Euclid's Elements of geometry, the first three books (the fourth, fifth, and sixth books) tr. from the Lat. To which is added, A compendium of algebra (A compendium of trigonometry).

PROPOSITION XI. PROBLEM.

To inscribe an equiangular and equilateral pentagon in a given circle (ABCDE).

Construct an isosceles triangle having each angle at the base double of the vertical angle (by Prop. 10, B. 4), and in the given circle inscribe ACE equiangular to it (by Prop. 2, B. 4), bisect the angles at the base, A and E, by the right lines AD and EB, and draw AB, BC, CD, De,

and EA. Therefore because each of the angles CAE and CEA is double of ECA (by Const.), and are bisected by the right lines AD and EB, the five angles CEB, BEA, ACE, CAD, and DAE shall be equal to one another, and therefore the arches upon which they stand are equal (by Schol. Prop. 29, B. 3), and therefore the right lines CB, BA, AE, ED, and DC, subtending these arches are also equal (by same Schol.), and therefore the pentagon ABCDE is equilateral. But since the arches AB and DE are equal, by adding the common arch BCD to both, the arch ABCD will be equal to the arch BCDE, therefore the angles AED and BAE standing upon them are also equal (by Schol. Prop. 29, B. 3), and it can be similarly demonstrated that all the remaining angles are equal, and therefore the pentagon ABCDE is also equiangular.

COR. 1.-Hence it appears that all equilateral figures inscribed in a circle, are also equiangular.

COR. 2. An equilateral and equiangular pentagon can be constructed upon a given line AE, by constructing upon it as a base, an isosceles triangle ACE, in which each of the angles at the base is double of the vertical angle (by Cor. Prop. 10, B. 4), and about it circumscribing a circle, the pentagon inscribed in this shall be constructed upon the given line.

PROPOSITION XII. PROBLEM.

To circumscribe an equiangular and equilateral pentagon about a given circle (ABCDE).

Let the points A, B, C, D, and E be the vertices of the angles of an equilateral and equiangular pentagon inscribed in the H given circle, and draw GH, HK, KL, LM, and MG, touching A the circle in these points, and GHKLM shall be an equilateral and equiangular pentagon circumscribed about the given circle. For draw FA, FG, FE, FM, and FD; and since in the triangles FGA, FGE, the sides GA and GE are equal, for they are touching the circle from the same point G, and the sides FA and FE are also equal, but FG is common to both, the angles FGA and FGE will be equal, and also AFG and EFG (by Schol. Prop. 8, B. 1), therefore AGE is double of FGE, and AFE double of GFE; it can be similarly demonstrated that DME is double of FME, and also DFE of MFE; but since the arches AE and ED are equal (by Const.), the angles AFE and EFD are equal (by Schol. Prop. 29, B. 3), and therefore the halves of them GFE and MFE are equal; but the angles FEG and FEM are also equal, and the side EF is common, therefore FGE and FME are equal, and the sides GE and EM (by Prop. 26, B. 1); therefore the right line GM is double of GE; it can be similarly demonstrated that GH is double of GA, but GE and GA are equal, and therefore GM and GH are also equal; it can be similarly demonstrated that the remaining sides are equal, therefore the pentagon GHKLM is equilateral: but since the angles DME

and AGE are double of FME and FGE, and FME is equal to FGE, DME will be also equal to AGE; and it can be similarly demonstrated that the remaining angles are equal, and therefore GHKLM is also equiangular.

SCHOL.-Any equilateral and equiangular figure can be similarly circumscribed about a given circle, by first inscribing such a figure, and then drawing tangents through its angles.

To inscribe a circle in a given equilateral and equiangular pentagon (ABCDE).

Bisect any two adjacent angles A and E by two right lines AF and EF meeting in F, and from F draw FG perpendicular to AE; the circle described from the centre F, with the interval FG, will be inscribed in the given pentagon. For draw FB, FC, and FD, and from F, to the sides, let fall the perpendiculars FH, FN, FM, and FL. Therefore since in the triangles AFB, AFE, the sides AB and AE are equal (by Hypoth.), and AF common, and the angles FAB and FAE equal (by Const.), the angles ABF and AEF shall be also equal (by Prop. 4, B. 1), but ABC and AED are equal (by Hypoth.), and therefore because AEF is half of AED (by Const.), ABF is also half of ABC; it can be similarly demonstrated that the remaining angles are bisected by right lines drawn from F; therefore in the triangles FBH and FBN, the angles FBH and FBN are equal, but the angles H and N are right angles, and the side FB, to which H and N are opposite, is

common to both, therefore the sides FH and FN are equal (by Prop. 26, B. 1); and it can, in the same manner, be demonstrated that all the perpendiculars are equal, therefore the circle described from the centre F, through G, shall pass through H, N, M, and L, and on account of the angles at G, H, N, M, and L being right angles, the sides of the given pentagon touch the circle at those points.

SCHOL.-Hence by the same method a circle can be inscribed in any equilateral and equiangular figure.

To describe a circle about a given equiangular and equilateral pentagon (ABCDE).

Bisect the angles A and E by the right lines AF and EF, and from the meeting point F, through either points A or E B describe a circle; this shall also pass through B, C, and D. For draw FB, FC, and FD, and in the triangles FAE and FAB, since the sides FA, AE, are

equal to the sides FA, AB, and the angle FAE is equal to the angle FAB (by Const.), the angles FBA and FEA shall be also equal (by Prop. 4, B. 1); therefore since the angles ABC and AED are equal (by Hypoth.), and FEA is half of AED (by Const.), FBA will be also half of ABC, and therefore ABC is bisected by FB; it can be similarly demonstrated that the remaining angles C and D are bisected; therefore because in the triangle AFE, the angles FAE and FEA are equal, for they are the halves of the equal angles BAE and AED, and therefore the sides FE and FA are equal (by Prop. 6, B. 1), and it can be similarly demonstrated

that all the remaining right lines FB, FC, and FD are equal, therefore the five lines FA, FB, FC, FD, and FE are equal; and therefore the circle described from the centre F, through A, shall pass through B, C, D, and E, and is circumscribed about the given pentagon.

SCHOL.-Hence by the same method a circle can be circumscribed about any equilateral and equiangular figure.

PROPOSITION XV. PROBLEM.

To inscribe an equilateral and equiangular hexagon in a given circle (ABCDEF).

Assume the centre G of the given circle, and draw the diameter AGD, from the centre A, describe a circle through G, and through the intersections B and F draw BE and FC diameters to the given circle drawing AB, BC, CD, DE, EF, and FA, the equi

lateral and equiangular hexagon

For

ABCDEF shall be inscribed in the given circle. because AB and AG are the radii of the same circle BGF, they are equal, and since GA and GB are the radii of the same circle ABCDEF, they are equal; the triangle BGA is therefore equilateral, and therefore the angle BGA is one third of two right angles (by Cor. 4, Prop. 32, B. 1); AGF is similarly equilateral, and the angle AGF is also a third part of two right angles; but BGA and AGF together with FGE are equal to two right angles (by Cor. 1, Prop. 13, B. 1), and therefore FGE is one third part of two right angles; therefore the three angles BGA, AGF, and FGE are equal, and therefore the vertical angles EGD, DGC, and CGB are equal, and therefore the six angles at G are equal, and therefore the arches on which they stand are also equal

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Euclid's Elements of geometry, the first three books (the fourth, fifth, and ...