PROP. I. THEOR. Book XI. ONE part of a straight line cannot be in a plane and see N. another part above it. If it be poffible, let AB, part of the the plane, and the part BC above it : line AB is in the plane, it can be produced in that plane: Let it be produced to D: And let any plane pafs thro' the ftraight line AD, and be turned about it until it pafs thro' the point C; and because the points B, C, are in this plane, the ftraight line BC is in it: a 7. def. 1. Therefore there are two ftraight lines ABC, ABD in the fame plane that have a common fegment AB, which is impoffible . b. Cor.11.1. Therefore one part, &c. Q. E. D. PROP. II. THEOR. WO ftraight lines which cut one another are in one See N. plane, and three ftraight lines which meet one ano T ther are in one plane. Let two ftraight lines AB, CD cut one another in E; AB, CD are one plane: And three ftraight lines EC, CB, BE which meet one another, are in one plane. Let any plane pass through the ftraight line EB, and let the plane be turned a. bout EB, produced, if neceffary, until it pass through the point C: Then because the points E, C are in this plane, the ftraight line EC is in it: For the fame reason, the straight line BC is in the fame; and, by the hypothefis, EB is in it: Therefore the three ftraight lines EC, CB, BE are in one plane: But in the plane in which EC, EB are, in the fame are b CD, AB: Therefore AB, CD are in one plane. Wherefore two straight lines, &c. N 4 E D a 7. def. f. B b. I. II. Q. E. D. PROP. Book X4. See N. PROP. III. THEOR. IF two planes cut one another, their common fection is a ftraight line. Let two planes AB, BC, cut one another, and let the line DB be their common fection: DB is a ftraight line: If it be not, from the point D to B, draw, in the plane AB, the ftraight line DEB, and in the plane BC the ftraight line DFB: Then two straight lines DEB, DFB have the fame extremities, and therefore include a space bea 10. Ax. r. twixt them; which is impoffible a: There fore BD the common fection of the planes B E C D See N. a 15. I. b 4. I. € 26. I. PROP. IV. THEOR. Fa ftraight line stand at right angles to each of two in point interfection, alfo be at right angles to the plane which paffes through them, that is, to the plane in which they are. Let the ftraight line EF ftand at right angles to each of the ftraight lines AB, CD in E, the point of their interfection: EF is alfo at right angles to the plane paffing through AB, CD. Take the ftraight lines AE, EB, CE, ED all equal to one another; and through E draw, in the plane in which are AB, CD, any ftraight line GFH; and join AD, CB; then, from any point F in EF, draw FA, FG, FD, FC, FH, FB: And because the two ftraight lines, AE, ED are equal to the two BE, EC, and that they contain equal angles a AED, BEC, the base AD is equal b to the bafe BC, and the angle DAE to the angle EBC: And the angle AEG is equal to the angle BEH; therefore the triangles AEG, BEH have two angles of one equal to two angles of the other, each to each, and the fides AE, EB, adjacent to the equal angles, equal to one another; wherefore they shall have their other fides equal: GE is therefore equal b 4. I. F d 8. L equal to EH, and AG to BH: And becaufe AE is equal to EB, Book XI. and FE common and at right angles to them, the base AF is equal b to the bafe FB; for the fame reason, CF is equal to FD: And becaufe AD is equal to BC, and AF to FB, the two fides FA, AD are equal to the two FB, BC, each to each; and the bafe DF was proved equal to the base FC; therefore the angle FAD is equal d to the angle FBC: Again, it was proved that GA is equal to BH, and alfo AF A to FB; FA, then, and AG, are equal to FB and BH, and the angle FAG has been proved equal to the angle FBH; therefore the bafe GF is equal b to the bafe 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 G C E H B ds is equal to the base FH; therefore the angle GEF is equal to the angle HEF; and confequently each of these angles is a с right angle. Therefore FE makes right angles with GH, e 10. def. 1. that is, with any ftraight line drawn through E in the plane paffing through AB, CD. In like manner, it may be proved, that FE makes right angles with every ftraight line which meets it in that plane. But a ftraight line is at right angles to a plane when it makes right angles with every ftraight line which meets it in that plane f: Therefore EF is at right angles to the plane f 3. def. 11. in which are AB, CD. Wherefore, if a straight line, &c. Q. E. D. PROP. V. THEOR. IF three ftraight lines meet all in one point, and a See N. ftraight line ftands at right angles to each of them in that point; these three ftraight lines are in one and the fame plane. Let the ftraight line AB ftand at right angles to each of the ftraight lines BC, BD, BE, in B the point where they meet; BC, BD, BE are in one and the fame plane. It not, let, if it be poffible, BD and BE be in one plane, and BC be above it; and let a plane país through AB, BC, the common fection of which with the plane, in which BD and BE are, Book XI. a 3. II. are, fhall be a ftraight a line; let this be BF: Therefore the three ftraight lines AB, BC, BF are all in one plane, viz. that which paffes through AB, BC; and because AB ftands at right angles to each of the ftraight lines BD, BE, It is alfo at right angles to the plane paffing through them; and therefore makes € 3. def 11. right angles with every ftraight A b4. II. a def. II. 64.1. line meeting in that plane; but C F -D E which are BD and BE: Wherefore the three ftraight lines BC, BD, BE are in one and the fame plane. Therefore, if three ftraight lines, &c. Q. E. D. PROP. VI. THEOR. IF two straight lines be at right angles to the fame plane, they shall be parallel to one another. Let the ftraight lines AB, CD be at right angles to the fame plane; AB is parallel to CD. Let them meet the plane in the points B, D, and draw the ftraight line BD, to which draw DE at right angles, in the fame plane; and make DE equal to AB, C D and join BÉ, AE, AD. Then, because A E to to DE, and BE to AD; AB, BE are equal to ED, DA; and, Book XI. in the triangles ABE, EDA, the bafe AE is common; there fore the angle ABE is equal to the angle EDA: But ABE is c 8. I. a right angle; therefore EDA is also a right angle, and ED perpendicular to DA: But it is alfo perpendicular to each of the two BD, DC: Wherefore ED is at right angles to each of the three straight lines BD, DA, DC in the point in, which they meet: Therefore thefe three ftraight lines are all in the fame plane: But AB is in the plane in which are BD, DA, & 5. 11. because any three ftraight lines which meet one another are in one plane: Therefore AB, BD, DC are in one plane: And e 2. 11. each of the angles ABD, BDC is a right angle; therefore AB is parallelf to CD. Wherefore, if two ftraight lines, &c. Q. E. D. f 28. I. PROP. VII. THEOR. d F two straight lines be parallel, the straight line drawn See N. from any point in the one to any point in the other, I' is in the fame plane with the parallels. Let AB, CD be parallel ftraight lines, and take any point E in the one, and the point F in the other: The ftraight line which joins E and F is in the fame plane with the parallels. A' If not, let it be, if poffible, above the plane, as EGF; and in the plane ABCD in which the parallels are, draw the ftraight line EHF from E to F; and fince EGF alfo is a ftraight line, the two ftraight lines EHF, EGF include a fpace between them, which is impoffible. Therefore the ftraight line joining the points E, F is not above the C E B G H 8 10. Από τ F D plane in which the parallels AB, CD are, and is therefore in that plane. Wherefore, if two straight lines, &c. Q. E. D. PROP. VIII. THEOR. IF two ftraight lines be parallel, and one of them is at See N. right angles to a plane; the other also shall be at right angles to the fame plane. Let |