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*Concerning How the Subtended Chords Are Sought in the Writings of the Ancients.*

- . Prop. Ptol.

Ptolemy^{1}showed First, following Euclid, the way the [lengths of the] sides of the particular figure to the inscribed circle ought to be found, namely of the equilateral triangle by Prop. 12. Book 13. Of the square by Prop. 6. Book 4. Of the Pentagon, Hexagon, and Decagon by Prop 9. and 10. Book 13.

- Prop. Ptol.

Secondly, by Prop. 30, Book 3: from Euclid, he teaches,*the Square of the Given Subtended Chord, taken from the Square of the given Diameter, to leave the Square of the Complement of the Subtended*[*Chord*]*to the Semi-circle*. And according to this method the Subtended Chords of 60, 90,120, 36, 72 and of the Complements 144 and 108 degrees have been found. - Thirdly, this
*Lemma*is proposed according to the need in subsequent proofs.*If the quadrilateral*[*lit. quadrangle*]*being inscribed in the Circle, the Rectangle comprising the Diagonals being equal to the*[*the sum of the*]*Rectangles comprising the opposite sides.*As for the Circle

*ABCD*, let the Diagonals be*DB*, 25;*AC*, 20. The Rectangle from the Diagonals 500 being equal to [the sum of] the Rectangles*AB*,*DC*, 360. And*AD*,*BC*, 140. For the line*BE*being drawn such that the angles*ABE*,*DBC*should be equal. The Triangles*DBC*,*ABE*are similar; because the Angles*CDB*,*CAB*being equal, since they shall be from the same section, by Prop. 21, Book 3. and*CBD*,*EBA*shall be equal from the construction; the remaining therefore being equal by Prop. 31, Book 1.*BD*,*DC*:*BA*,*AE*are therefore in proportion: and the Rectangles*BD*, AE;*DC*,*BA*equal from Prop.16, Book 6. Likewise, the Triangles*BDA*,*BCE*are similar, because the Angles*BCE*,*BDA*are on the same section, and therefore equal. And as*ABE*,*CBD*shall be equal from the construction, with the common part*DBE*taken away, the remaining [angles]*CBE*,*DBA*being equal; and therefore*CB*,*CE*:*BD*,*DA*, are proportionals, and the rectangles*CB*, DA;*CE*,*BD*being equal; but [the sum of] the rectangles*DB*, EA;*DB*,*CE*, was equal to the rectangle*DB*,*CA*by Prop. 1, Book 2, Euclid: and therefore the Rectangle with the Diagonals*DB*,*CA*being equal to [the sum of] the Rectangles*DC*,*BA*, and*BC*,*DA*, which had to be shown^{2}.

- Fourthly, as permitted from the above
*Lemma*, from two given arcs with unequal Subtended Chords, the Subtended Chord of the difference being found.Let the given Diameter [Figure 2-2] be

*AB*, 20.Of the Inscribed [Chords]

*AD*, 12. AC 53/5.*CD*being sought. Firstly,*DB*and*CB*should be found by the 2nd Prop. Since the angles*ADB*,*ACB*shall be in the semi-Circle, they shall be right, by Prop. 30, Book 3. and therefore if the Square*AD*, 144, being taken from the Square*AB*, 400, there will remain the Square*DB*, 256. By Prop. 47, Book 1.*DB*will be therefore 16. By the same method, by taking the Square*AC*, 319/25 from the Square*AB*, 400, there will remain the square*CB*, 36816/25, or 368__64__. And CB will be 19__2__. Therefore being given*AB*, 20;*AD*, 12;*AC*, 5__6__; and by finding*DB*, 16.*CB*, 19__2__.*CD*being sought by the preceding Lemma. The rectangle^{3}*AC*,*DB*, 89__6__, being taken from the rectangle*Ad*,*CB*, from the diagonals, 230__4__, will leave the rectangle 140__8__, being taken from the diameter*AB*, 20*CD*. Therefore with 140__8__being divided by 20, the Quotient 7__04__will be the length of the straight line*CD*, being sought.

By the same Lemma: From two given arcs with Subtended Chords, the Subtended Chord of the sum being found^{4}.Let the given diameter [Figure 2-3] be

*AB*, 20.*AC*, 12.*AD*, 5__6__.*DC*being sought.*CB*,*DB*especially being sought, as before^{5}: and*CB*will be 16,*DB*, 19__2__We have therefore*AB*,*AC*,*AD*, being given; and*CB*,*DB*being found. The rectangles*AD*,*CB*, 89__6__, and*AC*,*BD*, 230__4__being taken, of which the sum will be 320.__0__, to which the rectangle being comprised of the Diameter and*DC*, by the preceding Lemma, Prop. 3. Therefore given the Diameter*AB*, 20. Let it divide 320.__0__, the rectangle being taken from the Diameter and*DC*, the Quotient 16. will be the Subtended Chord*DC*being sought^{6}.

** Notes On Chapter Two**

^{1} The First Lemma looks at the problem of finding the lengths of the sides of some regular figures inscribed in a circle of unit radius. The Second Lemma applies the Theorem of Pythagoras to the sides of the appropriate right angled triangle to determine the ratios for the specified angles.

^{2} In modern terminology, we have *AE*.*BD* = *AB*.*CD*, and *CE*.*BD* = *CB*.*DA*; hence,

(*AE* + *EC*).*BD* = *AB*.*CD* + *CB*.*DA*, or *AC*.*BD* = *AB*.*CD* + *CB*.*DA*, as required. See the *CRC Handbook of Modern Mathematics*, p.84, for the 'whole story' on this theorem from a modern perspective. This theorem is the main 'workhorse' used by Ptolemy in the construction of his Table of Sines alluded to in the first chapter.

^{3} These are referred as 'oblongs' in the text always.

^{4} This should be called the 5^{th} Lemma, or the converse of Lemma 4.

^{5} That is, we find the remaining sides of the cyclic quadrilateral first.

^{6} This is a rather perfunctory and incomplete look at Ptolemy's method. Ptolemy uses these results to find the lengths of half chords in terms of the known chords of larger angles: in this way he subdivides 12 formed from the difference of 72 from the regular pentagon and 60 from the regular hexagon to find the half chords corresponding to 60, 30, ^{11}/_{20}, ^{3}/_{40} and eventually for ^{1}/_{20}, which he uses as the unit to build up his table. See his *Almagest* I for details. Briggs, however, wishes to move on to his own methods...

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