Almagest Book I: Circular Lemmas for Spherical Trigonometry

As noted in the previous Almagest post, I wanted to break this next set of lemma off because they use a new mathematical term: Crd arc.

The term is very simple. In fact, the name’s on the tin. It refers to the chord subtended by a specific arc. So let’s dive right in to the next proof and see how it’s used:

As always, the givens: $\overline{AZ}$ and $\overline{GH}$ are both perpendicular to $\overline{BD}$ which is a diameter.

Since they’re both perpendicular to $\overline{BD}$, they’re parallel to one another and since $\overline{AG}$ connects them, $\mathrm{\angle }EAZ$ and $\mathrm{\angle }EGH$ are congruent as are $\mathrm{\angle }AEZ$ and $GEH$. This makes $\mathrm{△}AEZ$ and $\mathrm{△}EGH$ similar triangles¹.

Similar triangles means we can take ratios of their sides to be equal so:

$$\frac{\overline{AZ}}{\overline{GH}}=\frac{\overline{AE}}{\overline{EG}}$$

And here’s where the new terminology comes in. Let’s start by noting that if we extend $\overline{AZ}$ all the way across the circle, that would be doubling it in length. In addition, there would now be another point lying on the perimeter directly across from A. Since we would now have a line that begins and ends on the circle, that’s the definition of a chord. And what arc would subtend it? It would be $arc2AB$.

Writing that out we can state:

$$2\overline{AZ}=Crdarc2AB$$

But, in this setup we don’t really have $2\overline{AZ}$. We have $\overline{AZ}$ so let’s clear that 2 out by dividing both sides by it:

$$\overline{AZ}=\frac{1}{2}Crdarc2AB$$

We can do the exact same for $\overline{GH}$ and state:

$$\overline{GH}=\frac{1}{2}Crdarc2BG$$

Jumping back a bit, we’d determined $\frac{\overline{AZ}}{\overline{GH}}=\frac{\overline{AE}}{\overline{EG}}$. And sinse we now have $\overline{AZ}$ and $\overline{GH}$, we can take their ratio to get something to substitute in. Since the halves will cancel, we get:

$$\frac{\overline{AZ}}{\overline{GH}}=\frac{Crdarc2AB}{Crdarc2BG}$$

Substituting that into our equation from above:

$$\frac{\overline{AE}}{\overline{EG}}=\frac{Crdarc2AB}{Crdarc2BG}$$

Which is the next of Ptolemy’s lemma. [13.3]

Next Ptolemy states that if we know $\frac{Crdarc2AB}{Crdarc2BG}$ and $arcAG$ we can determine $arcAB$ and $arcBG$. To demonstrate this, we examine another diagram.

Here, all points except Z remain unchanged (with the exception of H which has gone away since it is no longer needed for this proof). Now, $\overline{DZ}$ is drawn such that it is perpendicular to $\overline{AG}$. As such,  $\overline{DZ}$ also bisects $\overline{AG}$. Thus, if $\overline{AG}$ were given, we could determine $\overline{DZ}$ because $\mathrm{△}ADZ$ is a right triangle, $\overline{AD}$ is a radius, and thus, we could apply the Pythagorean theorem.

And as a bonus, $\overline{AZ}$ is the arc subtended by $\mathrm{\angle }ADZ$. Since we spent a bunch of time going through the math to create the table of chords, we could then look up the angle that corresponds to that chord length, and use that to determine all of the angles in $\mathrm{△}ADZ$.

We’ll still assume $\overline{AG}$ is given, and Ptolemy now assumes that $\frac{\overline{AE}}{\overline{EG}}$ is as well (or $\frac{Crdarc2AB}{Crdarc2BG}$. Ptolemy’s not exactly clear on which we’ll be having values for).

But if we know the total length of a line ($\overline{AG}$), and the ratio of its parts($\frac{\overline{AE}}{\overline{EG}}$), then we can determine each of the parts individually. That’s not immediately obvious as to how so let’s take an example.

Say we know the total length of the line to be 10 parts. We’re told the ratio of the parts is 0.42857. Stating that mathematically:

$$a+b=10$$

$$\frac{a}{b}=0.42857$$

Solve the first equation for a:

$$a=10–b$$

Substitute into the second equation:

$$\frac{10–b}{b}=0.42857$$

Solve for b:

$$\frac{10}{b}–\frac{b}{b}=0.42857$$

$$\frac{10}{b}–1=0.42857$$

$$10–b=0.42857b$$

$$10=1.42857b$$

$$b=7$$

Substitute that into the first equation and you’ll find that a = 3.

Thus, if you know the length of the full line and the ratio of its parts, then you can find the individual pieces.

But now since we could know $\overline{AE}$ and we already knew $\overline{AZ}$, we could subtract the two to find $\overline{EZ}$. Again, we could use our table of chords to look up the angle that subtends it ($\mathrm{\angle }EDZ$). Add that to $\mathrm{\angle }ADZ$ and you determine $\mathrm{\angle }ADB$. And since Ptolemy defines his circles to have a circumference of 360 parts, the arc subtended by an angle is equal to that angle. Which means if we know $\mathrm{\angle }ADZ$, we know $arcAB$.

Similarly, since we assumed we know the length of $\overline{AG}$, we can again look up the central angle that would subtend it and since the angle and arc are the same as we just stated, that means we’d know $arcAG$. Subtracting $arcAB$ we can then determine $arcBG$.

Next, we’ll derive another lemma similar to the last one.

Here $\overline{BZ}$ and $\overline{GH}$ are perpendicular to $\overline{AH}$. This means that $\mathrm{△}BEZ$ is similar to $\mathrm{△}GEH$. So again we can take the ratios of their sides to be equal:

$$\frac{\overline{GE}}{\overline{BE}}=\frac{\overline{GH}}{\overline{BZ}}$$

Again, we can notice that if we extend $\overline{BZ}$ and $\overline{GH}$ they would then be $Crdarc2AB$ and $Crdarc2GA$ (note we’re going the long way around with $arcGA$).

Writing that out gives us:

$$2(\overline{BZ})=Crdarc2AB$$

$$2(\overline{GH})=Crdarc2GA$$

Dividing both sides by 2:

$$\overline{BZ}=\frac{1}{2}(Crdarc2AB)$$

$$\overline{GH}=\frac{1}{2}(Crdarc2GA)$$

Taking the ratio $\frac{\overline{GH}}{\overline{BZ}}$ and cancelling the halves gives us:

$$\frac{\overline{GH}}{\overline{BZ}}=\frac{Crdarc2GA}{Crdarc2AB}$$

Substituting from where we noted $\frac{\overline{GE}}{\overline{BE}}=\frac{\overline{GH}}{\overline{BZ}}$:

$$\frac{\overline{GE}}{\overline{BE}}=\frac{Crdarc2GA}{Crdarc2AB}$$

That’s another lemma [13.4] we’ll keep on hand.

Ptolemy puts it to use with a remix on the previous diagram in order to prove that if we know $arcBG$ as well as $\frac{arc2GA}{arc2AB}$, we can determine $arcAB$.

Here $\overline{DZ}\perp \overline{BG}$ which means it bisects $\overline{BG}$. Therefore $\mathrm{\angle }BDZ$ subtends half of $arcBG$. So if we know $arcBG$ (which was one of the things Ptolemy assumes we’re going to know), since the central angle is the same and we’re dividing it in half, then we’ll know $\mathrm{\angle }BDZ$. If we know $arcBG$ we can look up the chord length for $\overline{BG}$ and half of that will be $\overline{BZ}$.

Taking stock, we’ve know two of the sides ($\overline{BD}$ since it’s a radius and $\overline{BZ}$ as we just noted) and an two angles ($\mathrm{\angle }BDZ$ and $\mathrm{\angle }BZD$ which is a right angle)  so we can determine $\overline{DZ}$ via the Pythagorean theorem and $\mathrm{\angle }DBZ$ via subtraction of the other two angles from the 180º in a triangle.

So $\mathrm{△}BDZ$ is completely known.

Jumping back to the proof setup, Ptolemy stated that it’s assumed we know $\frac{arc2GA}{arc2AB}$ which we previously proved is equal to $\frac{\overline{EG}}{\overline{BE}}$. Again, if we know $arcBG$ we can look up the chord length for $\overline{BG}$. With these, we can determine $\overline{EB}$.

Ptolemy simply says it’s true, but let’s make up some numbers to demonstrate this is true much the same as I did previously. Again, I’ll use 10 for the length of the total line since it’s nice and around. I’ll also use a ratio of $\frac{\overline{EG}}{\overline{BE}}=\frac{5}{3}$. I’ll also pick $\overline{BG}=4$. Writing all this out:

$$\frac{\overline{EG}}{\overline{BE}}=\frac{5}{3}$$

$$\overline{EG}=\overline{BE}+4$$

Plugging the second equation into the first since it doesn’t need to be solved:

$$\frac{\overline{BE}+4}{\overline{BE}}=\frac{5}{3}$$

Multiplying both sides by $\overline{BE}$:

$$\overline{BE}+4=\frac{5}{3}\overline{BE}$$

Subtract $\overline{BE}$ from both sides:

$$4=\frac{2}{3}\overline{BE}$$

Multiply by $\frac{3}{2}$:

$$\overline{BE}=6$$

So to restate what was said before the demonstration: if we know $arcBG$ and $\frac{\overline{EG}}{\overline{BE}}$ we can determine $\overline{BE}$. If we wanted to, we could add on $\overline{BZ}$ since we determined that as well, in order to get $\overline{EZ}$. We’d also found $\overline{DZ}$ which gives us two of the three sides in right triangle $\mathrm{△}BDE$, so we can naturally figure out $\overline{ED}$ via the Pythagorean theorem.

From this, Ptolemy states that this should allow us to figure out $\mathrm{\angle }EDZ$ but I can’t seem to figure out how. Looking up the angle that has a chord length corresponding to $\overline{DZ}$ doesn’t work because the table we derived only works if the chords ends are on one of Ptolemy’s unit circles and the angle is the central angle. If we used a bit of trigonometry it would be straightforward, but Ptolemy doesn’t use trig.

Regardless, the trig confirms that the angle is solvable which concludes the “preliminary theorems” Ptolemy wants to introduce in this section. There will still be one more theorem before moving on, but this is a special one known as Menelaus’ theorem and it’s notably different in character, so I’ll save that for another post.

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1. Angle-Angle similarity.