cone; if less, an obtuse angled cone; and if greater, an a. Book XI. cute angled cone. XIX, XX. XXI. revolving about one of the sides, containing the right angle, XXI. The axis of a cylinder is that fixed right line about which the parallelogram is moved. XXIII. The bases of a cylinder are the circles described by the motion of the two opposite fides of the parallelogram. XXIV. XXV. XXVI. XXVII. An octahedron is a solid figure contained by eight equilateral triangles. XXVIII. XXIX. quilateral triangles. ral figures, whereof those that are opposite are parallel. 0 NE part of a right line cannot be in a plain superficies, and For, Book XI. For, if possible, let the part AB of the right line ABC be in a plain superficies, and the part BC above the fame; there will be some right line in that plain which will make one right line with AB, which let be DB ; then the two right lines ABC, ABD, will have one common segment AB; which is imposa Cor. 14. 8. fiblea. Wherefore, &c. PRO P. II. THEO R. IF Ftwo right lines cut each other, they are both in one plain, and every triangle is in one plain. Let the two right lines AB, CD, cut each other in the point E, they are both in one plain. For, take any points F, G, in the right lines AB, CD, and join CB ; then the right lines AB, CD, are in one plain, and the triangle ECB is in one plain. For the parts DF, AG, cannot be in one plain, and FC, GB, above it"; therefore DC, AB, are in one plain ; and, because the points B, C, are in one plain, therefore the triangle ECB is in one plain. Where. fore, &c. PRO P. III. THE O R. IF F two plains cut each other, their common section will be a right line. Let the two plains be AB, BC, cutting each other; and let BD be their common fection ; then BD is a right line. For, if not, let the right line BED be drawn in the plain CB, and BFD in the plain BA; then two right lines bound a figure; AAX. Io. I. which cannot bea. Wherefore, &c. PRO P. IV. THE O R. IF F a right line stand in the common section of two right lines, cutting one another, and at right angles to the same ; then it fall be at right angles to the plain paling through these lines. Let the right line EF stand in the common section at right Book XI, angles to the two right lines AB, CD, then IF is likewise at right angles to the plain passing through AB, CD. For, take the right lines AE, ED, EB, EC, equal to one another, and join AD, BC ; through E draw GEH to the points G, H, in the right lines AD, BC ; join FD, FC, FA, FB, FG, FH ; then, because the two sides AF, ED, are equal to the two sides BE, EC ; and the angles AED, BEC“, equal; the base AD a 15. 1. is equal to the base BC ; and the remaining angles EAD, EDA, equal to the angles EBC, ECB, each to cach b; but the b 4. I, two angles ALG, EAG, in the triangle AGE, are equal to the two angles BEH, EBH, in the triangle HBE; and a fide AE in the one equal to a fide EB in the other ; the remaining fides AG, GE, in the one are equal to BH, HE, in the other, each to each“; but, because AE is equal EB, and FE com- c 26. 1, mon, and at right angles to AB, the base AF is equal to the bafe FB. For the same reason, FD is equal to FC; but FA, AD, are proved equal to FB, BC, each to each ; and the base FD equal to FC; therefore the angle FAD is equal to the angle FBC4 Again, because FA, AG, are proved equal to FB, d 8.1, BH, each to each, and the angle FAG equal to FBH ; the base FG is equal to FH. Now, fince FE, EH, are equal to FE, EG, and the base FH equal to FG; the angie FEH is equal to FEG; therefore each is a right angle €; therefore FE is at right c def. 10. 1o angles to all the lines passing through AB, BC; and therefore at right angles to the plain palling through AB, DC f: Where. f def. 3. fore, &c. PRO P. V. T H E O R. IF a right line siand in the common section of three right lines, and at right angles to them, these three right lines shall be in the same plain. Let the right line AB stand at right angles to the three right lines BC, BD, BE, in the point of contact B; these three lines shall be in the same plain. For, if not, let BD, BE, be in the same plain, and BC above it; and let the plain passing through AB, BC, be produced, till it meet the plain passing through BD, BE, and let BF be their common section, then BF is a right line?; then the three right a 3: lines BE, BD, BF, are in one plain; but AB is at right angles to BD, BE; therefore at right angles to BF, meeting BD, BE, in Bb; but the angle ABC is a right angle"; and the angles b 4. 0 ABC,c hyp: Boox XI. ABC, ABF, are in the same plain d; therefore the angle ABC is equal to ABF, a part to the whole; which is imposible ; d def. 3. therefore BC is in the same plain with BD, BE. Wherefore, &c. I there be two parallel lines, and a point taken in each of them, the right line joining these points fball be in the same plain with the parallel lines. Let AB, CD, be 'two parallel lines, and E, F, points taken in them ; then the right line EF joining these points is in the same plain with the parallels; if not, let it be elevated above the plain, as EGF ; through which let fome plain be drawn, whose common section with the plain in which the parallels are, let be EF a; then the two right lines EGF and EF bound a figure; b ax. 10. 1. which is impossible b; therefore the right line EF is not above, nor can it be below the plain, for the same reason ; therefore it is in the same plain. Wherefore, &c. PRO P. VII. and VIII. THEOR. IF two right lines be perpendicular to the same plain, thefe right lines are parallel; and, if two right lines are parallel, and one of them is perpendicular to some plain, then the other is perpendicular to the same plain. Let two right lines AB, CD, be perpendicular to the same plain, then AB is parallel to CD; and, if AB be parallel to CD, and AB be perpendicular to some plain, then CD is perpendicular to the same plain. First, let AB, CD, be perpendicular to some plain, and let them meer it in the points B, D; join BD; and, in the point D, draw ED at right angles to BD, and equal to AB; join BE, AE, AD; then, because AB is perpendicular to the plain in which BDE is, it will be at right angles to all the lines drawn in it, and touching AB”; but AB touches BD, BE, in the same plain ; therefore each of the angles ABD, ABE, is a right angle. For the same reason, each of the angles CDB, CDE, is a right angle; then, because AB is equal to DE, and BD common, the two lineş ED, DB, are equal to AB, BD; and the a def. 3 angle ABD equal to the angle EDB; for each is a right one ; Boo XI. therefore the base EB is equal to the base AD b; therefore EB, BA, are equal to ED, DA; and the base AE common; there. b 4. 8. fore the angles ABE, EDA, are equal“; but EBA is a right c 8. 1. angle; therefore EDA is likewise a right angle; therefore ED is perpendicular to AD, and likewise perpendicular to BD, DC; therefore BD, DA, DC, are in one plain d; but BD, DA, are d so in the same plain with AB ®; therefore AB, DC, are in' one e 2. plain ; and the angles ABD, CDB, right ones; therefore AB, DC, are parallelf. Second, If AB and CD are parallel, and AB perpendicular to fome plain, CD is perpendicular to the same plain. For, the same construction remaining, AB is proved at right angles to BD, BE ; and ED at right angles to DB, DA; and, because AB is parallel to CD, and DB joins them, CD, AB, are in the fame plain with DB 6; but ED is proved at right angles to DB, 8 6. DA, therefore at right angles to DC); for DC is in the same h 4. plain with DB, DA; therefore CD is at right angles to DE, DB Wherefore, &C f 28, I. i def. 3. R IGHT lines that are parallel to the same right line, altho' Let the right lines AB, CD, be each parallel to the right line EF, but not in the same plain with it, then AB will be parallel to CD. For, in EF, assume any point G, and draw GH at right angles to EF, in the same plain passing through EF, AB; and likewise GK at right angles to EF, in the plain pafsing through EF, CD; then, because EF is at right angles to GH, GK, it is also at right angles to the plain passing through GH, GK"; but EF is parallel to AB ; therefore AB is also at a 4. right angles to the plain passing through GH, GKb. For the bx fame reason, CD is perpendicular to the same plain ; therefore AB is parallel to CD); for each is at right angles to the same plain. Wherefore, &c. PRO P. X, TH E O R. IF two right lines touching one another, be parallel to two other right lines touching one another, but not in the same plain ; these right lines contain equal angles, Let |