XXIX. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. DEF. A. A parallelopiped is a solid figure contained by six quadrilateral figures, whereof every opposite two are parallel. PROPOSITION I. THEOR.-One part of a straight line cannot be in a plane, 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 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 it: therefore 7 Def. 1. there are two straight lines ABC, ABD in the same plane that have a common segment AB; which is† impossible. + Cor. 11. 1. Therefore, one part, &c. Q. E. D. R * 7 Def. 1. + 1. 11. PROPOSITION II. THEOR. Two straight lines which cut one another are in 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, CD are in one plane; and three straight lines EC, CB, BE, which meet one another are in one plane. Let any plane pass through the straight line EB, and let the plane be turned about EB, produced if C D B it; therefore the three straight lines EC, CB, BE are in one plane: but in the plane in which EC, EB are, in the same are† CD, AB: therefore AB, CD are in one plane. Wherefore, two straight lines, &c. Q. E. D. PROPOSITION III. THEOR.-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 DB be their common section: DB shall be 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 DFB: then two straight lines DEB, DFB have the same extremities, and therefore include a space betwixt * 10 Ax. 1. them; which is* impossible: therefore BD, the common section of the planes AB, BC, cannot but be a straight line. Wherefore, if two planes, &c. Q. E. D. PROPOSITION IV. THEOR.-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 shall also be 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 F, in EF, draw FA, G FG, FD, FC, FH, FB: and because the two straight lines AE, ED are equal to the two D B 15. 1. BE, EC, each to each, and that they contain equal angles* AED, BEC, the base AD is equal to the base BC, and + 4. 1. the angle DAE to the angle EBC: and the angle AEG is equal to the angle BEH; therefore the triangles AEG, ‡ 15. 1. 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 have their other sides equal; therefore GE is equal to EH, || 26. 1. and AG to BH: and because AE is equal to EB, and FE common and at right angles to them, the base AF is equal§ § 4. 1. 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 equal to the base FC; therefore the angle FAD is equal to the angle FBC: ¶ 8. 1. Again, it was proved that GA is equal to BH, and also AF to FB; FA, then, and AG, are equal to FB and BH, each to each; and the angle FAG has been proved equal to the angle FBH; therefore the base GF is equal* to the 4.1. base FH: Again, because it was proved that GE is equal to EH, and 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 equal to the angle HEF; and consequently each 10 Def. 1. of these angles is a right angle: + 8.1. Therefore 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 3 Def. 13. it in that plane: therefore EF is at right angles to the plane in which are AB, CD. Wherefore, if a straight line, &c. Q. E. D. • 3. 11. + 4. 11. PROPOSITION V. THEOR.-If three straight lines meet all in one point, and a straight line stand 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 and BE are, shall be a straight* 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 lines BD, BE, it is also at right angles† to the plane passing through them; and therefore makes right 13 Def. 11. angles with every straight line meeting it in that plane: but BF, which is in that plane, meets it; therefore the angle ABF is a right angle: But the angle ABC, by the hypothesis, is also a right angle; therefore the angle ABF is equal to the angle ABC, and they are 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. PROPOSITION 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 equal to AB, and join BE, AE, AD. because Then, AB is perpendicular to the plane, it shall make right* angles with every straight line which meets it, and is in that plane: but BD, F B A E BE, which are in that plane, do each of them meet AB; therefore each of the angles ABD, ABE is a right angle: For the same reason, each of the angles CDB, CDE is a right angle: And because AB is equal to DE, and BD common, the two sides AB, BD are equal to the two ED, BD, each to each; and they contain right angles; therefore the base AD is equal to the base BE: Again, because AB is equal to DE, and BE to AD; AB, BE are equal to ED, DA, each to each; and, in the triangles ABE, EDA, the base AE is common; therefore the angle ABE is equal† to the angle EDA: But ABE is a right angle; therefore EDA is also a right angle, and ED perpendicular to DA: But it is also perpendicular to each of the two BD, DC; wherefore ED is at right angles to each of the three straight * 3 Def. 11. † 4. 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