5.11. § 2.11. 28. 1. lines BD, DA, DC, in the point in which they meet; therefore these three straight lines are all in the same plane: But AB is in the plane in which are BD, DA, because any three straight lines which meet one another, are in one plane§; therefore AB, BD, DC, are in one plane: and each of the angles ABD, BDC is a right angle; therefore AB is parallel to CD. Wherefore, if two straight lines, &c. Q. E. D. PROPOSITION VII. THEOR.-If two straight lines be parallel, the straight line drawn from any point in the one to any point in the other, is in the same plane with the parallels. Let AB, CD be parallel straight lines, and take any point E in the one, and the point F in the other: the straight line which joins E and F, is in the same plane with the parallels. If not, let it be, if possible, above the plane, as EGF; and in the plane ABCD, in which the parallels are, draw the straight line EHF from E to F: and since EGF also is a straight line, the two straight lines EHF, EGF include a space between them, which is impos 10. Ax. 1. sible*: therefore the straight line joining the points E, F * 7. 11. is not above the plane in which the parallels AB, CD are, and is therefore in that plane. Wherefore, if two straight lines, &c. Q. E. D. PROPOSITION VIII. THEOR.-If two straight lines be parallel, and one of them is at right angles to a plane, the other also shall be at right angles to the same plane. Let AB, CD be two parallel straight lines, and let one of them AB be at right angles to a plane: the other CD is at right angles to the same plane. Let AB, CD meet the plane in the points B, D, and join BD: therefore* AB, CD, BD are in one plane. In the plane to which AB is at right angles, draw DE at right angles to BD, and make DE equal to AB, and join BE, AE, AD. And because AB is perpendicular to the plane, it is per pendicular to every straight line which meets it, and is in that planet; therefore each of the angles ABD, ABE † 3 Def. 11. is a right angle: And because the straight line BD meets the parallel straight lines AB, CD, the angles ABD, CDB are together equal to two right angles: and ABD is a right angle; ‡ 29. 1. therefore also CDB is a right angle, and CD perpendicular to BD: And because AB is equal to DE, and BD common, the two AB, BD are equal to the two ED, DB, each to each; and the angle ABD is equal to the angle EDB, because each of them is a right angle; therefore the base AD is equal|| to the base BE: 4. 1. Again, because AB is equal to DE, and BE to AD; the two AB, BE are equal to the two ED, DA, each to each; and the base AE is common to the triangles ABE, EDA; wherefore the angle ABE is equal§ to the angle EDA: but § 8. 1. ABE is a right angle; and therefore EDA is a right angle, and ED perpendicular to DA: But it is also perpendicular to BD; therefore ED is perpendicular to the plane which passes through BD, DA, 4.11. and shall* make right angles with every straight line meet- * 3 Def. 11. ing it in that plane: But DC is in the plane passing through BD, DA, because all three are in the plane in which are the parallels AB, CD; wherefore ED is at right angles to DC; and therefore CD is at right angles to DE: but CD is also at † 4. 11. * 4. 11. + 8. 11. I 6. 11. right angles to DB; CD then is at right angles to the two straight lines DE, DB, in the point of their intersection D; and therefore is at right angles† to the plane passing through DE, DB, which is the same plane to which AB is at right angles. Therefore, if two straight lines, &c. Q. E.D. PROPOSITION IX. THEOR. Two straight lines, which are each of them parallel to the same straight line, and not in the same plane with it, are parallel to one another. Let AB, CD be each of them parallel to EF, and not in the same plane with it: AB shall be parallel to CD. In EF take any point G, from which draw, in the plane passing through EF, AB, the straight line GH at right angles to EF; and in the plane passing through EF, CD, draw GK at right angles to the same EF. And because EF is perpendicular both to GH and GK, EF is per F pendicular to the plane HGK passing through them: and EF is parallel to AB; therefore AB is at right angles† to the plane HGK: For the same reason, CD is likewise at right angles to the plane HGK; Therefore AB, CD are each of them at right angles to the plane HGK. But if two straight lines are at right angles to the same plane, they shall be parallel to one another; therefore AB is parallel to CD. Wherefore, two straight lines, &c. Q. E. D. PROPOSITION X. THEOR.-If two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two, the first two and the other two shall contain equal angles. Let the two straight lines AB, BC, which meet one another, be parallel to the two straight lines DE, EF, that meet one another, and are not in the same plane with AB, BC: the angle ABC is equal to the angle DEF. Take BA, BC, ED, EF all equal to one another; and join AD, CF, BE, AC, DF: Because BA is equal and parallel to ED, therefore AD is* both equal and parallel to BE: For the same reason, CF is equal and parallel to BE; Therefore AD and CF are each of them equal and parallel to BE. But straight lines that are parallel to the same straight line, and not in the same plane with it, are parallel to one another; therefore AD is parallel to CF; and it is equal to it; and AC, DF join them towards D B 33. 1. + 9. 11. 1 Ax. 1. F the same parts; and therefore || AC is equal and parallel to || 33. 1. DF. And because AB, BC are equal to DE, EF, each to each, and the base AC to the base DF, the angle ABC is equal§ to the angle DEF. Therefore, if two straight lines, § 8. 1. &c. Q. E. D. PROPOSITION XI. PROB.-To draw a straight line perpendicular to a plane, from a given point above it. Let A be the given point above the plane BH; it is required to draw from the point A, a straight line perpendicular to the plane BH. In the plane, draw any straight line BC, and from the point A* draw AD perpendicular to BC. If then AD be * 12. 1. also perpendicular to the plane BH, the thing required is already done: but if it be not, from the point D draw†, † 11. 1. in the plane BH, the straight line DE at right angles to BC; and from the point A draw AF perpendicular to DE; and through F draw‡ GH parallel to BC: 31. 1. And because BC is at right angles to ED and DA, BC is at right angles to the plane passing through ED, DA; || 4. 11. and GH is parallel to BC: but, if two straight lines be parallel, one of which is at right angles to a plane, the 8. 11. other shall be at right angles to the same plane; wherefore GH is at right angles to the plane through ED, DA, 3 Def. 11. and is perpendicular to every straight line meeting it in that plane: 4. 11. * 11. 11. + 31. 1. But AF, which is in the plane through ED, DA, meets it; therefore GH is perpendicular to AF; and consequently AF is perpendicular to GH; and AF is perpendicular to DE; therefore AF is perpendicular to each of the straight lines GH, DE. But 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 passing through them: but the plane passing through ED, GH, is the plane BH;' therefore AF is perpendicular to the plane BH: therefore, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Which was to be done. PROPOSITION XII. PROB. To erect a straight line at right angles to a given plane, from a point given in the plane. Let A be the point given in the plane; it is required to erect a straight line from the point A, at right angles to the plane. From any point B above the plane, draw* BC perpendicular to it; and from A draw† AD parallel to BC. Because, therefore, AD, CB are two parallel straight |