rallel to one another; and the others parallelograms.

XIV.

A sphere is a solid figure described by the revolution of a semicircle about its diameter, which remains unmoved.

XV.

The axis of a sphere is the fixed straight line about which the semicircle revolves. XVI. The centre of a sphere is the same with that of the semicircle. XVII. The diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the superficies of the sphere. XVIII.

A cone is a solid figure described by the revolution of a right angled triangle about one of the sides containing the right angle, which side remains fixed.

If the fixed side be equal to the other side containing the right angle, the cone is called a right angled cone; if it be less than the other side, an obtuse angled, and if greater, an acute angled cone.

XIX. The axis of a cone is the fixed straight line about which the triangle revolves. XX. The base of a cone is the circle described by that side containing the right angle, which revolves.

XXI.

A cylinder is a solid figure described

If it be possible, let AB, part of the straight line ABC, be in the plane, and the part BC above it: And since the straight line AB is in the plane, it can be produced in that plane: Let it

be produced to D: And let any plane pass through the straight line AD, A B D and be turned about

straight lines AE, ED are equal to the two BE, EC, and that they contain equal angles (10. Ax. 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 (10. Ax. 1.); therefore the triangles AEG, BEH have two angles of one equal to two angles of the other, each to each, and the sides AF, 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, A are equal to the two a FB, BC, each to each; and the base DF was proved equal to theD base FC; therefore

E

C

H

B

the angle FAD is equal (8. 1.) to the

angle FBC: Again it was proved that AG is equal to BH, and also AF to FB; FA, then, and AG, are equal to FB and BH, and the angle FAG has been proved equal to the angle FBH; therefore the base GF is equal (4. 1.) to the base FH: Again, because it was proved that GE is equal to EH, and EF is common; GE, EF are equal to HE, EE; and the base GF is equal to the base FH: therefore the angle GEF is equal (8. 1.) to the angle HEF; and consequently each of these angles is a right (10. def.) 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 manuer, 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.

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 (3. 11.) line; let this be BF: Therefore the three straight lines AB, BC, BF, are all in one plane, viz. that which passes thro' AB, BC; and because AB stands at

fore the angle ABF is equal to the angle ABC, and they are both in the same plane, which is impossible: Therefore the straight line BC is not above the plane in which are BD and

BE: Wherefore the three straight lines BC, BD, BE are in one and the same plane. Therefore, if three straight lines, &c. Q. E. D.

base AD is equal (4. 1.) to the 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. ̧

between them, which is impossible: (10. Ax. 1.) Therefore the straight line joining the 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.

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, CD, 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 A to DE, and BD common, the two AB, BD are equal to the two ED, DB, and the angle ABD B is equal to the angle EDB, because each of them is a right angle;

E

C

therefore the base AD is equal (4. 1.) to the base BE: Again, because AB is equal to DE, and BE 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 perpendicular to BD; therefore ED is perpendicular (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 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, DB 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 plane to which AB is at right angles. Therefore, if two straight lines, &c. Q. E. D.

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