The Elements of Euclid; viz. the first six books, together with the eleventh and twelfth. Also the book of Euclid's Data. By R. Simson. To which is added, A treatise on the construction of the trigonometrical canon [by J. Christison] and A concise account of logarithms [by A. Robertson].

8. 1.

the triangles BFK, CFK, the two BF, FK are equal to the Book IV. two CF, FK; and the base BK is equal to the base KC; therefore the angle BFK is equal to the angle KFC, and the angle BKF to FKC; wherefore the angle BFC is double of the angle KFC, and BKC double of FKC: For the same reason, the angle CFD is double of the angle CFL, and CLD double of CLF: And because the circumference BC is equal to the circumference CD, the angle BFC is equalf 27. 3, to the angle CFD; and BFC is double of the angle KFC, and CFD double of CFL; therefore the angle KFC is equal to the angle CFL; and the right angle FCK is equal to the right angle FCL: Therefore, in the two triangles FKC, FLC, there are two angles of one equal to

which is adjacent to the equal angles in each, is common to both; therefore the other sides shall be equals to the 26. 1. other sides, and the third angle to the third angle: Therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC: And because KC is equal to CL, KL is double of KC; In the same manner it may be shown that HK is double of BK: And because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK shall be equal to KL: In like manner, it may be shown that GH, GM, ML are each of them equal to HK, or KL: Therefore the pentagon GHKLM is equilateral. It is also equiangular; for, since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and KLM double of FLC, as was before demonstrated, the angle HKL is equal to KLM: And in like manner it may be shown, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM: Therefore the five angles GHK, HKL, KLM, LMG, MGH, being equal to one another, the pentagon GHKLM is equiangular: And it is equilateral, as was demonstrated; and it is described about the circle ABCDE. Which was to be done.

Book IV.

9. 1.

PROP. XIII. PROB.

To inscribe a circle in a given equilateral and equiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to inscribe a circle in the pentagon ABCDE.

Bisect the angles BCD, CDE by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE: Therefore, since BC is equal to CD, and CF common to the triangles BCF, DCF the two sides BC, CF are equal to the two DC, CF; and the angle BCF is equal to the angle DCF; therefore the base 4. 1. BF is equal to the base FD, and the other angles to the other angles, to which the equal sides are opposite; therefore the angle CBF is equal to the angle CDF: And because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF; CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bisected by the B straight line BF: In the same manner it may be demonstrated, that the angles BAE, AED, are bisected by the straight lines 'AF, FE: From 12. 1. the point F, draw FG, FH,

K D FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA: And because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; in the triangles FHC, FKC there are two angles of one equal to two angles of the other, and the side FC, which is opposite to one of the equal angles in each, is 26. 1. common to both; therefore the other sides shall be equal, each to each; wherefore the perpendicular FH is equal to the perpendicular FK: In the same manner it may be demonstrated; that FL, FM, FG are each of them equal to FH, or FK: Therefore the five straight lines FG, FH, FK, FL, FM are equal to one another: Wherefore the circle described from the centre F, at the distance of one

of these five, shall pass through the extremities of the other Book IV. four, and touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right angles; and that a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches the 16. 3. circle: Therefore each of the straight lines AB, BC, CD, DE, EA touches the circle: wherefore it is inscribed in the pentagon ABCDE. Which was to be done.

PROP. XIV. PROB.

To describe a circle about a given equilateral and equiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to describe a circle about it.

Bisect the angles BCD, CDE by the straight lines CF, . 9. 1. FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E. It may be demonstrated, in the same manner, as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA, FE: And because the angle BCD is equal to the angle CDE, and that

FCD is the half of the angle BCD, and CDF, the half of CBE; the angle FCD is equal to FDC; wherefore the side CF is equal to the side FD: In like manner 6.1. it may be demonstrated that FB, FA, FE, are each of them equal to FC or FD: Therefore the five straight lines FA, FB, FC, FD, FE are equal to one another; and the circle described from the centre F, at the distance of one of them, shall pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done.

Book IV.

PROP. XV. PROB.

See N. To inscribe an equilateral and equiangular hexagon in a given circle.

Let ABCDEF be the given circle; it is required to inscribe an equilateral and equiangular hexagon in it.

Find the centre G of the circle ABCDEF, and draw the diameter AGD: and from D, as a centre, at the distance DG, describe the circle EGCH, join EG, CG, and produce them to the points B, F; and join AB, BC, CD, DE, EF, FA; The hexagon ABCDEF is equilateral and equiangular.

Because G is the centre of the circle ABCDEF, GE is equal to GD: And because D is the centre of the circle EGCH, DE is equal to DG; wherefore GE is equal to ED, and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG, are equal to one another, because the angles at the base of an isosceles 5.1. triangle are equala; and the three angles of a triangle 32. 1. are equal to two right angles; therefore the angle EGD is the third part of two right angles: In the same manner it may be demonstrated, that the angle DGC is also the third part of two right angles: And because the straight line GC makes with EB

the adjacent angles EGC, CGB 13. 1. equal to two right angles: the F remaining angle CGB is the third part of two right angles: therefore the angles EGD, DGC, CGB are equal to one another: And to 415. 1. these are equal the vertical oppoE site angles BGA, AGF, FGE: Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another: But 26. 3. equal angles stand upon equal

circumferences; therefore the six

circumferences AB, BC, CD, DE, EF, FA are equal to one another: And equal circumferences are subf 29.3. tended by equalf straight lines; therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular; for, since the cir

cumference AF is equal to ED, to each of these add the Book IV. circumference ABCD; therefore the whole circumference FABCD shall be equal to the whole EDCBA: And the angle FED stands upon the circumference FABCD, and the angle AFE upon EDCBA; therefore the angle AFE is equal to FED: In the same manner it may be demonstrated that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED: Therefore the hexagon is equiangular; and it is equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Which was to be done.

COR. From this it is manifest, that the side of the hexagon is equal to the straight line from the centre, that is, to the semidiameter of the circle.

And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonstrated from what has been said of the pentagon; and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon.

PROP. XVI. PROB.

To inscribe an equilateral and equiangular quin- See N. decagon in a given circle.

Let ABCD be the given circle; it is required to inscribe an equilateral and equiangular quindecagon in the circle. ABCD.

Let AC be the side of an equilateral triangle inscribed a • 2. 4. in the circle, and AB the side of an equilateral and equiangular pentagon inscribed in the same; therefore, of such * 11.4. equal parts as the whole circumference ABCDF contains fifteen, the circumference ABC,

being the third part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three; therefore BC their difference contains two of the same parts: Bisect BC in E; therefore BE, EC are, each of them, the fifteenth part of the whole circumference ABCD :

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The Elements of Euclid; viz. the first six books, together with the eleventh ...