A Snippet of History: Application of Ptolemy's Theorem

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The discussion focuses on the application of Ptolemy's Theorem as presented in "De Revolutionibus Orbium Coelestium: Liber Primus," highlighting its visual proof of a trigonometric identity. Two main questions arise: the reasoning behind the upper chord being represented as ##\sin(\beta-\alpha)## and the quest for a purely geometrical proof of the relationship between chords and diameters. A construction involving a unit diameter and a right triangle is suggested to explore these concepts. The translation of a Latin passage elaborates on the relationships between the chords and diagonals within a semicircle. The conversation emphasizes the mathematical intricacies and historical significance of Ptolemy's work.
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Thought following might be of interest - application of Ptolemy's Theorem from "De Revolutionibus Orbium Coelestium: Liber Primus".

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Amazing visual proof of the trigonometric identity, but 2 things are not clear to me.

1. Why is the upper chord ##\sin(\beta-\alpha)##?
2. What would a purely geometrical (i.e. trigonometry-free) proof of the equality "chord times diameter equals diagonals' product minus the product of the other chords" be?
 
dextercioby said:
Amazing visual proof of the trigonometric identity, but 2 things are not clear to me.

1. Why is the upper chord ##\sin(\beta-\alpha)##?
2. What would a purely geometrical (i.e. trigonometry-free) proof of the equality "chord times diameter equals diagonals' product minus the product of the other chords" be?
1. Construct a unit diameter from either end of the chord. Complete right triangle by joining end of diameter to other end of chord.
2. Translate the Latin :wink: (don't worry - I'll type out a translation from Stephen Hawking's book "On the Shoulders of Giants")
 
In the triangle ABCD with diameter AD, let the straight lines AB and AC subtending unequal arcs be given. To us, who wish to discover the chord subtending BC, there are given by means of the aforesaid (Porism aka Pythagoras Thm) the chords BD and CD subtending the remaining arcs of the semi-circle and these chords bound the quadrilateral ABCD in the semicircle. The diagonals AC and BD have been given together with the three sides AB, AD and CD. And as has already been shown:

rect AC,BD = rect AB,CD + rect AD,BC

Therefore

rect AD,BC = rect AC,BD - rect AB,CD

Accordingly in so far as the division may be carried out:

(rect AC,BD - rect AB,CD)/AD = BC

Further when, for example, the sides of the pentagon and the hexagon are given from the above, by this computation a line is given subtending 12 degrees - and it is equal to 20791 parts of the diameter.
 
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