Because the two parallel planes KL, MN are cut by the plane EBDX, the cominon sections EX, B D are parallel : (XI. 16.) for the same reason, because the two parallel planes GH, KL are cut by the plane XXFC, the common sections AC, XF are parallel : and because EX is parallel to BD, a side of the triangle ABD; as AE to EB, so is AX to XD: (VI. 2.) again, because XF is parallel to AC, a side of the triangle ADC: as AX to XD, so is CF to FD: Wherefore, if two straight lines, &c. 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 A Let any plane D E pass through AB, and let CE be the common section of the planes DE, CK; take any point F in 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: therefore AB is parallel to FG: (I. 28.) 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. Therefore, if a straight line, &c. Q.E.D. PROPOSITION XIX. THEOREM. 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. B EF 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 B C 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. (XI. 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. PROPOSITION XX. THEOREM. 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. B Ed 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, in the plane which passes through BA, AC, make 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 BA 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 B D 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 DĂ 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: (1. 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. 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 ingles CBA, ABD, DBC, any two of them are greater than the third; (XI. 20.) therefore the angles CBA, ABD are greater than the angle DBC: 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; 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, DAE, 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, ABF, FBC, of which any two are greater than the third, (XI. 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 : 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.) 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. NOTES TO BOOK XI. The solids considered in the eleventh and twelfth books are Geometrical solids, portions of space bounded by surfaces which are supposed capable of penetrating and intersecting one another. In the first six books, all the diagrams employed in the demonstrations, are supposed to be in the same plane which may lie in any position whatever, and be extended in every direction, and there is no difficulty in representing them roughly on any plane surface ; this, however, is not the case with the diagrams employed in the demonstrations in the eleventh and twelfth books, which cannot be so intelligibly represented on a plane surface on account of the perspective. A more exact conception may be attained, by adjusting pieces of paper to represent the different planes, and drawing lines upon them as the constructions may require, and by fixing pins to represent the lines which are perpendicular to, or inclined to any planes. Any plane may be conceived to move round any fixed point in that plane, either in its own plane, or in any direction whatever ; and if there be two fixed points in the plane, the plane cannot move in its own plane, but may move round the straight line which passes through the two fixed points in the plane, and may assume every possible position of the planes which pass through that line, and every different position of the plane will represent a different plane ; thus, an indefinite number of planes may be conceived to pass through a straight line which will be the common intersection of all the planes. Hence, it is manifest, that though two points fix the position of a straight line in a plane, neither do two points nor a straight line fix the position of a plane in space. If, however, three points, not in the same straight line, be conceived to be fixed in the plane, it will be manifest, that the plane cannot be moved round, either in its own plane or in any other direction, and therefore is fixed. Also, any conditions which involve the consideration of three fixed points not in the same straight line, will fix the position of a plane in space; as also two straight lines which meet or intersect one another, or two parallel straight lines in the plane. Def. v. When a straight line meets a plane, it is inclined at different angles to the different lines in that plane which may meet it; and it is manifest that the inclination of the line to the plane is not determined by its meeting any line in that plane. The inclination of the line to the plane can only be determined by its inclination to some fixed line in the plane. If a point be taken in the line different from that point where the line meets the plane, and a perpendicular be drawn to meet the plane in another point; then these two points in the plane will fix the position of the line which passes through them in that plane, and the angle contained by this line and the given line, will measure the inclination of the line to the plane; and it will be found to be the least angle which can be formed with the given line and any other straight line in the plane. If two perpendiculars be drawn upon a plane from the extremities of a straight line which is inclined to that plane, the straight line in the plane intercepted between the perpendiculars is called the projection of the line on that plane; and it is obvious that the inclination of a straight line to |