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* AD perpendicular to BC: if then * 12. 1. AD be also perpendicular to the plane BH, the thing required is already done; but if it be not, from the point D draw,* in the plane BH, the straight line DE * 11. 1. at right angles to BC; and from the point A draw AF perpendicular to DE: AF shall be perpendicular to the plane BH. E D H 8. 11. * 3 Def. Through F draw* GH parallel to BC: and because * 31. 1. BC is at right angles to ED and DA, BC is at right angles to the plane passing through ED, DA: and * 4. 11. GH is parallel to BC: but, if two straight lines be parallel, one of which is at right angles to a plane, the other is at right angles to the same plane; wherefore GH is at right angles to the plane through ED, DA; and is perpendicular to every straight line meeting it in that plane: but AF, which is in the 11. 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 is also at right angles to the plane * 4. 11. passing through them: but the plane passing through ED, GH is the plane BH; therefore AF is perpen * dicular 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. * 11. 11. * 31. 1. PROP. 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 lines, and one of them BC is at right angles to the *8. 11. given plane, the other AD is also at right angles to it: therefore a straight line has been erected at right angles to a given plane from a point given in it. Which was to be done. PROP. XIII. THEOR. From the same point in a given plane, there cannot be two straight lines at right angles to the plane, upon the same side of it: and there can be but one perpendicular to a plane from a point above the plane. For, if it be possible, let the two straight lines AB, AC, be at right angles to a given plane from the same point A in the plane, and upon the same side of it; and let a plane pass through BA, AC; the common section of this with the given plane is a *3. 11. straight line passing through A: let DAE be their common section: therefore the straight lines AB, AC, * * DAE are in one plane: and because CA is at right * * 3 Def 11. same reason BAE is a right angle: wherefore the angle CAE is equal to the angle BAE; but this is 11 Ax. impossible, because they are in one plane. Also, from a point above a plane, there can be but one perpendicular to that plane; for, if there could be two, they would be parallel to one another, which is absurd. * 6. 11. Therefore, from the same point, &c. Q. E. D. PROP. XIV. THEOR. Planes to which the same straight line is perpendicular, are parallel to one another. Let the straight line AB be perpendicular to each of the planes CD, EF; these planes shall be parallel to one another. * * 3 Def. If not, they will meet one another when produced; let them meet; then their common section is a straight line GH, in which take any point K, and join AK, BK. Then, because AB is perpendicular to the plane EF, it is perpendicular to the straight line BK which is in that plane: therefore 11. ABK is a right angle. For the same reason, BAK is a right angle; wherefore the two angles ABK, BAK of the triangle ABK are equal to two right angles, which is impossible: therefore the planes CD, EF, * 17. 1. though produced, do not meet one another; that is, they are parallel. Therefore planes, &c. Q. E. D. X 11. 8 Def. 11. 11. 11. PROP. XV. THEOR. If two straight lines meeting one another, be parallel to two other straight lines which meet one another, but are not in the same plane with the first two; the plane which passes through these is parallel to the plane passing through the others. Let AB, BC, two straight lines meeting one another, be parallel to DE, EF two other straight lines that meet one another, but are not in the same plane with AB, BC: the planes through AB, BC, and DE, EF shall not meet, though produced. From the point B draw BG perpendicular to the plane which passes through DE, EF, and let it meet 31. 1. that plane in G; and through G draw GH parallel to ED, and GK parallel to EF. And because BG is *3 Def. perpendicular to the plane through DE, EF, it makes right angles with every straight line meeting it in that plane: but the straight lines GH, GK in that plane meet it; therefore each of the angles BGH, BGK is a right angle: and because *9. 11. BA is parallel to GH (for each of them is parallel to DE, and they are not both in the same plane with it) the angles GBA, BGH are together 29. 1. equal to two right angles: and BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA. For the same reason, GB is perpendicular to BC: therefore since the straight line GB stands at right angles to the two straight lines BA, 4. 11. BC, which cut one another in B; GB is perpendicular *Const. to the plane through BA, BC: and it is perpendicular* to the plane through DE, EF; therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: but planes to which the same straight line is perpendicular, are parallel to one another: there- * 14. 11. fore the plane through AB, BC is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D. PROP. XVI. THEOR. If two parallel planes be cut by another plane, their common sections with it are parallels. Let the parallel planes AB, CD be cut by the plane EFHG, and let their common sections with it be EF, GH: EF shall be parallel to GH. For, if it is not, EF, GH shall meet, if produced, either on the side of FH, or EG. First, let them be produced on the side of FH, and meet in the point K: therefore, since EFK is in the plane AB, every point** 1. 11. in EFK is in that plane; and K is a point in EFK; therefore K is in the plane AB. For the same reason K is also in the plane CD: wherefore the planes AB, CD produced meet one another; but they do not meet, since they are parallel by the hypothesis: therefore the straight lines EF, GH do not meet when produced on the side of FH. In the same manner it may be proved, that EF, GH do not meet when produced on the side of EG: but straight lines which are in the same plane and do not meet, though produced either way, are parallel: therefore EF is parallel to GH. Wherefore, if two parallel planes, &c. Q. E. D. |