Epitome of Ptolemy's Almagest

First [Book]

But $b.d$ is equal to $d.g$ by original: "per.28.tertij" Proposition 28 of the third book [of Euclid's Elements]. Therefore, triangle $e.d.g$ becomes an isosceles triangle (having two equal sides). Thus, by Proposition 4 of the first book, angle $d.e.g$ is equal to angle $d.g.e$. However, both angles at $a.d.z$ are right angles because $d.z$ is a perpendicular. Therefore, triangle $e.d.z$ is equiangular to triangle $g.d.z$. Hence, by Proposition 4 of the first book, $e.z$ becomes equal to $z.g$. But $e.g$ is the difference excessusThe "excess" or the remaining length when one segment is subtracted from another. of $a.g$ over $a.b$; therefore, $z.g$ is half of that difference. Through the corollary of the first [proposition] of this book, from the given chord $b.g$, the chord $a.b$ will become known. Therefore, $e.g$ will be known as the difference; so its half, namely $z.g$, will be given. Furthermore, since in the right-angled triangle $a.d.g$, by Proposition 30 of the third book, the perpendicular $d.z$ descends from the right angle to the base; therefore, by Proposition 8 of the sixth book, $d.g$ is the mean proportional between $a.g$ and $g.z$. Thus, by Proposition 16 of the sixth book, the product of $a.g$ and $g.z$ is equal to the square of $d.g$. Since $a.g$ and $g.z$ are given, $d.g$ (which was sought) will become known. By this doctrine, you will find the chords of many arcs: just as the chord of an arc of twelve degrees is known from the above, the chord of an arc of six degrees will now become known; from this, the chord of an arc of three degrees; from this, the chord of an arc of one and a half degrees; from this, the chord of an arc of a half and a quarter [three-quarters]; and so on for others.

Geometric diagram of a circle with inscribed lines and triangles labeled b, d, g, z, e.

Proposition VI.

Geometric diagram of a circle with an inscribed quadrilateral labeled b, g, d, e and center z. Handwritten annotations: "note", "to be made", "known", "motion".

Given the chords of two arcs in a semicircle, the chord of the arc composed of these two will also be known.
Let there be in circle $a.b.d$ (with center $z$ and diameter $a.z.d$) two arcs $a.b$ and $b.g$ with known chords $a.b$ and $b.g$ given. I say that the chord of the whole arc $a.g$ becomes known. For when the lines $a.g, b.d, g.d$ are drawn, as well as the diameter $b.z.e$ and lines $g.e$ and $d.e$, then by the corollary of the first [proposition] of this book, $b.d$ will be known from $a.b$, and $g.e$ will be known from $b.g$. Therefore, in the quadrilateral $b.g.d.e$, the diagonals $b.d$ and $g.e$ are given; and the two sides $b.g$ and $a.b$ are equal to $d.e$. The side $b.e$ is also known because it is the diameter of the circle. Therefore, by the first [proposition] of this book, the square of the side—namely $d.g$—will become known. Hence, from the corollary of the first [proposition] of this book, $a.g$ will be known, which is what was proposed. From these premises, the chords of all arcs in a semicircle increasing by one and a half degrees are made clear.

Proposition VII.

Geometric diagram of a circle with complex intersecting lines and arcs labeled b, g, z, d, a, e, t, s.

For unequal arcs in a semicircle, the ratio of the larger arc to the smaller is greater than the ratio of the chord of the larger to the chord of the smaller.
In a semicircle, let arc $b.g$ be larger than arc $a.b$. Let the chord of the larger be $b.g$ and the chord of the smaller be $a.b$. I say that the ratio of arc $b.g$ to arc $a.b$ is greater than the ratio of chord $b.g$ to chord $a.b$. For I shall divide angle $a.b.g$ into equal parts by line $b.d$ by Proposition 9 of the first book; and I shall extend $a.g$ intersecting $b.d$ at $e$. Also, $a.d$ and $d.g$ [are drawn], and by Propositions 28 and 25 of the third book, $a.d$ becomes equal to $d.g$. Since, by Proposition 3 of the sixth book, the ratio of chord $b.g$ to chord $a.b$ is as $g.e$ to $e.a$, and $g.b$ is greater than $a.b$, therefore $g.e$ is greater than $e.a$. Thus, point $z$ dividing $a.g$ into equal parts will be within $e.g$, and when $d.z$ is drawn, by Proposition 8 of the first book, both angles at $a.d.z$ will be right angles. Therefore, in triangle $e.z.d$, by Propositions 18 and 32 of the first book, side $d.e$ is greater than side $d.z$; and likewise in triangle $a.e.d$, side $d.a$ is longer than side $d.e$. Therefore, if we set $d$ as the center of a circle whose circumference passes through $e$, it is necessary that this periphery cuts $d.a$ passing below $a$ and does not reach $d.z$ passing above $z$. Let it therefore cut $d.a$ at $b$ and let the continued $d.z$ meet the periphery at $t$. Because, therefore, sector $e.d.t$ is larger than triangle $e.d.z$.

Series of geometric sketches at the bottom of the page. Left: an arc with labels. Center: a triangle inscribed in an arc with value "28 2/3". Right: a triangle in a circle with value "52 36" and additional line segments labeled "76.6" and "b", "c".