses.
$$\overline{AD} \cdot \overline{BG} + \overline{DG} \cdot \overline{AB} = \overline{BD} \cdot \overline{AG}$$
And that’s it. That’s Ptolemy’s Theorem. Right now, it probably looks like a bunch of letters so let’s break it down in terms of what it means in the diagram. What it comes down to is: “the sum of the product of opposing sides is equal to the product of the diagonals.”
Phrasing it slightly less formally, “multiply the opposing sides, and then add them together. It’s the same as multiplying the two diagonals together.”
While other astronomers prior to Ptolemy had chord tables (what we’re still working up to completing with all this), it isn’t certain how they came by them. So it appears Ptolemy was the first to take this route.
However, this theorem alone doesn’t really get very far. What we really want to do is take the angle-chord relationships that we’ve derived in the previous post and being able to do things like add or subtract them or cut them in half to get the lengths of new chords. Fortunately, some applications of Ptolemy’s theorem will help us get there. But we’ll save those for the next post.
SIDE NOTE: If you didn’t catch it when reading through this post, I’ve started trying out a new footnotes plug-in. I noticed I was running into a lot of cases in which I wanted to add side notes that got rather long and distracted from the flow. So I decided footnotes would be a better route. If you hover over the superscript, it will display a pop-up bubble of the footnote. If you click on it, it will take you to the footnotes section which has a little button to return you to where you left off for the corresponding footnote!
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