on which they stand are equal, and the lines subtending 267 those arcs (III. 29), therefore the hexagon ABCDEF is equilateral, and also, since it is inscribed in a circle, equiangular. ΠΟΡΙΣΜΑ. COROLLARY.-The side of a hexagon is equal to the radius of the circle in which it is inscribed. Annotations. Corollary 1.-Hereby an equilateral triangle DBF may very easily be inscribed in a given circle. Scholium. To make a true hexagon upon a given right line AB. Make an equilateral triangle BGA upon the given line, and from the centre G describe a circle through A and B; that circle shall contain the hexagon made upon the given line AB.-Andr. Tacquet. Corollary 2.-The area of a regular hexagon is six times that of the equilateral triangle described on the same straight line. Corollary 3.-Every equilateral figure inscribed in a circle is also equiangular; for its angles are formed by the chords of equal arcs, and, therefore, stand on equal arcs (III. 28). PROPOSITION XVI.—PROBLEM. Πρότασις ις'.—Εἰς τὸν δοθέντα κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι. 400 To inscribe an equilateral and equiangular quindecagon/ in a given circle CAD. D Let CD be the side of an equilateral triangle inscribed in the circle CAD, and CA the side of an equilateral pentagon also inscribed in the circle CAD; bisect the arch AD; the right line joining AB is the side of the B required quindecagon. For if the whole circumference be divided A into fifteen parts, the arch CD, since it is the third part of the whole circumference, contains five of these parts; in like manner the arch CA contains three of them, therefore the arch AD contains two; and therefore the arch AB is the fifteenth part of the whole circumference, and AB is the side of the required equilateral quindecagon. Any other way of dividing the circumference into any given parts is as yet unknown; wherefore, in the construction of ordinate figures, we are forced to have recourse to mechanic artifices.-Isaac Barrow. Corollary.-Only three kinds of ordinate polygons can be so placed at a point as to make a continuous plane surface. These polygons are the equilateral triangle, the square, and the hexagon.-Proclus. APPENDIX ON THE INSCRIBED, CIRCUMSCRIBED, AND EXSCRIBED CIRCLES OF A GIVEN TRIANGLE. To find the centres of the inscribed and exscribed circles of a triangle BAC. Construction.-Join L' and L the middle points of the arcs above and below the base with the vertex A by indefinite right lines. From L' cut off parts L'o" and L'o", equal to L'C or L'B; and from L cut off parts Lo and Lo', equal to LC or LB. Draw the remaining lines as indicated in the figure. Proof. The points o, o, o", o" are the centres of the circles. For, since Lo and LC are equal, the angles LoC and LCo are equal, and LoC is equal to the sum of oCA and oAC, or of oCA and OAB (since the arc BC is bisected in L), or of oCA and LCB, but LCo is equal to the sum of LCB and oCB; therefore oCA is equal oCB; the angle ACB is therefore bisected by Co, and as Co' is at right angles to oC; Co' bisects the angle below the base. But the angle BAC is bisected by the line Aoo'; hence it is evident that o is the centre of the inscribed circle, and o' of the circle exscribed to the base. Since L'o" and L'C are equal, the angles L'o'C and L'Co" are equal but L'o'C is equal to the sum of o′′CG, and o"GC and L'Co" is equal to the sum of L'CA and ACo", but o"GC and L'CA are each equal to L'LA, therefore ACo" is equal to o'CG, therefore Co" bisects the angle ACG, and, therefore, o" is the centre of the circle exscribed to the side AC; in the same way it may be proved that o" is the centre of the circle exscribed to the side BA. Corollary. It is evident from the Construction that the six lines which join the centre of the inscribed and exscribed circles are bisected by the circumference of the circumscribed circle. PROPOSITION B. If a right line drawn through the centre of the inscribed circle, be terminated by the circumference of the circumscribed circle, the rectangle under the segments shall be equal to twice the rectangle under the radii of the inscribed and circumscribed circles; and similarly if a line drawn through the centre of an exscribed circle be terminated by the circumference of the circumscribed circle, the rectangle under the segments shall be equal to twice the rectangle under the radius of the circumscribed and of that exscribed circle. Proof. Since the angles LLC and oAy stand on the same segment LC, they are equal, and therefore the right-angled triangles LL'C and oAy are equiangular, therefore the rectangles LC × oA and LL'xoy are equal (Exercise 112, on Books 1. II. and III.); but Lo=LC, LL' = twice the radius of circumscribed circle, and oy is the radius of the inscribed circle; therefore LoxoA=LL' x oy = twice the rectangle, &c. If any line be drawn through o, terminated by the circumscribed circle, the rectangle under its segments will be equal to LoxoA (III. 35), hence the truth of the Proposition is manifest. In a similar manner it follows in the equiangular triangles LLC and o’Ay', that LC × o'A=LL' x o'y', therefore, &c. PROPOSITION C. To express the distances of the centres of the inscribed and exscribed circles from the centre of the circumscribed circle in terms of the radii of these circles. Let R be the radius of the circumscribed circle r of the inscribed circle, and r', r", r"" of the circles exscribed to the sides a, b, and c. In the isosceles triangle OAL (II. 6), the rectangle LoA is equal to the difference of the squares of OA and Oo; from which, and the last proposition, it follows, that if D be the distance of the centres D2 R2-2 Rr. = From the same Proposition, it follows, that Oo's - oL2 = Lo' × o'A ; therefore, Corollary. By adding these equations, it follows, that D2 + D2 + D′′2 + D""2 = 4 R2+2R (r'" + r” + r' − r). PROPOSITION D. To find the relation between the radii of the inscribed, circumscribed, and exscribed circles. Since o"o" is bisected in L', it follows, that o"x" + o′′x" = 2L'D; and since o'o is bisected in L, o'x' - ox = 2LD, therefore, by addition, o"x" + o′′x" + o'x' · — 0x = 2 (L'D + LD) = 2LL', or r"+r"+r'-r=4R. Corollary. By combining this equation with that contained in the Corollary of the last Proposition, it follows, that D2 + D'2 + D"2+ D"2 = 12 R2. PROPOSITION E. To express the area of a triangle in terms of the radii of the inscribed and erscribed circles, and the segments of the sides into which they are divided by the points of contact. From Cors. I and 4, Prop. IV. it appears that if the area of the triangle be denoted by A, we shall have the equations PROPOSITION F. To investigate the relations which exist between the rectangles under the radii of the inscribed and exscribed circles, and the rectangles under the segments of the sides. From the equiangular right-angled triangles Box and Bo'x', it follows, that Bxx Br' =ox × o'x', and similarly for the other sides; therefore, -b) (s—c)=rr', From the equiangular triangles Bo"x" and Bo"x", it follows, that Bx" × Bx"= 0"x" xo"x", and similarly for the other sides, therefore, s (s—a) = r' 8 (s—b) =r" PROPOSITION G. To investigate an expression for the area of a triangle in terms of the segments of the sides, and also in terms of the radii of the inscribed and exscribed circles. By multiplying together the first two equations of Prop. E, we have To investigate the relation which exists between the reciprocals of the radii of the inscribed and exscribed circles. From the equations of Prop. E, it follows, by division, that |