Epitome of Ptolemy's Almagest

page 23. p
header: Book

...from the side of the pentagon you will find the chord of an arc of 108 degrees; and so similarly for the others. The chord of an arc is a straight line segment whose endpoints both lie on the circle. In modern terms, it is related to the sine of half the angle.

A small geometric diagram at the top left showing a triangle with sides labeled 'a' and 'b' and several intersecting arcs and lines below it.
A given straight line above the boundary of a circle at a proposed distance equidistant. Mark this.

Proposition II.

A circular diagram showing a quadrilateral inscribed within a circle. Points are labeled a, b, g, d. Diagonals intersect at e. Handwritten numbers 4, 4, 2 are written inside the segments.
Teaches how to find chords of 12 degrees.

If a quadrilateral inscribed in a circle is a rectangle, the rectangle In this context, "rectangle" refers to the product of two lengths, or the area formed by them. which is contained under its two diameters diagonals is equal to the two rectangles which are contained under its opposite sides, when taken together.
¶ Let there be a quadrilateral $a.b.g.d$ inscribed in a circle $a.b.g.d$, whose diameters are $a.g$ and $b.d$. I say that the product of $b.d$ times $a.g$ is equal to the sum of the two rectangles made from $a.d$ times $b.g$ and from $a.b$ times $d.g$. ¶ For, by the 23rd proposition of the first book referring to Euclid’s Elements, I shall place angle $a.b.e$ equal to angle $d.b.g$. By adding the common angle $e.b.d$ to both, angle $a.b.d$ becomes equal to angle $e.b.g$. Moreover, angle $b.d.a$ is equal to angle $b.g.e$ by the 20th proposition of the third book. Therefore, by the 32nd of the first book, the third angle, namely $b.a.d$, will be equal to the third angle $b.e.g$. Therefore, the triangles $a.b.d$ and $e.b.g$ are similar or equiangular. Thus, by the 6th proposition of the sixth book, the proportion of $a.d$ to $e.g$ is as the proportion of $b.d$ to $b.g$. Wherefore, by the 17th of the sixth book, the product of $a.d$ times $b.g$ is equal to that which is made from $b.d$ times $e.g$. Likewise, angle $a.b.e$ is equal to angle $d.b.g$ by hypothesis, and by the 20th of the third book, angle $b.a.e$ is equal to angle $b.d.g$. Therefore, by the 32nd of the first book, the third angle is equal to the third. The triangles $a.b.e$ and $d.b.g$ are therefore equiangular. Thus, by the 4th of the sixth book, $a.b$ is to $b.d$ as $a.e$ is to $d.g$. Wherefore, by the 17th of the sixth book, the product of $a.b$ times $d.g$ is equal to that which is made from $b.d$ times $a.e$. Now, however, it was shown that the product of $a.d$ times $b.g$ is equal to that which is made from $b.d$ times $e.g$. But by the first proposition of the second book, the products of $b.d$ times $e.g$ and $b.d$ times $a.e$ together are equal to the product of $b.d$ times $a.g$. Therefore, the product of $b.d$ times $a.g$ is equal to those which are made from $a.d$ times $b.g$ and from $a.b$ times $d.g$, which was to be demonstrated. This is the classic proof of Ptolemy's Theorem.

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Proposition IV.

A diagram of a circle with chords labeled a, b, g, d. Lines connect various points to form triangles and a quadrilateral within the circle.
Teaches how to find the chord of 6 degrees first.

1/5 of 360 is 72; 1/6 of 360 is 60. The difference is 12.

With the chords of unequal arcs in a semicircle being known: the chord of the arc by which the greater exceeds the lesser shall become known.
¶ As in the semicircle $a.b.d$ upon the diameter $a.d$, let the chords $a.b$ and $a.g$ be known. I say that the chord $b.g$ becomes known. For by the corollary of the first [proposition] of this book, the chords $b.d$ and $g.d$ also become known. ¶ In the quadrilateral $a.b.g.d$, let the diameters $a.g$ and $b.d$ be known; the opposite sides $a.b$ and $g.d$ are also known. Therefore, by the preceding proposition, the product of $a.d$ times $b.g$ will become known. But $a.d$ is known because it is the diameter of the circle; therefore $b.g$ becomes known, which was sought. Through this, you will know the chords of many arcs. For you will find the chord of the arc by which the fifth part of the circumference exceeds the sixth part: namely, the chord of an arc of 12 degrees; and so for others.

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Proposition V.

A diagram of a semicircle with diameter a-g and a chord b-g. A point d bisects the arc b-g, and perpendicular line d-e is drawn to the diameter.
Teaches how to find the chord of 6 degrees first.

For whatever arc in a semicircle a chord has been given: the chord of half of such an arc becomes known.
¶ Let the arc $b.g$ be placed in the semicircle $a.b.g$ upon the diameter $a.g$, and let its chord be given; and let point $d$ divide the arc $b.g$ into equal parts by the 29th proposition of the third book. I say that the chord $b.d$ or $d.g$ becomes given. ¶ For having drawn the chords $a.b, b.d,$ and $d.g$, and by the 12th proposition of the first book, let the perpendicular line $d.e$ go from point $d$ onto the diameter $a.g$. It must first be shown that $e.g$ is half of the excess of line $a.g$ over $a.b$, thus: Let $a.e$ be equal to $a.b$ by the third of the first book. With $d.e$ drawn, the two sides $d.a$ and $a.b$ of triangle $d.a.b$ are equal to the two sides $d.a$ and $a.e$ by the last of the sixth book, or by the 26th of the third book, because the arcs subtending the said angles are equal. Therefore, by the fourth of the first book, the base $b.d$ is equal to the base $d.e$.

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A circular diagram with an inscribed triangle, labeled with handwritten numbers "29. 3." and "12. p". Below it is text in a different hand: "D. To divide three arcs into equal parts".

A circular diagram on the bottom right with an inscribed triangle and segments, labeled with the number "26. 3."

From a point marked outside to a given line
infinitely [extended], draw a perpendicular line

According to equal arcs, equal [chords] are taken.