DEMONSTRATION. If it be not, from the point D, draw in the plane AB, the straight line DE at right angles to AD (a), the common section of the plane AB with the third plane; 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 (b): 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 (c): 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. PROPOSITION XX. THEOREM.-If a solid angle (A) be contained by three plane angles (BAC, CAD, DAB), any two of them are greater than the third. DEMONSTRATION. 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 (a); 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, EAC (e): and, by the construction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC (f), that is, than the angle BAC: but BAC is not less than either of the angles DAB, DAC; therefore BAC with either of them is greater than the other. PROPOSITION XXI. THEOREM.-Every solid angle is contained by plane angles, which together are less than four right angles. DEMONSTRATION. 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, AČ, 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 (a); 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, DCB, BDC: but the three angles DBC, DCB, BDC are equal to two right angles (6): 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, DAC, BAD, 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, DAE, EAF, FAB: these shall together be less than four right angles. F 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 (a), 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 BCDEFB; 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 (6), that is, as there are sides in the polygon BCDEFB; 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 (c); therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles (d): but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has een proved; wherefore the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at ▲, are less than four right angles. SCHOLIUM. This proposition does not hold good if any of the angles of the rectilineal figure BCDEFB be re-entrant, the reason of which will be seen from the Scholia to Corollaries 7 and 8, Prop. 32 в, Book 1. PROPOSITION XXII. THEOREM.-If every two of three plane angles (B, E, H) be greater than the third, and if the straight lines (AB, BC, DE, EF, GH, HK) which contain them be all equal, a triangle may be made of the straight lines (AC, DF, GK) that join the extremities of those equal straight lines. DEMONSTRATION. If the angles B, E, H are equal, AC, DF, GK are also equal (a), and any two of them greater than the third: but if the angles are not all equal, let the angle ABC be not less than either of the two E, H; therefore the straight line AC is not less than either of the other two DF, GK (6): and therefore it is plain that AC, together with either of the other two, must be greater than the third: also DF, with GK, shall be greater than AC; for at the point B, in the straight line AB, form the angle ABL equal to the angle H (c), and make BL equal to one of the straight lines AB, BC, DE, EF, GH, HK, and join AL, LC. Then, because AB, BL are equal to GH, HK, each to each, and the angle ABL to the angle GHK, the base AL is equal to the base GK (a): and because the angles E, H are greater than the angle ABC (d), of which the angle H is equal to ABL, therefore the remaining angle E is greater than the angle LBC (e): and because the two sides LB, BC are equal to the two DE, EF, each to each, and that the angle E is greater than the angle LBC, the base DF is greater than the base LC (f): and it has been proved that GK is equal to AL; therefore DF and GK are greater than AL and LC (g): but AL and LC are greater than AC (h); much more than are DF and GK greater than AC. Wherefore, every two of these straight lines AC, DF, GK are greater than the third; and, therefore, a triangle may be made (i), the sides of which shall be equal to AC, DF, GK. PROPOSITION XXIII. PROBLEM. To make a solid angle which shall be contained by three given plane angles (B, E, H), any two of them being greater than the third, and all three together less than four right angles. about the triangle LMN describe a (c) VI. 5. (d) III. 1. I. 8. circle (c), and find its center X (d), which will be either within the triangle, or in one of its sides, or without it. N First, let the center X be within the triangle, and join LX, MX, NX: AB shall be greater than LX. If not, AB must either be equal to, or less than LX: first let it be equal: then, because AB is equal to LX, and that AB is also equal to BC, and LX to XM, AB and BC are equal to LX and XM, each to each; and the base AC is, by construction, equal to the base LM; wherefore the angle B is equal to the angle LXM (e): for the same reason, the angle E is equal to the angle MXN, and the angle H to the angle NXL; therefore the three angles B, E, H are equal to the three angles LXM, MXN, NXL: but the three angles LXM, MXN, NXL are equal to four right angles (f); therefore also the three angles B, E, H are equal to four right angles: but, by the hypothesis, they are less than four right angles; which is absurd: therefore AB is not equal to LX. But neither can AB be less than LX: for, if possible, let it be less: and upon the straight line LM, on the side of it on which is the center X, describe the triangle LOM (b), (f) I. 13, Cor. 3. I. 21. |