cause bd touches the circle, and dc is drawn from the point of contact d, the angle b dc is equal (iii. 32) to the angle dac in the alternate segment of the circle ; to each of these add the angle cda; therefore the whole angle bda is equal to the two angles cda, dac; but the exterior angle bed is equal (i. 32) to the angles cda, dac; therefore also bda is equal to bcd; but bda is equal (i. 5) to the angle cbd because the side ad is equal to the side a b; therefore cbd, or dba is equal to bcd; and consequently, the three angles bda, dba, bcd, are equal to one another ; and because the angle dbc is equal to the angle bed, the side bd is equal (i. 6) to the side dc; but bd was made equal to ca, therefore also ca is equal to cd, and the angle cda equal (i. 5) to the angle dac; therefore the angles cda, dac together, are double of the angle dac: but bcd is equal to the angles cda, dac, therefore also bed is double of d ac, and bod is equal to each of the angles bda, dba ; each therefore of the angles bda, dba is double of the angle da b, wherefore an isosceles triangle abd is described, having each of the angles at the base double of the third angle. Which was to be done. PROPOSITION XI.- PROBLEM. Describe (iv. 10) an isosceles triangle fg h, having each of the angles at g, h double of the angle at f; and in the circle abcde inscribe (iv. 2) the triangle acd equiangular to the tri- the angle at f, and each of the angles acd, cda equal to the angle at g db; and join a b b c, g de, e der abcde is the pentagon required. Because each of the angles a cd, cda is double of cad, and are bisected by the straight lines ce, db, the five angles dac, a ce, ecd, cdb, b da are equal to one another ; but equal angles stand upon equal circumferences (iii . 26); therefore the five circumferences, a b, bc, cd, de, e a are equal to one another : and equal circumferences are subtended by equal straight lines (iii. 29); therefore the five straight lines a b, bc, cd, de a e are equal to one another. Wherefore the pentagon a bcde is equilateral. It is also equiangular; because the circumference a b is equal to the circumference de; if to each be added bcd, the whole abcd is equal to the whole ed cb: and the angle a ed stands on the circumference a b c d, and the angle ba e on the circumference e dcb; therefore the angle bae is equal (iii. 27) to the angle aed. For the same reason, each of the angles a b c, b c d, cde is equal to the angle ba e, or a ed; therefore the pentagon a bede is equiangular; and it has been shewn that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done. PROPOSITION XII.-PROBLEM. To describe an equilateral and equiangular pentagon about a given circle. LET a b c d e be the given circle ; it is required to describe an equilateral and equiangular pentagon about the circle a bcde. Let the angles of a pentagon, inscribed in the circle, by the last proposition, be in the points a, b, c, d, e, so that the circumferences a b b c, cd, de, ea are equal (iv. 11); and through the points a, b, c, d, e, draw gh, hk, kl, lm, mg, touching (iii. 17) the circle ; take the centre f, and join fb, fk, fc, fl. fd; and because the straight line kl touches the circle abcde in the point c, to which fc is drawn from the centre f, fc is perpendicular (iii. 18) to kl, therefore each of the angles at c is a right angle: for the same reason, the angles at the points b, d are right angles : and because fok is a right angle, the square of fk is equal (i. 47) to the squares of fc, ck: for the same reason, the square of fk is equal to the squares of fb, bk: therefore the squares of fc, ck are equal to the squares of fb, bk, of which the square of fc is equal to the square of fb; the remaining square of ck is therefore equal to the remaining square of bk, and the straight line ek equal to bk : and because fb is equal to fc, and fk common to the triangles bfk, cfk, the two bf, fk are equal to the two cf, fk, and the base bk is equal to the base kc; therefore the angle bfk is equal (i. 8) to the angle k fc, 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 old double of clf: and because the circum 8 ference b c is equal to the circumference ed, the angle bfc is equal (iii. 27) to the angle cfd; and bfc is double of the angle kfc, and cfd double of cfl; therefore h the angle kfc is equal to the angle cfl; and the right angle fek is equal to the right angle fcl: therefore, in the two triangles fkc, flc, there are two angles of d one equal to two angles of the other, each to each, and the side fc, which is adjacent to the equal angles in each, is common to с 1 both; therefore the other sides shall be equal (i. 26) to the other sides, and the third angle to the third angle : m : therefore the straight line kc is equal to cl, and the angle fkc to the angle flc: and because k c is equal to cl, kl is double of k c. In the same manner it may be shewn that hk is double of bk : and because bk is equal to kc, as was demonstrated, and that kl is double of kc, and kh double of bk, hk shall be equal to kl. In like manner, it may be shewn 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 fk c is equal to the angle flc, and that the angle h kl is double of the angle fkc, and klm double of flc, as was before demonstrated, the angle hk) is equal to klm: and in like manner it may be shewn that each of the angles k hg, hgm, g ml is equal to the angles hkl or klm: therefore the five angles ghk, h kl, klm, 1mg, 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. PROPOSITION XIII.- PROBLEM. To inscribe a circle in a given equilateral and equiangular pentagon. LET a bcde be the given equilateral and equiangular pentagon ; it is required to inscribe a circle in the pentagon a bcde. Bisect (i. 9) the angles bcd, cde by the straight lines cf, d f, 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 b cf, dcf the two sides bc, cfare equal to the two dc, cf; and the angle bcf is equal to the angle dcf; therefore the base bf is equal (i. 4) 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 cd f, and that cde is equal to cba, a and cdf to cbf; cba is also double of the angle cbf; therefore the angle a bf is equal to the angle cbf; wherefore the angle a b c is bisected by the straight line bf. In the same manner, it may be demonstrated that the angles bae, aed, are bisected by the straight lines a f, fe: h from the point f, draw (i. 12) fg, fh, fk, fl,fm perpendiculars to the straight lines a b, bc, cd, de, e a: and because the angle h cf is equal to k cf, and the right d k angle fhc equal to the right angle fkc; in the triangles fhc, fk c 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 common to both; therefore the other sides shall be equal (i. 26), 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 g : the centre f, at the distance of one of these five, shall pass through the extremities of the other four, and touch the straight lines a b, bc, cd, de, e a, 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 (iii. 16) the circle ; therefore each of the straight lines ab, bc, cd, de, e a touches the circle : wherefore it is inscribed in the pentagon abcde. Which was to be done. PROPOSITION XIV.- PROBLEM. 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 (i. 9) the angles bcd, cde by the straight lines cf, 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 pro a position, that the angles cba, ba e, a ed are bisected by the straight lines fb, fa, fe. And because the angle bed is equal to the angle by cd e, and that fcd is the half of the angle bcd, and cdf the half of cde; the angle fcd is equal to fdc; wherefore the side of is equal (i. 6) to the side fd. In like manner, 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 a bcde. Which was to be done. PROPOSITION XV.—PROBLEM. 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 a b c d ef is equilateral and equiangular. Because is the centre of the circle abcdef, ge is equal to gd. And becauseod is the centre of the circle egch, đe is equal to ag; wherefore ge is equal to ed, and the triangle egd is equilateral ; and therefore its three angles eg d, gde, deg are equal to one another, because the angles at the base of an isosceles triangle are equal (i. 5); and the three angles of a triangle are equal (i. 32) 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 equal (i. 13) to two right angles; the remaining angle cgb is the third d part of two right angles : therefore the angles egd, dgc, cg b are equal to one another. And to these are equal (i. 15) the vertical opposite angles bga, agf, fge, therefore the six angles egd, dgc, cgb, bga, agf, fge are equal h to one another. But equal angles stand upon equal (iii. 26) circumferences; therefore the six circumferences ab, bc, cd, de, ef, fa are equal to one another. And equal circumferences are subtended by equal (iii. 29) 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 circumference af is equal to ed, to each of these add the circumference abcd; therefore the whole circumference fa bed shall be equal to the whole edcba. And the angle fed stands upon the circumference fabcd, and the angle a fe upon edcba; therefore the angle a fe 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 a fe or fed: therefore the hexagon is equianglar; and it is equilateral, as was shewn; 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. : PROPOSITION XVI.- PROBLEM. To inscribe an equilateral and equiangular quindecagon in a given circle. LEt abcd be the given cirele; it is required to inscribe an equilateral and equiangular quindecagon in the circle abcd. Let ac be the side of an equilateral triangle inscribed (ii. 4) in the circle, and a b the side of an equilateral and equiangular pentagon inscribed (iv. 11) in the same; therefore, of such equal parts as the whole circumference abcdf contains fifteen, the circumference a b c, being the |