and it was proved that AX is to XD, as AE to EB; Q.E.D. PROPOSITION XVIII. THEOREM. If a straight line be at right angles to a plane, every plane which passes through it shall be at right angles to that plane. Let the straight line AB be at right angles to the plane CK. Every plane which passes through AB shall be at right angles to the plane CK. D G A Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; take any point Fin CE, from which draw FG in the plane DE at right angles to CE. (1. 11.) And because AB is perpendicular to the plane CK, therefore it is also perpendicular to every straight line in that plane meeting it; (xi. def. 3.) and consequently it is perpendicular to CE: wherefore ABF is a right angle: but GFB is likewise a right angle; (constr.) and AB is at right angles to the plane CK; therefore FG is also at right angles to the same plane. (XI. 8.) But one plane is at right angles to another plane, when the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other plane; (XI. def. 4.) and any straight line FG in the plane DE, which is at right angles to CE, the common section of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all planes which pass through AB, are at right angles to the plane CK. If two planes which cut one another be each of them perpendicular to a third plane; their common section shall be perpendicular to the same plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common section of the first two. Then BD shall be perpendicular to the third plane. If it be not, from the point D draw, in the plane AB, the straight line DE at right angles to AD, the common section of the plane AB with the third plane; (1. 11.) and in the plane BC draw DF at right angles to CD the common section of the plane BC with the third plane. And because the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD, their common section, DE is perpendicular to the third plane. (xı. def. 4.) In the same manner, it may be proved, that DF is perpendicular to the third plane. Wherefore, from the point D two straight lines stand at right angles to the third plane, upon the same side of it, which is impossible: (XI. 13.) therefore, from the point D there cannot be any straight line at right angles to the third plane, except BD the common section of the planes AB, BC: therefore BD is perpendicular to the third plane. Wherefore, if two planes, &c. Q.E.D. If a solid angle be contained by three plane angles, any two of them are greater than the third. Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB. Any two of them shall be greater than the third. If the angles BAC, CAD, DAB be all equal, it is evident, that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A in the straight line AB, make, in the plane which passes through BA, AC, the angle BAE equal to the angle DAB; (1. 23.) and make AE equal to AD, and through E draw BEC cutting AB, AC, in the points B, C, and join DB, DC. And because DA is equal to AE, and AB is common, the two DA, AB, are equal to the two EA, AB, each to each; and the angle DAB is equal to the angle EAB; therefore the base DB is equal to the base BE: (1. 4.) and because BD, DC are greater than CB, (1. 20.) and one of them BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC: (1. ax. 5.) and because DA is equal to AE, and AC common, but the base DC greater than the base EC; therefore the angle DAC is greater than the angle EAC; (1. 25.) and, by the construction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC: (I. ax. 4.) but BAC is not less than either of the angles DAB, DAC: therefore BAC, with either of them, is greater than the other. Wherefore, if a solid angle, &c. Q.E.D. PROPOSITION XXI. THEOREM. Every solid angle is contained by plane angles, which together are less than four right angles. First, let the solid angle at A be contained by three plane angles BAC, CAD, DAB. These three together shall be less than four right angles. D B Take in each of the straight lines AB, AC, AD, any points B, C, D, and join BC, CD, DB. Then, because the solid angle at B is contained by the three plane. angles CBA, ABD, DBC, any two of them are greater than the third; (x1. 20.) therefore the angles CBA, ABD are greater than the angle DBC: BC: for the same reason, the angles BCA, ACD are greater than the angle DCB; and the angles CDA, ADB greater than BDC: wherefore the six angles CBA, ABD, BCA, ACD, CDA, ADB, are greater than the three angles DBC, BCD, CDB: but the three angles DBC, BCD, CDB are equal to two right angles; (1. 32.) therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles : and because the three angles of each of the triangles ABC, ACD, ADB are equal to two right angles, therefore the nine angles of these three triangles, viz. the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD are equal to six right angles ; of these the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles: therefore the remaining three angles BAC, CAD, DAB, which contain the solid angle at A, are less than four right angles. Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAĚ, EAF, FAB. These shall together be less than four right angles. Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB. And because the solid angle at B is contained by three plane angles CBA, ÀBF, FBC, of which any two are greater than the third, (x1. 20.) the angles CBA, ABF, are greater than the angle FBC: for the same reason, the two plane angles at each of the points C, D, E, F, viz. those angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the angles of the polygon BCDEF: therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon: and because all the angles of the triangles are together equal to twice as many right angles as there are triangles; (1. 32.) that is, as there are sides in the polygon BCDEF; and that all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon; (1. 32. Cor. 1.) therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles: (1. ax. 1.) but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved; wherefore the remaining angles of the triangles, viz. those of the vertex, which contain the solid angle at A, are less than four right angles. Therefore, every solid angle, &C. ́ Q.E.D. PROPOSITION XXII. THEOREM. If every two of three plane angles be greater than the third, and if the straight lines which contain them be all equal; a triangle may be made of the straight lines that join the extremities of those equal straight lines. PROPOSITION XXIII. PROBLEM. To make a solid angle which shall be contained by three given plane angles, any two of them being greater than the third, and all three together less than four right angles. PROPOSITION A. THEOREM. If each of two solid angles be contained by three plane angles, which are equal to one another, each to each; the planes in which the equal angles are, have the same inclination to one another. PROPOSITION B. THEOREM. If two solid angles be contained, each by three plane angles which are equal to one another, each to each, and alike situated; these solid angles are equal to one another. PROPOSITION C. THEOREM. Solid figures which are contained by the same number of equal and similar planes alike situated, and having none of their solid angles contained by more than three plane angles, are equal and similar to one another. PROPOSITION XXIV. THEOREM. If a solid be contained by six planes, two and two of which are parallel; the opposite planes are similar and equal parallelograms. PROPOSITION XXV. THEOREM. If a solid parallelopiped be cut by a plane parallel to two of its opposite planes; it divides the whole into two solids, the base of one of which shall be to the base of the other, as the one solid is to the other. PROPOSITION XXVI. PROBLEM. At a given point in a given straight line, to make a solid angle equal to a given solid angle contained by three plane angles. PROPOSITION XXVII. PROBLEM. To describe from a given straight line a solid parallelopiped similar,. and similarly situated, to one given. PROPOSITION XXVIII. THEOREM. If a solid parallelopiped be cut by a plane passing through the diagonals of two of the opposite planes; it shall be cut in two equal parts. PROPOSITION XXIX. THEOREM. Solid parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another. PROPOSITION XXX. THEOREM. Solid parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another. PROPOSITION XXXI. THEOREM. Solid parallelopipeds, which are upon equal bases, and of the same altitude, are equal to one another." Solid parallelopipeds which have the same altitude are to one another as their bases. COR. From this it is manifest, that prisms upon triangular bases, of the same altitude, are to one another as their bases. PROPOSITION XXXIII. THEOREM. Similar solid parallelopipeds are one to another in the triplicate ratio of their homologous sides. COR. From this it is manifest, that, if four straight lines be continual proportionals, as the first is to the fourth, so is the solid parallelopiped described from the first to the similar solid similarly described from the second; because the first straight line has to the fourth the triplicate ratio of that which it has to the second. |