For the above figures will be added on and these following, not inconveniently: but with some of these there may be a certain one with a little more work, we will be including these sides to the extent of showing them the most; for from those which are being propounded here, it will be possible also for the rest to be completed, by those who will have wished to investigate that [aspect] precisely.

If from four lines in continuing proportion, with the [sum of the] biggest and the smallest being equal to [the sum of] double the larger mean and the smaller mean: the larger sides of a triangle being equal, with the base being equal to the larger mean, and each angle to the base will be the triple of the remaining one¹. And the base will be the side of the heptagon inscribed in the circle with the triangle².

Let ANFBCPQ be a regular heptagon inscribed in the circle³; and let AB, BC, BD, DE, or HO be continued proportionals: The sum of AB and HO will be equal to the sum of the lines AO, HD, BD; that is the double of the larger mean BC, and of the smaller mean BD. For the triangles BAC, BCD are equiangular ; therefore BC and CD are equal; likewise AF, AD, and FC are equal: and if making the angles GAD, GDA equal [choose G to insure this], the triangles AGD, FBC will have equal sides; and HO, DE are equal.

Setting⁴ AB, 1. BC will be 1 (1): BD, 1 (2):
DE, 1 (3); And 1 + 1 (3) will be equal to 2 (1) + 1 (2).

If the radius of the circumscribed circle XC is being set to 1:
BC will be the side of the heptagon, 867767478235;
CF the chord subtending two sides will be 1563662964936;
AB the chord subtending three sides will be 1949855824364 .

If from four lines with continuing proportions, [the sum of] the maximum and the minimum is equal to three times the larger mean: with the largest sides of the triangle being equal, the base equal to the larger mean, and both the base angles are four times the other angle: and the base will be the side of a regular nonagon inscribed in a circle with the triangle.

Let APCBR be the nonagon, and let AB,BC,BD,DE be continuing proportionals. [The sum of] AC and HO will be equal to the [sum of] the lines AO, HG, GC: that is the triplicate of the line BC⁵. For the angles BAC, BCD are equal and therefore BC, CD are equal. And if RM be drawn parallel to FC, the angles FCG, CGM will be equal; from which also [angles] GMC and GCM are equal; [triangles] APM, ARF, RBC are equal as they stand on the periphery. And drawing the line AM, the triangles AMG, AMP are equal angled and with equal sides. And the triangles AQP, ANG, FBC have equal sides: and the lines AO, HG, CB; likewise HO, DE are equal.
Setting⁶ AB, 1; BC, 1.(1). BD, 2. (2); DE, 1(3). 3 (2) and 1 + 1(3) will be equal.

The side⁷ of the [regular] Pentadecagon inscribed in the circle with radius 1, the side [is found from root of the three numbers] ⁷/₄ - .⁵/₁₆ - . bin ¹⁵/₈ - . ⁴⁵/₆₄, which is equal to the chord subtended by an angle of 24, 41582338164.
The side of the 24-gon is . bin 2. - . bin. 2 + . 3. Or, . bin. ¹/₂ +.¹/₂ - .bin.³/₂ - . ⁹/₈. Or the root of the three numbers [. trin.] 2 - .¹/₂ - .³/₂. Which is equal to the length of the chord subtended by the angle15, 26105238444 .
The side of the 30-gon is . bin ⁵/₈ - . bin. ⁴⁵/₆₄ - . ⁵/₁₆ - ¹/₄ . Or, . trin. ⁹/₄ - .⁵/₁₆ - .bin.¹⁵/₈ + . ⁴⁵/₆₄. Which is equal to the length of the chord subtended by the angle12< degrees>, 2090569265353.

In these figures, as with those above, the sides could be found, if the radius of the circumscribed circle had been given, or the side of any polygon (from these which are included in this chapter or the previous one) inscribed in the same circle as that, of which the side is sought, that with a single example should suffice to show the most.
Let the side of the octagon be given as 6: the side of the pentadecagon is required, inscribed in the same circle as the given octagon.

¹AB/BC = BC/BD = BD/DE = 1/. The larger mean BC = AB, and the smaller mean
BD = ²AB; while AB is the largest side, and DE = ³AB is the smallest side.

²The isosceles triangle Briggs has in mind has base BC, equal base angles of 77¹/₇, and an apex angle of 25⁵/₇.

³Having noted that ABC is isosceles. AOG is constructed similar to ABC with AO= AB, and DGH is congruent to this triangle, but inverted. The triangle GHO is congruent to BDE, and CBD is congruent to AOG and DGH. The proof of all of this can be left as an exercise, but it can be seen that these smaller triangles have sides parallel to particular sides or diagonals of the figure.

⁴It can be seen from the diagram that AB can be dissected up into lengths of sides of these lesser triangles; thus :-
AB = AO + DH + BD - OH = AB + AB + ² - ³AB. Hence, 1 + 3 = 2 + ².
we will solve the cubic x³ - x² - 2x + 1 = 0, where x = .
[Note the notation in use at the time to describe the powers of a variable, which is not itself written down: thus the cubic would be written as 1(3) - 1(2) - 2(1) +1 to be equal to zero].
The reduced cubic, by making the substitution z = x + ¹/₃, is :
z³ - ⁷/₃z + ⁷/₂₇ = 0. The standard substitution, due originally to Viète, for the general reduced cubic: z³ + pz + q= 0, sets z = w - p/(3w), to give the 6^th power equation:

(w³)² - (w³)q - p³/27 = 0, for which w³ = {q (q² + 4 p³ /27)}/2.

i/3

i(/32p/3)

It seems more likely than not that Briggs made use of his iterative method for solving polynomial equations, as set out in the Trigonometria Britannica, Chapters III, IV, V, which is now called Newton's Method.

⁵In the construction, the point N is located to make the angles AGN = BAG; the triangles FBC and ANG are congruent, as the triangles BDE and HNO are also.

⁶ In modern terms, for the first set of proportions, we have
AB/BC = BC/CD = CD/DE = 1/,
and initially set AB = 1, then BC = , BD = BC² = ², DE = BD²/BC = ³.
Briggs' equation then becomes: AO + HO = 1 + ³; while 3BC = 3.
Hence: 1 + ³ = 3.
we will solve the cubic x³ - 3x + 1 = 0, where x = .
The cubic is already in the correct form so x = z. The standard substitution used above,
with p = -3, and q = -1, sets
z = w + 1/w, to give the 6^th power equation:

(w³)² + (w³) +1 = 0, for which w³ = {-1 i(3)}/2 = eⁱ.

i/3

i(/32p/3)

⁷The detailed working of these results, and similar ones in succeeding chapters, have not been included in these brief notes.