Unit II · Ptolemy’s Theorem on Cyclic Quadrilaterals & Chord Calculus¶
Almagest Book I, Chapter 10b (Heiberg 1898, Vol. I.1, pp. 36–46)¶
1. Ptolemy’s Theorem on Cyclic Quadrilaterals (Heiberg I, pp. 36–37)¶
p. 36 Having obtained these primary chords, the next step is to set forth the corollary that follows directly through the theorem we are about to write down, being most useful on every occasion:
Proposition (Ptolemy’s Theorem). Let there be any quadrilateral inscribed in a circle, ΑΒΓΔ, and let its diagonals ΑΓ–Γ and ΒΔ be joined. We must prove that the rectangle contained by ΑΓ and ΒΔ is equal to the sum of the rectangles contained by the opposite sides, ΑΒ · ΔΓ and ΑΔ · ΒΓ.
Construction. Let angle ∠ΑΒΕ be set out equal to angle ∠ΔΒΓ by drawing ΒΕ to meet diagonal ΑΓ at Ε (Eucl. I.23).
p. 37 Demonstration. If to each of the equal angles ∠ΑΒΕ and ∠ΔΒΓ we add the common angle ∠ΕΒΔ, then the whole angle ∠ΑΒΔ is equal to the whole angle ∠ΕΒΓ. Yet angle ∠ΒΔΑ is also equal to angle ∠ΒΓΕ, for they stand upon the same circumference ΑΒ (Eucl. III.21). Therefore triangle ΑΒΔ is equiangular with triangle ΕΒΓ! Consequently, by Eucl. VI.4, their sides about the equal angles are proportional—as ΒΓ is to ΓΕ, so is ΒΔ to ΔΑ:
$$\frac{B\Gamma}{\Gamma E} = \frac{B\Delta}{\Delta A} \quad\Longrightarrow\quad B\Gamma \cdot \Delta A = B\Delta \cdot \Gamma E \qquad\text{(}:just:`Eucl. VI.16 <VI.16>`\text{)}.$$
Again, since angle ∠ΑΒΕ is equal to angle ∠ΔΒΓ, and angle ∠ΒΑΕ is equal to angle ∠ΒΔΓ (subtending the same arc ΒΓ, Eucl. III.21), triangle ΑΒΕ is equiangular with triangle ΔΒΓ. Therefore proportionally, as ΒΑ is to ΑΕ, so is ΒΔ to ΔΓ:
$$\frac{BA}{AE} = \frac{B\Delta}{\Delta\Gamma} \quad\Longrightarrow\quad BA \cdot \Delta\Gamma = B\Delta \cdot AE \qquad\text{(}:just:`Eucl. VI.16 <VI.16>`\text{)}.$$
Now it was proved above that $B\Gamma \cdot \Delta A = B\Delta \cdot \Gamma E$. Adding the two equalities together and invoking Eucl. II.1 ($B\Delta \cdot AE + B\Delta \cdot \Gamma E = B\Delta \cdot A\Gamma$), the whole rectangle contained by $A\Gamma$ and $B\Delta$ is equal to the sum of the rectangle $AB \cdot \Delta\Gamma$ and the rectangle $A\Delta \cdot B\Gamma$. Which was to be proved.
2. Chord of the Difference (Heiberg I, p. 38)¶
p. 38 Corollary (Difference of Two Arcs). Given a semicircle ΑΒΓΔ on diameter ΑΔ, suppose the chords ΑΒ and ΑΓ from Α are given in magnitude. Joining ΒΓ, ΒΔ, and ΓΔ, the supplementary chords ΒΔ and ΓΔ are immediately given via Eucl. III.31 and Eucl. I.47 ($\operatorname{crd}^2(180^\circ - \theta) = D^2 - \operatorname{crd}^2\theta$). Since ΑΒΓΔ is a cyclic quadrilateral in the circle, Ptolemy’s Theorem gives:
$$AB \cdot \Gamma\Delta + A\Delta \cdot B\Gamma = A\Gamma \cdot B\Delta \quad\Longrightarrow\quad B\Gamma = \frac{A\Gamma \cdot B\Delta - AB \cdot \Gamma\Delta}{A\Delta}.$$
Applying this immediately to the pentagon arc ($72^\circ$) and hexagon arc ($60^\circ$), Ptolemy obtains the chord of their difference ($12^\circ$):
$$\operatorname{crd}(12^\circ) = \frac{\operatorname{crd}(72^\circ)\operatorname{crd}(120^\circ) - \operatorname{crd}(60^\circ)\operatorname{crd}(108^\circ)}{120} \approx 12;32,36^p.$$
3. Half-Arc Chord Lemma (Heiberg I, pp. 39–40)¶
p. 39 Next, if any chord in the circle be given, how do we find the chord subtending half the same arc?
Let ΑΒΓ be a semicircle on diameter ΑΓ, let chord ΓΒ be given, and let its arc be bisected at Δ, joining ΑΒ, ΑΔ, ΒΔ, and ΔΓ. Cut off ΑΕ along ΑΓ equal to ΑΒ, and drop perpendicular ΔΖ onto ΑΓ (∠ΔΖΓ).
Demonstration. Since ΑΒ = ΑΕ, ΑΔ is common, and ∠ΒΑΔ = ∠ΕΑΔ (standing on equal half-arcs ΒΔ and ΔΓ), triangles ΒΑΔ and ΕΑΔ are congruent (Eucl. I.4), so ΒΔ = ΔΕ. Yet ΒΔ = ΔΓ since the arcs are equal; therefore triangle ΔΕΓ is isosceles with ΔΕ = ΔΓ, and the altitude ΔΖ bisects the base so ΕΖ = ΖΓ = ½ ΕΓ. Now in the right triangle ΑΔΓ inscribed in the semicircle (Eucl. III.31), triangle ΑΔΓ is similar to triangle ΔΖΓ (Eucl. VI.8), whence:
$$\frac{A\Gamma}{\Gamma\Delta} = \frac{\Gamma\Delta}{\Gamma Z} \quad\Longrightarrow\quad \Gamma\Delta^2 = A\Gamma \cdot \Gamma Z = \frac{1}{2}\,A\Gamma \cdot (A\Gamma - AB) = 60 \cdot \bigl(120 - \operatorname{crd}(180^\circ - \theta)\bigr).$$
p. 40 Iterating this half-arc formula four times from $\operatorname{crd}(12^\circ) = 12;32,36^p$, Ptolemy derives:
$\operatorname{crd}(6^\circ) \approx 6;16,49^p$
$\operatorname{crd}(3^\circ) \approx 3;8,28^p$
$\operatorname{crd}(1\tfrac{1}{2}^\circ) \approx 1;34,15^p$
$\operatorname{crd}(\tfrac{3}{4}^\circ) \approx 0;47,8^p$
4. Chord of the Sum & Impossibility of Trisection by Lines (Heiberg I, pp. 41–42)¶
p. 41 Again, let arcs ΑΒ and ΒΓ be given in a circle around center Ε (Panel II of Figure 4 above), and draw diameter ΒΕΔ. Because ΑΒΓΔ is cyclic with diagonal ΒΔ equal to the diameter $D = 120^p$, Ptolemy’s Theorem yields the chord of the combined arc ΑΓ:
$$\operatorname{crd}(\alpha + \beta) = \frac{\operatorname{crd}(\alpha)\operatorname{crd}(180^\circ - \beta) + \operatorname{crd}(\beta)\operatorname{crd}(180^\circ - \alpha)}{120}.$$
p. 42 By combining $1\frac{1}{2}^\circ$ with all preceding chords, one immediately fills in every multiple of $1\frac{1}{2}^\circ$—yet two intermediates remain inside each $1\frac{1}{2}^\circ$ interval if the table is to advance by half-degrees! For if we possessed $\operatorname{crd}(\frac{1}{2}^\circ)$, both missing entries would follow immediately by sum and difference with $1\frac{1}{2}^\circ$. However, when a chord such as that of $1\frac{1}{2}^\circ$ is given, the chord subtending one-third of the same arc can in no way be found geometrically by means of lines (οὐδαμῶς γίνεται διὰ τῶν γραμμῶν)—though if that were possible, we would immediately have $\operatorname{crd}(\frac{1}{2}^\circ)$ itself!
5. Ptolemy’s Inequality Lemma & $\operatorname{crd}(1^\circ) \approx 1;2,50^p$ (Heiberg I, pp. 43–46)¶
p. 43 Instead, Ptolemy establishes a rigorous geometric sandwich for $\operatorname{crd}(1^\circ)$ from the known chords of $1\frac{1}{2}^\circ$ and $\frac{3}{4}^\circ$:
Lemma (Ptolemy’s Chord–Arc Inequality). In circle ΑΒΓΔ, let two unequal consecutive chords ΑΒ $<$ ΒΓ be drawn. Bisect angle ∠ΑΒΓ by ΒΔ meeting chord ΑΓ at Ε and the circumference at Δ, join ΑΔ and ΓΔ, and drop perpendicular ΔΖ onto ΑΓ (∠ΔΖΑ).
Demonstration. Since ∠ΑΒΔ = ∠ΓΒΔ, chords ΑΔ and ΓΔ are equal, and since ΓΒ $>$ ΒΑ, by Eucl. VI.3 we have ΓΕ > ΕΑ, so the foot Ζ of the altitude in isosceles triangle ΑΔΓ falls between Ε and Γ. Consequently ΔΑ > ΔΕ > ΔΖ. Describing the circular arc ΗΕΘ centered at Δ with radius ΔΕ (meeting ΔΑ produced at Η and ΔΖ at Θ), triangle ΔΕΖ is strictly larger than sector ΔΕΘ, whereas triangle ΔΕΑ is strictly smaller than sector ΔΕΗ:
$$\frac{EZ}{EA} = \frac{\operatorname{area}(\triangle \Delta EZ)}{\operatorname{area}(\triangle \Delta EA)} > \frac{\operatorname{area}(\text{sector }\Delta E\Theta)}{\operatorname{area}(\text{sector }\Delta EH)} = \frac{\angle E\Delta Z}{\angle E\Delta A}.$$
p. 44–45 By componendo and doubling (ΓΑ = 2 ΓΖ), it follows that $\frac{\Gamma E}{EA} < \frac{\angle\Gamma\Delta B}{\angle B\Delta A}$, and since $\frac{\Gamma B}{BA} = \frac{\Gamma E}{EA}$ (Eucl. VI.3) while central/inscribed angles stand as their arcs (Eucl. VI.33):
$$\frac{\operatorname{crd}(B\Gamma)}{\operatorname{crd}(BA)} < \frac{\operatorname{arc}(B\Gamma)}{\operatorname{arc}(BA)}.$$
p. 45–46 Pinching $\operatorname{crd}(1^\circ)$ to Sexagesimal Seconds:
Setting $\operatorname{arc}(BA) = \frac{3}{4}^\circ$ and $\operatorname{arc}(B\Gamma) = 1^\circ$ gives $\operatorname{crd}(1^\circ) < \frac{4}{3}\operatorname{crd}(\frac{3}{4}^\circ) = \frac{4}{3}(0;47,8^p) = 1;2,50,40^p \approx 1;2,50^p$.
Setting $\operatorname{arc}(BA) = 1^\circ$ and $\operatorname{arc}(B\Gamma) = 1\frac{1}{2}^\circ$ gives $\operatorname{crd}(1^\circ) > \frac{2}{3}\operatorname{crd}(1\frac{1}{2}^\circ) = \frac{2}{3}(1;34,15^p) = 1;2,50,0^p = 1;2,50^p$.
Since $\operatorname{crd}(1^\circ)$ has been proved strictly both smaller and larger than quantities that agree to $1;2,50^p$, Ptolemy concludes with complete rigor:
$$\operatorname{crd}(1^\circ) = 1;2,50^p = \frac{377}{360}, \qquad \operatorname{crd}(\tfrac{1}{2}^\circ) \approx 0;31,25^p,$$
from which every remaining entry in the $360$-row Table of Chords follows!