cone; if lefs, an obtufe angled cone; and if greater, an a- Book XI.

cute angled cone.

XIX,

The axis of a cone is that fixed right line about which the tri

angle is moved.

XX.

The base of a cone is the circle defcribed by the revolving line.

XXI.

A cylinder is, a figure described by a right angled parallelogram, revolving about one of the fides, containing the right angle, remaining fixed.

XXII.

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 defcribed by the motion of the two oppofite fides of the parallelogram.

XXIV.

Similar cones and cylinders are fuch, whofe axes and diameters of their bafes are proportional.

XXV.

A cube is a folid figure contained by fix equal squares.

XXVI.

A tetrahedron is a folid figure contained by four equal equilateral triangles.

XXVII.

An octahedron is a folid figure contained by eight equilateral triangles.

XXVIII.

A dodecahedron is a folid figure contained by twelve equal equilateral and equiangular pentagons.

XXIX.

An icofahedron is a folid figure contained by twenty equal equilateral triangles.

XXX.

A parallelopipedon is a folid figure contained by fix quadrilateral figures, whereof those that are oppofite are parallel.

PROP. I. THE OR.

NE part of a right line cannot be in a plain fuperficies, and
another part above it.

For,

BOOK XI. For, if poffible, let the part AB of the right line ABC be in a plain fuperficies, and the part BC above the fame; there will be fome 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 fegment AB; which is impofa Cor. 14. 1. fible. Wherefore, &c.

a I.

def, 4. 11.

PRO P. II. THE OR.

IF
F two 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 plàin, 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. Wherefore, &c.

Ax. Io. I.

.

PROP. III. THE OR.

F two plains cut each other, their common feclion will be a right line.

IF

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; which cannot be a. Wherefore, &c.

PROP. IV. THE OR.

IF a right line ftand in the common fection of two right lines, cutting one another, and at right angles to the fame; then it Shall be at right angles to the plain paffing through thefe lines.

Book XI,

Let the right line EF ftand in the common fection at right angles to the two right lines AB, CD; then EF is likewife at right angles to the plain paffing 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 fides AE, ED, are equal to the two fides BE, EC; and the angles AED, BEC, equal; the bafe AD a 15. I. is equal to the base BC; and the remaining angles EAD, EDA, equal to the angles EBC, ECB, each to each ; but the b 4. I, two angles AEG, 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, cach to each; but, because AE is equal to EB, and FE com- c 26. 1, mon, and at right angles to AB, the bafe AF is equal to the bafe FBb. For the fame reafon, FD is equal to FC; but FA, AD, are proved equal to FB, BC, each to each; and the bafe FD equal to FC; therefore the angle FAD is equal to the angle FBC. Again, because FA, AG, are proved equal to FB, d 8. I, BH, each to each, and the angle FAG equal to FBH; the bafe FG is equal to FH. Now, fince FE, EH, are equal to FE, EG, and the bafe FH equal to FG; the angle FEH is equal to FEG; therefore each is a right angle; therefore FE is at right e def. 10. 1, angles to all the lines paffing through AB, BC; and therefore at right angles to the plain paffing through AB, DC, Where f def. 3. fore, &c.

PRO P. V. THE O R.

a right line fand in the common fection of three right lines, and at right angles to them, these three right lines fhall be in the fame plain.

Let the right line AB ftand at right angles to the three right lines BC, BD, BE, in the point of contact B; these three lines fhall be in the fame plain.

For, if not, let BD, BE, be in the fame plain, and BC above it; and let the plain paffing through AB, BC, be produced, till it meet the plain paffing through BD, BE; and let BF be their common fection, 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 angles; and the angles b 4. ABC, c hyp:

Q

Book XI.ABC, ABF, are in the fame plain ; therefore the angle ABC is equal to ABF, a part to the whole; which is impoffible; d def. 3. therefore BC is in the fame plain with BD, BE. Wherefore,

&c.

IF there be two parallel lines, and a point taken in each of them, the right line joining thefe points fhall be in the fame 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 fame plain with the parallels; if not, let it be elevated above the plain, as EGF; through which let fome plain be drawn, whofe common fection with the plain in which the parallels are, let be EF; then the two right lines EGF and EF bound a figure; bax. 10. 1.which is impoffible 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 fame plain. Wherefore, &c.

a 3.

PRO P. VII. and VIII. THE OR.

IF two right lines be perpendicular to the fame plain, these right lines are parallel; and, if two right lines are parallel, and one of them is perpendicular to fome plain, then the other is perpendicular to the fame plain.

Let two right lines AB, CD, be perpendicular to the fame plain, then AB is parallel to CD; and, if AB be parallel to CD, and AB be perpendicular to fome plain, then CD is perpendicular to the fame plain.

First, let AB, CD, be perpendicular to fome plain, and let them meet it in the points B, D; join BD; and, in the point D, draw ED at right angles to BĎ, and equal to AB; join BE, AE, AD; then, becaufe AB is perpendicular to the plain in which BDE is, it will be at right angles to all the lines drawn in a def. 3. it, and touching AB"; but AB touches BD, BE, in the fame plain; therefore each of the angles ABD, ABE, is a right angle. For the fame reafon, each of the angles CDB, CDE, is a right angle; then, because AB is equal to DE, and BD common, the two lines ED, DB, are equal to AB, BD; and the

Boo XI.

angle ABD equal to the angle EDB; for each is a right one; therefore the base EB is equal to the base AD b; therefore EB, BA, are equal to ED, DA; and the bafe AE common; there- b 4. 1. fore the angles ABE, EDA, are equal; but EBA is a right c 3. 1. angle; therefore EDA is likewife a right angle; therefore ED is perpendicular to AD, and likewife perpendicular to BD, DC; therefore BD, DA, DC, are in one plain ; but BD, DA, are d s. in the fame plain with AB; therefore AB, DC, are in one € 2. plain; and the angles ABD, CDB, right ones; therefore AB, DC, are parallel f.

f 28. I.

Second, If AB and CD are parallel, and AB perpendicular to fome plain, CD is perpendicular to the fame plain. For, the fame 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; but ED is proved at right angles to DB, 8 6. DA; therefore at right angles to DC; for DC is in the fame h 4. plain with DB, DA; therefore CD is at right angles to DE, DB. Wherefore, &c.

i def. 3.

R

PROP. IX. THEOR.

IGHT lines that are parallel to the fame right line, altho
not in the fame plain with it, are parallel to one onother.

Let the right lines AB, CD, be each parallel to the right line EF, but not in the fame plain with it; then AB will be parallel to CD. For, in EF, affume any point G, and draw GH at right angles to EF, in the fame plain paffing through EF, AB; and likewife GK at right angles to EF, in the plain paffing through EF, CD; then, because EF is at right angles to GH, GK, it is also at right angles to the plain paffing through GH, GK; but EF is parallel to AB; therefore AB is also at a 4. right angles to the plain paffing through GH, GK. For the fame reason, CD is perpendicular to the fame plain; therefore AB is parallel to CD; for each is at right angles to the fame plain. Wherefore, &c.

IF

PRO P. X. THE O R.

F two right lines touching one another, be parallel to two other right lines touching one another, but not in the fame plain ; thefe right lines contain equal angles.

Let

b 70