a. 33. I. Book XI. fore AD is a both equal and parallel to b. 9. 11. d. 8. I. c d B C E F to DF. and because AB, BC are equal to DE, EF, and the base AC to the bafe DF; the angle ABC is equal to the angle DEF. Therefore if two ftraight lines, &c. Q. E. D. 2. 12. 1. b. II. I. C. 31. I. e. 8. II. T AO draw a ftraight 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 ftraight line perpendicular to the plane BH. In the plane draw any straight line BC, and from the point A draw a AD perpendicular to BC. If then AD be alfo perpendi cular 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 at right angles to BC; and from the point A draw AF perpendicular G to DE; and thro' F draw GH parallel to BC. and because BC is at right angles to ED and DA, BC is A E H B D e d. 4. II. at right angles to the plane paffing thro' ED, DA. And GH is parallel to BC; but if two straight lines be parallel, one of which is at right angles to a plane, the other shall be at right angles to the fame plane; wherefore GH is at right angles to the plane thro' f.3.Def. 11. ED, DA, and is perpendicular to every ftraight line meeting it in that plane. But AF, which is in the plane thro' ED, DA meets it. therefore GH is perpendicular to AF, and confequently AF is Book XI. 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 stands 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 paffing through them. but the plane paffing 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 ftraight line AF is drawn perpendicular to that plane. Which was to be done. To PROP. XII. PROB. O erect a ftraight line at right angles to a given Let A be the point given in the plane; it is required to erect a straight line from the point A at right an gles to the plane. From any point B above the plane draw a BC perpendicular to it; and from A draw b AD parallel to BC. because therefore AD, CB are two parallel straight lines, and one of them BC is at right angles to the given plane, the other AD is D B also at right angles to it. therefore a straight line has been erect- c. 8. 11e ed at right angles to a given plane from a point given in it. Which was to be done. FRO ROM the fame point in a given plane there cannot be two ftraight lines at right angles to the plane, the fame fide of it. and there can be but one perupon pendicular 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 fame point A in the plane, and upon the fame fide of it; and let a plane pass thro' BA, AC; the common section of this with the given plane is a straight a line a. 3. II. Book XI. paffing through A. let DAE be their common fection. therefore the straight lines AB, AC, DAE are in one plane. and because CA is at right angles to the given plane, it shall make right angles with every straight line meeting it in B C that plane. but DAE which is in that plane meets CA; therefore CAE is a right angle. for the fame reason BAE is a right angle. wherefore the angle CAE is equal to the angle BAE; and they are in one plane, which is impoffible. Also, from a point above a plane there can be but one perpendicular to that plane; for if there could be two, b. 6. 11. they would be parallel to one another, which is abfurd. There fore from the fame point, &c. Q. E. D. PLAN D PROP. XIV. THEOR. which A E LANES to which the fame ftraight line is perpendicular, are parallel to one another. Let the ftraight line AB be perpendicular to each of the planes CD, EF; thefe planes are parallel to one another. If not, they shall meet one another when produced; let them meet; their common section shall be a ftraight line GH, in which take any point K, and join AK, BK. then because AB is perpendicular to the plane 3.3.Def.11. EF, it is perpendicular to the ftraight C a line BK which is in that plane. there- fore the two angles ABK, BAK of the the triangle ABK are equal to two right b. 17. 1. angles, which is impoffible ". therefore the planes CD, EF though produced do G H c. 8.Def.11. not meet one another; that is, they are parallel. Therefore planes, &c. Q E. D. IF PROP. XV. THEOR. Book XI. F two ftraight lines meeting one another, be parallel See N. to two straight lines which meet one another, but are not in the fame plane with the first two; the plane which paffes through these is parallel to the plane pasfing through the others. Let AB, BC two straight lines meeting one another, be parallel to DE, EF that meet one another, but are not in the fame plane with AB, BC. the planes through AB, BC, and DE, EF fhall not meet though produced. b. 31. I. From the point B draw BG perpendicular to the plane which a. 11. II. paffes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED, and GK parallel to EF. and because BG is perpendicular to the plane through DE, EF, it fhall make right angles with every straight line meeting it in that plane. but the ftraight B E lines GH, GK in that plane meet it. therefore each of the angles BGH, BGK is a right angle. and because BA is paral- A D lel to GH (for each of them is H d. 9. tt. parallel to DE, and they are not both in the fame plane with it) the angles GBA, BGH are to e gether equal to two right angles. and BGH is a right angle, e. 29. 1. therefore alfo GBA is a right angle, and GB perpendicular to BA. for the fame reason, GB is perpendicular to BC. fince therefore the ftraight line GB ftands at right angles to the two straight lines BA, BC, that cut one another in B; GB is perpendicular f f. 4. II. 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 fame straight line is perpendicular, are parallel to one another. g. 14. 11. therefore the plane thro' AB, BC is parallel to the plane thro' DE, EF. Wherefore if two straight lines, &c. QE. D. Book XI. See N. PROP. XVI. THEOR. F two parallel planes be cut by another plane, their common fections with it are parallels. Let the parallel planes AB, CD be cut by the plane EFHG, and let their common fections with it be EF, GH. EF is parallel to GH. For, if it is not, EF, GH fhall meet, if produced, either on the the Hypothefis. therefore the A meet when produced on the fide K F H B D E of FH. in the fame manner it may be proved that EF, GH do not meet when produced on the fide of EG. but ftraight lines which are in the fame 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. F two ftraight lines be cut by parallel planes, they fhall be cut in the fame ratio. Let the ftraight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D. as AE is to EB, fo is CF to FD. Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are paral |