a Book XI. be less than the other side, an obtuse angled, and if greater, an acute angled cone. XIX. The axis of a cone is the fixed straight line about which the triangle revolves. XX. XXI. right angled parallelogram about one of its sides which XXII. The axis of a cylinder is the fixed straight line about which the parallelogram revolves. XXILI. The bases of a cylinder are the circles described by the two revolving opposite sides of the parallelogram. XXIV. XXV. XXVI. XXVII. An octahedron is a solid figure contained by eight equal and equilateral triangles. XXVIII. XXIX. a Def. A. teral figures, whereof every opposite two are parallel. a a Book XI. PROP. I. THEOR. One part of a straight line cannot be in a plane, See N. and another part above it. If it be possible, let AB, part of the straight line ABC, be in the plane, and the part BC above it: And since the straight line AB is in the plane, it can be produced in that plane: Let it be produced to D: And let any plane pass through the A B straight line AD, and be turned about it until it pass through the point C: and because the points B, C are in this plane, the straight line BC is in ita : Therefore there are two straight lines ABC, ABD •7 Def, 1. in the same plane that have a common segment AB; which is impossible. Therefore, one part, &c. Q.E.D. Cor. 11.1. PROP. II. THEOR. Two straight lines which cut one another are in Let two straight lines AB, CD, cut one another in E; A E. the straight line EC is in ita : For the a 7 Def. l. same reason, the straight line BC is in the same; and by the hypothesis, EB is in it: Therefore the three straight lines EC, CB, BE are in one plane : B But in the plane in which EC, EB are, in the same are bb 1, 11. CD, AB: Therefore AB, CD, are in one plane. Wherefore two straight lines, &c. Q. E.D. 1 : Book XI. PROP. III. THEOR. tion is a straight line. Let two planes AB, BC, cut one another, and let the line DB. be their common seetion : DB is a straight line: If it be ' not, from the point D to B, draw, in the plane AB, the straight line DEB, and in the plane BC, the straight line E F therefore include a space betwixt them; • 10 Ax. 1. which is impossiblea : Therefore BD the common section of the planes AB, PROP. IV. THEOR. two straight lines in the point of their intersection, Let the straight line EF stand at right angles to each of the straight lines AB, CD, in E, the point of their intersection : EF is also at right angles to the plane passing through AB, CD. Take the straight lines AE, EB, CE, ED all equal to one another; and through E draw, in the plane in which are AB, CD, any straight line GEH; and join AD, CB; then, from any point Fin EF, draw FA, FG, FD, FC, FH, FB: And because the two straight lines AE, ED are equal to • 15.1. the two BE, EC, and that they contain equal anglesa AED, 4.1. BEC, the base AD is equal to the base BC, and the angle DAE to the angle EBC: And the angle AEG is equal to the angle BEHa: therefore the triangles AEG, BEH have two angles of one equal to two angles of the other, each to each, and the sides AE, EB, adjacent to the equal angles, equal to one another: wherefore they shall have their other « 26. 1. sides equal“: GE is therefore equal to EH, and AG to BH: b 4.1. с and because AE is equal to EB, and FE common and-at Boox XI. right angles to them, the base AF is equal to the base FB; for the same reason, CF is equal to FD: And because AD is equal to BC, and AF to FB, the two sides FA, AD are equal to the two FB, BC, each to each ; -and the base DF was proved E equal to the base FC; therefore the angle FAD is equal to the angle 68.1. FBC: Again, it was proved that GA is equal to BH, and also AF to FB; C FA, then, and AG, are equal to FB and BH, and the angle FAG has beenG proved equal to the angle FBH; therefore the base GF is equal to Έ B the base FH: Again, because it was proved, that GE is equal to EH, and D EF is common; GE, EF are equal В. 06 to HE, EF; and the base GF is equal to the base FH: therefore the angle GEF is equal to the angle HEF; and consequently each of these angles is a right d angle. There- d 10 Def 1. fore FE makes right angles with GH, that is, with any straight line drawn through E in the plane passing through AB, CD. In like manner, it may be proved, that FE makes right angles with every straight line which meets it in that plane. But a straight line is at right angles to, a plane when it makes right angles with every straight line which meets it in that planee: Therefore EF is at right e 3 Def. 1l. angles to the plane in which are AB, CD. Wherefore, if a straight line, &c. Q.E.D. PROP. V. THEOR. IF three straight lines meet all in one point, and a See N. AB stand at right angles to each of the straight lines BC, BD, BE, in B the point where they meet; BC, BD, BE, are in one and the same plane. If not, let, if it be possible, BD and BE be in one plane, and BC be above it; and let a plane pass through AB, BC, the common section of which, with the plane in which BD 1 * 3. 11. Book XI. and BE are, shall be a straight a line; let this be BF: Therefore the three straight lines AB, BC, BF, are all in one plane, viz. that which passes through AB, BC; and because AB stands at right angles to each of the straight 5 4. 11. lines BD, BE, it is also at right angles to the plane pass<3 Def. 11. ing through them; and therefore makes right angles c with every straight line meeting it in Ai с D B both in the same plane, which is impossible; Therefore the straight line BC is not above the plane in which are BD and BE: Wherefore the three straight lines BC, BD, BE are in one and the same plane. Therefore, if three straight lines, &c. Q.E.D. equal a PROP. VI. THEOR. If two straight lines be at right angles to the same plane, they shall be parallel to one another. Let the straight lines AB, CD be at right angles to the same plane; AB is parallel to CD. Let them meet the plane in the points B, D, and draw the straight line BD, to which draw DE at right angles, in the same plane; and make DE to AB, and join BE, AE, AD. Then, c because AB is perpendicular to the 13 Def. 11. plane, it shall make righta angles with every straight line which meets it, and E to the base BE: Again, because AB is equal to DE, and |