and they are upon equal straight lines BC, BK: But similar seg- Book VI.' ments of circles upon equal straight lines, are equal 8 to one ano. m ther : Therefore the segment BXC is equal to the segment COK: 8 24. 3. And the triangle BGC is equal to the triangle CGK; therefore the whole, the sector BGC, is equal to the whole, the sector CGK: For the same reason, the sector KGL is equal to each of the sectors BGC, CGK: In the same manner, the sectors EHF, FHM, MHN may be proved equal to one another : Therefore, what multiple foever the circumference BL is of the circumference BC, the fame multiple is the sector BGL of the sector BGC: For the same reason, whatever multiple the circumfe. rence EN is of EF, the same multiple is the sector EHN of the sector EHF: And if the circumference BL be equal to EN, the feltor BGL is equal to the sector EHN; and if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if less, lefs : Since then, there are four magnitudes, the two circumferences BC, EF, and the two fectors BGC, EHF, and of the circumference BC, and sector BGC, the circumference BL and sector BGL are any equi. multiples whatever; and of the circumference EF and sector EHF, the circumference EN and sector EHN are any equimultiples whatever ; and that it has been proved, if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if equal, equal ; and if less, less. Therefore, as the circumference BC is to the circumference EF, fo b 5. def. s. is the sector BGC to the sector EHF. Wherefore, in equal circles, &c. Q. E. D. Book VI. See N. PRO P. B. THE OR. which likewife cuts the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the fegments of the base, together with the square of the straight line bisecting the angle. a Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD, the rectangle BĂ, AC is equal to the rectangle BD, DC, together with the square of AD. Describe the circle a ACB about the triangle, and produce A с same segment; the triangles ABD, B D E See N. 7F from any angle of a triangle a straight line be drawn perpendicular to the base ; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle. Let AEC be a triangle, and AD the perpendicular from the angle Authe kae: PC; the rectangle BA, AC is equal to the redan, a coi ained or AD and the diameter of the circle de scrised at the triangic. Describe Describe the circle ACB about the triangle, and draw its diameter AE, and join EC: Because the . right angle BDA is equal to the angle ECA in a semicircle, and the angle ABD to the angle’AEC B in the same fegment; the triangles ABD, AEC are equiangular : Therefore as d BA to AD, fo is EA to AC; and consequently the rectangle BA, AC is equal E to the rectangle EA, AD. If therefore, from an angle, &c. Q. E. D. D C 21. 3. d 4. 6. € 16. 6. PROP. D. THEOR. drilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides. Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD; the rectangle contained by AC, BD is equal to the two rectangles contained by AB, CD, and by AD, BC+. Make the angle ABE equal to the angle DBC ; add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC: And the angle BDA is.equal to the a 21. 3. angle BCE, because they are in the same segment; therefore the triangle ABD is equiangular to the triangle BCE: Wherefore b as b 4. 0. BC is to CE, fo is BD to DA; and consequently the rectangle BC, AD is equal to the rectangle BD, CE: Again, because the angle ABE is equal to the angle DBC, and the angle · BAE to the angle BDC, the triangle ABE is equiangular to the triangle BCD: As therefore BA to AE, To is BD to DC; wherefore the rectangle BA, DC is equal to the rectangle BD, AE: But the rectangle BC, AD has been shewn equal to the rectangle BD, CE; therefore the whole rectangle AC, BD d is equal to the rectangle AB, DC, together with the rectangle AD, BC. Therefore the rectangle, &c. Q. E D. N2 THE C 16. S. † This is a Lemma of al Poplomagus, io page 9. of his μεγαλη συνταξις. E Ε L Ε Μ Ε Ν Τ S M T S A . I. II. NII, when it makes right angles with every straight line meeting IV. drawn in one of the planes perpendicularly to the common V. contained by that straight line, and another drawn from the VI. ed by two straight lines drawn from any the same point of VII. Two VII. Book XI. Two planes are said to have the same, or a like inclination to one another, which two other planes have, when the said VIII. IX. than two plane angles, which are not in the same plane, in X. •The tenth definition is omitted for reasons given in the notes.' See N. XI. See N. Similar folid figures are such as have all their solid angles equal, each to each, and which are contained by the same number XII. stituted betwixt one plane and one point above it in which XUI. two that are opposite are equal, similar, and parallel to one XIV. XV. XVI. XVII. the center, and is terminated both ways by the superficies of XVIII. angled triangle about one of the sides containing the right angle, which fide remains fixed. angle, the cone is called a right angled cone; it it be less N 3 XIX. The |