XVIII.

A right cone is the solid figure generated by the revolution of a right-angled triangle about one of the sides including the right angle, which remains fixed.

The cone is called a right-angled, an obtuse-angled, or an acute-angled cone, according as the fixed side of the generating triangle is equal to, less than, or greater than the other side including the right angle.

OBS. Unless the contrary be expressly stated, whenever a cone is spoken of, a right cone is to be understood.

XIX.

The axis of a cone is the fixed straight line about which the generating triangle revolves.

XX.

The base of a cone is the circle swept out by that side of the generating triangle including the right angle, which revolves.

XXI.

A cylinder is the solid figure generated by the revolution of a rectangle about one of its sides, which remains fixed.

XXII.

The axis of a cylinder is the fixed straight line about which the generating rectangle revolves.

XXIII.

The bases of a cylinder are the two circles swept out by the opposite sides of the generating rectangle, which revolve.

XXIV.

Cones and cylinders are defined to be similar when they have their axes and the radii of their bases proportional.

XXV.

A cube is a polyhedron bounded by six equal

squares.

XXVI.

A regular tetrahedron is a polyhedron bounded by four equal equilateral triangles.

XXVII.

A regular octahedron is a polyhedron bounded by eight equal equilateral triangles.

XXVIII.

A regular dodecahedron is a polyhedron bounded by twelve equal regular pentagons.

XXIX.

A regular icosahedron is a polyhedron bounded by twenty equal equilateral triangles.

OBS. The polyhedrons defined in the five preceding defns are regular polyhedrons (Bk. xi. Def. 10. Obs. 2); and it can be proved that these five are the only regular polyhedrons which exist. Unless the contrary be expressly stated, when a tetrahedron, octahedron, dodecahedron, or icosahedron is spoken of, a regular one is to be understood.

A.

A parallelopiped is a polyhedron bounded by six parallelograms, each opposite pair of which are equal and parallel.

PROPOSITIONS.

PROP. I. THEOR.

One part of a straight line cannot be in a plane and another part without the plane.

For if there can, let. if possible, one part AB of a straight line ABC be in the plane EF, and another part BC without the plane. Then since the straight line AB is in the plane EF, it can be produced to any length required in a straight line in this plane (Post. 2); let it be produced to a point D in the plane EF. Also take some

plane passing through the straight line AD; and let it revolve about AD, remaining fixed, until it pass through the point c, and let it then have arrived into the position AGD, C being a point, in the plane agd.

Then because the points B, C are both in the plane AGD, the straight line BC is in this plane, by the def" of a plane (Def. 7). Hence there are two straight lines ABC, ABD in the same plane AGD which have a common

segment AB; which is impossible (i. 11. Cor.). Therefore one part of a straight line cannot be in a plane, and another part without the plane. Which was to be proved.

PROP. II. THEOR.

Two straight lines which cut one another shall be in one plane. And three straight lines, each of which cuts the other two, shall be in one plane.

E.

I. Let the two straight lines AB, CD cut one another in Then AB, CD shall be in one plane.

Take some plane passing through AB; and let it revolve about AB, remaining fixed, until it pass through the point c, and let it then have arrived into the position BAF, c being a point in the plane BAF.

Then because the points E, o are both in the plane BAF, the straight line EC which joins them is in this plane by the def" of a plane (Def. 7). Therefore AB, EC are in the plane BAF; but CD is in the same plane BAF that CE is in, since one part of a straight line cannot be in a plane, and another part without it (xi. 1): therefore AB, CD are both in one plane ABF. proved.

F

C

E

X

B

D

Which was to be

II. Let EC, CB, BE be three straight lines, each of which cuts the other two, the points of intersection being E, C, B. Then they shall all three be in one plane.

For it may be shewn as in Part I. of the prop" that the straight line EC is in a plane BEF which passes through BE, and that the straight line BC is in the plane BEF. Therefore EB, BC, CE are all three in one plane BEF. Which was to be proved.

PROP. III. THEOR.

If two planes cut one another: then their common section shall be a straight line.

Then the line DB

Let the two planes AG, CK cut one another, and let their common section be the line DB. shall be a straight line.

A a

Then G

For if not: let, if possible, DB be not a straight line. since B, D are two points in the plane AG, the straight line which joins them is in this plane (Def. 7), and it does not coincide with BD; let it be BED. Also since B, D are two points in the plane CK, the straight line which joins them is in this plane, and it does not coincide with BD; let it be BFD.

K

Because BED, BFD are in two different planes, and do not coincide with BD, they are different straight lines, and they have the same points B, D for their extremities; therefore the two straight lines BED, BFD enclose a space: which is impossible (Ax. 10). Therefore BD is a straight line. Which was to be proved.

If a straight line be at right angles to each of two straight lines cutting one another at the point of their intersection: then it shall be perpendicular to the plane, which passes through them, or in which they

are.

Let the straight line AB be at right angles to each of the two straight lines CD, EF which cut one another at the point of their intersection B. Then AB shall be perpendicular to the plane (xi. 2) passing through EF, CD. In BC take any point &, and from BF, BD, BE cut off (i. 3) BH, BK, BL each equal to BG. Join GH, LK; and through B draw in the plane, in which CD, EF are, any straight line MBN, cutting GH in мM and LK in N. Lastly, join ag, am, ah, AK, AN, AL.

By const" the four straight lines BG, BH, BK, BL are equal. And the angles GBH, LBK are equal, since they are opposite vertical angles (i. 15); therefore the two triangles GBH, KBL have the two sides GB, BH respectively equal to the two sides KB, BL and the included angle GBH equal to the included angle KBL. Therefore these two triangles are equal in every respect (i. 4); and hence the