22. SIMILAR CONES AND CYLINDERS are those which have their axes and the diameters of their bases proportionals.

23. A PARALLELOPIPED is a solid figure contained by six quadrilateral figures, whereof every opposite two are parallel.

SCHOLIUM. A parallelopiped is a prism with parallelograms for its base. When its sides are rectangles it is said to be right, if otherwise, oblique.

24. A POLYHEDRON is a solid figure contained by plane figures.

SCHOLIUM. When all the plane figures are equal and similar, the polyhedron is said to be regular.

25. A CUBE, or HEXAHEDRON, is a solid figure contained by six equal squares.

26. A TETRAHEDRON is a solid figure contained by four equal and equilateral triangles.

27. An OCTAHEDRON is a solid figure contained by eight equal and equilateral triangles.

28. A DODECAHEDRON is a solid figure contained by twelve equal pentagons which are equilateral and equiangular.

29. An ICOSAHEDRON is a solid figure contained by twenty equal and equilateral triangles.

PROPOSITION I.

THEOREM.-One part of a straight line cannot be in a

plane, if another part is above it.

DEMONSTRATION. 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, and be turned about

(b) I. 11 cor.

it until it pass through the point C; and because the points B, C are in this plane, the straight line (a) BC is in it: therefore there are two straight lines ABC, ABD in the same plane that have a common segment AB: which is impossible (b). Therefore AB and CD are in the same plane.

PROPOSITION II.

THEOREM.-If two straight lines (AB, CD) cut one another, they are in one plane; and if three straight lines (EC, CB, BE) meet one another, they are in one plane.

DEMONSTRATION. Let any

plane pass through the straight line EB, and let the plane be turned about EB, produced if necessary, until it pass through the point C: then, because the points E, C are in this plane, the straight line EC is in it (a); for the same reason, the straight line BC is in the same; and by the hypothesis, EB is in it; therefore the three straight lines EC, CB, BE are in one plane: but in the plane in

which EC, EB are, in the same are CD, AB (6); therefore AB, C'D

are in one plane.

PROPOSITION III.

THEOREM.-If two planes (AB, BC) cut one another, their common section (DB) is a straight line.

DEMONSTRATION. If it be not, from the point D to B, draw, in the plane AB, the straight line DEB, and in the plane BC, the straight line DFB: then two straight lines DEB, DFB have the same extremities, and therefore include a space betwixt them; which is impossible (a); therefore BD, the common section of the planes AB, BC, cannot but be a straight line.

(a) I. Ax. 10.

PROPOSITION IV.

THEOREM.-If a straight line (EF) stand at right angles to each of two straight lines (AB, CD) in the point of their intersection (E), it shall also be at right angles to the plane which passes through them, that is, to the plane in which they are.

DEMONSTRATION. Take

the straight lines AE, EB, CE, ED, all equal to one another; and through E, draw, in the plane in which are AB, CD, any straight line GEH, and join AD, CB: then from any point F, in EF, draw FA, FG, FD, FC, FH, FB: and because the two straight lines AE, ED are equal to the two BE, EC, each to each, and that they contain equal angles AED, BEC (a), the base AD is

B

equal to the base BC, and the angle DAE to the angle EBC (6): and the angle AEG is equal to the angle BEH (a); therefore the triangles AEG, BEH have two angles of the one, equal to two angles of the other, each to each, and the sides AE, EB, adjacent to the equal angles, equal to one another; wherefore they have their other sides equal (c); therefore GE is 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 to the base FB (6); 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 are equal to the two FB, BC, each to each; and the base DF was proved equal to the base FC; therefore the angle FAD is equal to the angle FBC (d); again, it was proved that GA is equal to BH, and also AF to FB; therefore FA and AG, are equal to FB and BH, each to each; and the angle FAG has been proved equal to the angle FBH; therefore the base GF is equal to the base FH (6): again, because it was proved that GE is equal to EH, and EF is common, therefore GE, EF are equal to HE, EF, each to each; and the base GF is equal to the base FH; therefore the angle GEF is equal to the angle HEF (d); and consequently each of these angles is a right angle (e); 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 manner 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 (f): therefore EF is at right angles to the plane in which are AB, CD.

PROPOSITION V.

THEOREM. If three straight lines (BC, BD, BE) meet all in one point (B), and a straight line (AB) stand at right angles to each of them in that point, these three straight lines are in one and the same plane.

DEMONSTRATION. 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, is a straight line (a); let this be BF: therefore the three straight lines AB, BC, BF are all in one plane, viz. that which passes through AB, BC: and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles to the plane passing through them (b); and therefore makes right angles with every straight line in that plane which meets it (c): but BF, which is in that plane, meets it; therefore the angle ABF is a right angle: but the angle ABC, by the hypothesis, is also a right angle; there

fore the angle ABF is equal to the angle ABC, and they are both in the same plane; which is impossible (d): 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.