With regular figures applied to a circle. If two figures of the same kind are being ascribed to the same circle, the one inside, the other outside: a third, of which the sides are of number equal to the remaining sides being taken together, with the same inscribed circle, [the area] will be the mean proportional between the rest.

Let two triangles DCB, GEF be adscribed to the same triangle, to which is being inscribed the hexagon DICHBR : I assert the hexagon¹ to be the mean proportional between the adscribed triangles.

For the triangles ACM, ACF are similar: and AM, AC ( or AH), AF are in continued proportion. And therefore the triangles ACM, ACH, ACF as they have the same altitude, they are as their bases: that is they are in continued proportion, as their bases, and for the figure, of these triangles of the hexagon, they are also in continued proportion. Let the radius of the circle AD be 1. BC will be the side of the inscribed triangle .3, by Prop.12, Book13, Euclid. The perimeter of the triangle .27. The semi-perimeter .6³/₄. AM ¹/₂. The area of the triangle . ²⁷/₁₆.
AO . ³/₄. DR 1. Triangle DAR . ³/₁₆. The area of the hexagon DICHBR is .²⁷/₄ ; EF .12; AB, 1; EAF .3; and EGF .27.


The line DR of the inscribed hexagon 1. The circumscribed line NS .⁴/₃. For the lengths AO .³/₄, DR 1, AR, 1, NS .⁴/₃ are in proportion. and the area of the triangle ANS .¹/₃, and of the circumscribed hexagon .12. Again, the area of the inscribed dodecagon is 3. For XO is 1 - .¹/₄ and the square of XO 1³/₄ - .3, and the square of DO ¹/₄. It is therefore the square of DX, 2 - .3; and the square of half the line DX is ¹/₂ - .³/₁₆. and the square of the perpendicular from the point A to the line DX is ¹/₂ + .³/₁₆. The area of the triangle ADX is .¹/₁₆ , or ¹/₄, and the whole area of the inscribed dodecagon is 3. Which is the mean proportional between the inscribed hexagon .27/4 and the circumscribed hexagon .12.

With the inscribed square BCDE the side has length .2, area 2; the circumscribed square FGHK has side of length 2, and area 4.
With the inscribed octagon² the side DP, .bin.2 - .2 . (For AQ is .¹/₂. PQ 1 - .¹/₂, and the square of PQ 1¹/₂ - .2. The square of DQ: ¹/₂; therefore the square of DP: 2 - .2 ). The square DO (¹/₄ of the square DP) will be ¹/₂ - .¹/₈, which being taken from the square of the radius AD 1, will give the square AO ¹/₂ + .¹/₈ and the product with the line AO .bin.¹/₂ + .¹/₈ times OD .bin.¹/₂ - .¹/₈ will be .¹/₈ the area of the triangle³ ADP. The area of the inscribed octagon will be .8. Which is the mean proportional between the area of the inscribed square 2, and the circumscribed square 4.


The line MN, of the circumscribed octagon .8 - 2. (because HA, HN being equal, and therefore DN is .2 - 1, and NM twice the same, is .8 - 2. But as HA, HN being equal, it is evident by Prop.5, Book 1, Euclid; because HAN has the value ³/₄ of the line by the construction, and HNA being equal to the angles NGA, NAG by Prop.32, Book 1, Euclid.) The radius AD is 1; the product of the radius by DN .2 - 1 is .2 - 1, being equal to the area of the triangle AMN, of which the octuplet
.128 - 8 being equal to the area of the circumscribed octagon.

With the inscribed hexadecagon [16-gon], the length of the side is .bin.2 - .bin.2 + .2
The length of the side of the inscribed pentagon is .bin.⁵/₂ - .⁵/₄, with the perpendicular to the centre multiplied by the line .bin.³/₈ + .⁵/₆₄ the product of this perpendicular with the half of the line of the pentagon will be the area of the triangle .bin.⁵/₈ + .⁵/₁₀₂₄. and the total area of the inscribed pentagon⁴ will be .bin.¹²⁵/₈ + .³¹²⁵/₁₀₂₄. The side of the circumscribed pentagon will be .bin.20 - .320. The product of the radius by the half of this side will be .bin.5 - .20, the area of the triangle AGH, and the total area of the circum-scribed pentagon will be .bin.125 - .12500.

The side of the inscribed decagon BP is .⁵/₄ - 1/2, of which the square is ³/₂ - .⁵/₄ . The square of the line BX is ³/₈ - .⁵/₆₄ and the square of AX is ³/₈ + .⁵/₆₄, the area of triangle ABP is
.bin.3/8 - .⁵/₁₀₂₄, and the total area of the decagon.bin.¹²⁵/₈ - .³¹²⁵/₆₄, which is the mean proportional between the areas of the inscribed and the circumscribed pentagons.

If there is shall be given a circle with diameter 9,and the sides of the inscribed and circumscribed triangles are being sought.

The sides of the triangle will be: inscribed 7794228 ; circumscribed 15588457.
If there shall be given the side 6 of a triangle: and being sought the diameters of the inscribed and circum-scribed circles.

The diameters of the circles will be: inscribed 3464203; circumscribed 6928406.
If there shall be given the area .243 of a triangle: and being sought the areas of the inscribed and circum-scribed circles.

The areas of the circles will be: inscribed 9424778; circumscribed (four times that of the inscribed) 37699112.

2. For the Circle and the Square.

For the circle and adscribed regular many-sided figures.

With these logarithms being found, if it is proposed of any these figures, being given either their side, or perimeter, or area; we will be able to find any of the other terms of this. Because we showed the triangle more spread out, so we will show the same with another single example from the rest one after another.
Let the given side of the regular octagon be of 7 parts. Being sought is the side, perimeter, and area of the pentagon, with the same circle as the inscribed octagon.
The side being sought thus may be found.
With the circle of which the radius is unity, the side of the inscribed octagon is 7653668647. But the side of the pentagon is 1175579504. And if the side of the octagon being taken as 7 parts, the side of the pentagon being sought will be the fourth proportional. Therefore the logarithms of the given terms are taken, so that the logarithm of the term being sought may be found (as in Ch. 15).

Where it is being observed, the logarithm of the first term(as this is less than unity) to be defective [ i.e. a negative number], as we showed in Ch. 10; and, because of this, is not to be taken from the sum of the means, but rather being added; and the logarithm of the fourth proportional will be 1,03147706, and the side sought 107516982.
This side, if we neglect the rule of proportion with logarithms, will be the root of the four numbers⁵ 122¹/₄ + . 7503¹/₈ - . 3001¹/₄ - .1500⁵/₈

By the same method the perimeter of the pentagon will be found, if for the third term the perimeter of the given pentagon is taken, and of this, the logarithm. As:

If we seek the area of the pentagon, with nothing other being given except the side of the octagon inscribed in the same circle: we should remember that similar plane figures, to be in the duplicate ratio with the logarithms of the same sides; and therefore the side of the octagon squared ought to be taken, as duly being able to establish the comparison. Nevertheless, with this matter, it is not necessary that we should be carefully with the squares themselves, because by Ch. 16 it will be easiest, as we will have the logarithms of that method,. As you see here:

If the area of the given decagon is 6, and the side of the octagon circumscribing the same circle is sought, of which the decagon is inscribed; they will be:

And thus in this way with a circle, and with these figures adscribed to the same circle, from a single term given, any other can be found by [using] logarithms.

¹Note: Briggs refers to figures according to the number of angles they contain; thus, 'sexangulum' for hexagon.

² . means 'latus' or side, being one of the abbreviations for the square root sign at this time, while bin. is short for 'bina', meaning 'of the pair', and indicates in this case that the square root of both terms is included. The expression thus means (2 - 2).

³Here we have a good example of the nature of algebraic manipulations at the time. In modern terms, we have
DO² = ¹/₄DP² = ¹/₂ - (¹/₈). Then AO² = ¹/₂ + (¹/₈), and the product
AO.OD = ( ¹/₂ + (¹/₈)) (¹/₂ - (¹/₈)) = (¹/₈), the area of ADP.

⁴As this development may not be so obvious as the previous constructions have been, we give a construction of the pentagon, as shown by Henry E. Dudeney in his Amusements in Mathematics, Nelson, 1917, P. 38. In this construction, the radius of the circumscribed circle is taken as unity. The mid-point A of the line BC is found, and the length AD used to mark off with compasses the equal length AE: of size (⁵/₂). The length EC is (5 -1)/2, and hence ED = (5/2 - 5/2), and the equal length FD marked off with the compasses as a side of the pentagon. The other sides are produced by drawing equal arcs around the circle. It suffices to show that this is indeed the length of side of the inscribed pentagon: for if the radius of the circumscribed circle is R, then the length of the side a is given by a = 2Rsin(/5), where sin(/5) = {(5 - 5)/8}, This expression for sin(/5) in terms of radicals follows from the expansion of sin 5 in terms of a sum of powers of sin , which can be factorised and the equation solved for sin : see e.g. CRC Concise Encyclopedia of Mathematics, p.1853. Thus, on setting R = 1, the radius of the in-circle to the pentagon is {(3+5)/8}, while the area of ABC is {(5 +5)/32}, etc. Thus, apart from a typographical error in the original, Briggs' working is correct, as we expect.


⁵This can be readily found from the result s₅ = s₈ sin(/5)/sin(/8) = 1075199.