Book XI. 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 re- XXII. The axis of a cylinder is the fixed straight line about which the parallelogram revolves. XXIII. 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. A dodecahedron is a solid figure contained by twelve equal pentagons, which are equilateral and equiangular. XXIX. DEF. A. figures, whereof every opposite two are parallel. Book XI. PROP. I. THEOR. One part of a straight line cannot be in a plane, See N. and another part above it. C 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 straight line AD, and be turned about it A B D until it pass through the point C; and because the points B, C, are in this plane, the straight line BC is in it a : Therefore a 7. def. 1. there are two straight lines ABC, ABD in the same plane, that have a common segment AB, which is impossible 6.b Cor. 11. 1. Therefore, “ one part,” &c. Q. E. D. PROP. II. THEOR. Two straight lines which cut one another are in See N. one plane, and three straight lines which meet one another are in one plane. Let two straight lines AB, CD cut one another in E; AB, D E the straight line EC is in it a : For the a 7. def. 1. 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 : But in C B the plane in which EC, EB are, in the same are b CD, AB: Therefore AB, b). ll. CD, are in one plane. Wherefore, “ iwo straight lines,” &c. Q. E. D. Book XI. PROP. III. THEOR. See N. If two planes cut one another, their common section is a straight line. Let two planes AB, BC, cut one another, and let the line E F C С D PROP. IV. THEOR. See N. If a straight line stand at right angles to each of two straight lines in the point of their intersection, it shall also be at right angles to the plane which passes through them, that is, to the plane in which they are. 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 the two BE, EC, and that they contain equal anglesa AED, 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 BEHạ; therefore the triangles AEG, BEH have two angles of the 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 sides equal“: GE is therefore a 15. 1. b 4. 1. c 26. 1. equal to EH, and AG to BH: and because AE is equal to Book XI. EB, and FE common, and at right angles to them, the base AF is equal to the base FB; for the same reason, CF is b 4. 1. 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 equal to the base FC; therefore the angle FAD is equald to the angle FBC: d 8.). Again, it was proved, that AG is A с equal to BH, and also AF to FB; FA, then, and AG are equal to FB and BH, and the angle FAG has G been proved equal to the angle FBH; therefore the base GF is equal to E H the base FH: Again, because it was proved that GE is equal to EH, and D B EF is common; GE, EF are equal to HE, EF; and the base GF is equal to the base FH; therefore the angle GEF is equald to the angle HEF; and consequently each of these angles is a righte angle. Therefore FE makes right angles with GH, e 10.def. 1. 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 planef: Therefore EF is at right f 3.def. 11. angles to the plane in which are AB, CD. Wherefore, if 6 a straight line,” &c. Q. E. D. PROP. V. THEOR. If three straight lines meet all in one point, and a See N. straight line stands at right angles to each of them in that point ; these three straight lines are in one and the same plane. Let the straight line 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 Book XI, and BE are, shall be a straight a line; let this be BF: There fore the three straight lines AB, BC, BF, are all in one plane, a 3. 11. viz. that which passes through AB, BC, and because AB stands at right angles to each of the straight lines BD, BE, b 4. II. it is also at right angles 6 to the plane passing through them; c 3. def. 1 1. and therefore makes right angles with every straight line meeting A C F D E 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. PROP. VI. THEOR. Two straight lines, which are at right angles to the same plane, are parallel to one another. Let the straight lines AB, CD be at right angles to the same plane BDE; 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 equal to A B, and join BE, AE, AD. Then, A, because AB is perpendicular to the a 3. def. 11. plane, it makes right a angles with every straight line which meets it, and D the two ED, DB; and they contain right angles; therefore, b 4. 1. the base AD is equal b to the base BE: Again, because AB |