two straight lines, AE, ED are equal to the two BE, EC, and that they contain equal angles (15. 1.) AED, BEC, the base AD is equal (4. 1.) to the base BC, and the angle DAE to the angle EBC: and the angle AEG is equal to the angle BEH (15. 1.); 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 sides equal (26. 1.): GE is therefore equal to EH, 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 F the base EC; therefore the angle FAD is equal (8. 1.) to the angle FBC: again, it was proved that GA is equal to BH, and also AF to FB; FA, then, and AG are equal to FB С and BH, and the angle FAG has been proved A equal to the angle FBH; therefore the base GF (4. 1.) to the base FH: again, because G it was proved, that GE is equal to EH, and EF is common; GE, EF are equal to HE, H E D B of these angles is a right (10. def. 1.) angle. 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 it in that plane (3. def. 11.): therefore EF is at right 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 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 and BE are, shall be a straight * See Note. 1 (3. 11.) 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 (4. 11.) to the plane passing through them; and therefore makes right angles (3. def. 11.) with every straight line meeting it in that plane; but BF, which is in that plane, meets it: there. A fore the angle ABF is a right angle; but the angle ABC, by the hypothesis, is also a right с angle; therefore the angle ABF is equal to F the angle ABC, and they are both in the same plane which is impossible: therefore the straight line BC is not above the plane D in which are BD and BE: wherefore the B 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. 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 B, D, to which draw DE at right angles, in the same plane; and make DE equal to AB, and join BE, AE, AD. A C Then because AB is perpendicular to the plane, it shall make right (3. def. 11.) angles with every straight line which meets it, and is in that plane: but BD, 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 B 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, DB: and they contain right angles; therefore the base AD is equal (4. 1.) to the E base BE: again, because AB is equal to DE, and BE to AD; AB, BE are equal to ED, DA; and, in the triangles ABE, EDA, the base AE is common; therefore the angle ABE is equal (8. 1.) 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 lines BD, DA, DC, in the point in which they meet: therefore these three straight lines are all in the same plane (5. 11.): but AB is in the plane in which are BD, DA, because any three straight lines which meet one another are in one plane (2. 11.): therefore AB, BD, DC are in one plane: and each of the angles ABD, BDC is a right angle; therefore AB is parallel (28. 1.) to CD. Wherefore, if two straight lines, &c. Q. E. D. PROP. 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 A E B are, draw the straight line EHF from E to F; and since EGF also is a straight line, the two straight lines G H С F D points E, F 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. PROP. VIII. THEOR. If two straight lines be parallel, and one of them be 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 (7. 11.) 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 perpendicular to every straight line which meets it, and is in that plane (3. def. 11.); therefore each of the angles ABD, ABE is a right angle; and because the straight line BD meets the parallel straight lines AB, OD, the angles ABD, CDB are together equal (29. 1.) to two right angles: and ABD is a right angle; 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, and the angle ADB is equal to the angle EDB, because each of * See Note. 21 them is a right angle; therefore the base AD С is equal (4. 1.) to the base BE: again, because A AB is equal to DE, and EB to AD; the two AB, BE are equal to the two ED, DA; and the base AE is common to the triangles ABE, EDA; wherefore the angle ABE is equal (8. 1.) to the angle EDA, and ABE is a right angle; and therefore EDA, is a right angle, and ED perpendicular to DA: but it is also perpendicu- B lar to BD: therefore ED is perpendicular D (4. 11.) to the plane which passes through BD, DA, and shall (3. def. 11.) make right angles with every straight line meeting it in that E 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 right angles to DB; CD then is at right angles to the two straight lines DE, BD in the point of their intersection D; and therefore is at right angles (4. 11.) to the plane passing through DE, DB, which is the same pláne to which AB is at right angles. Therefore, if two straight lines, &c. Q. E. D. PROP. 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 perpendicular (4. 11.) to the A H B plane HGK passing through them: and EF is parallel to AB, therefore AB is at right angles (8. 11.) to the plane HGK. G For the same reason, CD is likewise E at right angles to the plane HGK. Therefore, AB, CD are each of them at right angles to the plane HGK. But С K D if two straight lines be at right angles to the same plane, they shall be parallel (6. 11.) to one another. Therefore AB is parallel to CD. Wherefore, two straight lines, &c. Q. E. D. PROP. X. THEOR. If two straight lines meeting one another be parallel to two other 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 (33. 1.) both equal and parallel to BE. For B 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 A С that are parallel to the same straight line, and not in the same plane with it, are parallel (9. 11.) to one another. Therefore AD is parallel to CF; and it is equal (1. Ax. 1.) to it, and AC, DF join them towards the same parts; and therefore (33. 1.) AC is equal and parallel to DF. And because AB, BC are equal to E DE, EF, and the base AC to the base DF; D F the angle ABC is equal (8. 1.) to the angle DEF. Therefore, if two straight lines, &c. Q. E. D. PROP. 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 (12. 1.) AD perpendicular to BC. If then AD be 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 A from the point A draw AF perpen E dicular to DE; and through F draw (31. 1.) GH parallel to BC: and be F cause BC is at right angles to ED and H G DA: BC is at right angles (4. 11.) to the plane passing through ED, DA. And GH is parallel to BC : but, if two straight lines be parallel, one of wh is at right angles to a plane, the other B D С shall be at right (8. 11.) angles to the same plane; wherefore GH is at right angles to the plane through ED, DA, and is perpendicular (3. def. 11.) to every straight line meeting it in that plane. But AF, which is in the plane through ED, DA meets it: therefore GH is |