B D E C equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, and the angle DAB is equal to the angle EAB: therefore the base DB is equal (4. 1.) to the base BE. And because BD, DC are greater (20. 1.) than CB, 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. And because DA is equal to EA, and AC common, but the base DC greater than the base EC: therefore the angle DAC is greater (25. 1.) than the angle EAC: 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. But BAC is not less than either of the angles DAB, BAC: therefore BAC, with either of them, is greater than the other. Wherefore, if a solid angle, &c. Q. E. D. PROP. XXI. THEOR. 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 are 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 angles CBA, ABD, DBC, any two of them are greater (20. 11.) than the third; 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 anD gles DBC, BCD, CDB; but the three angles DBC, BCD, CDB, are equal to two right angles (32. 1.): 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. B A Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; these together are less than four right angles. Let the planes in which the angles are, be cut by a plane, and let the common section 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 (20. 11.) than the third, 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. the 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 (32. 1.); that is, as there are sides in the polygon BCDEF: E D F 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. Cor. 32. 1.): therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles. 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 at the vertex, which contain the solid angle at A, are less than four right angles. Therefore every solid angle, &c. Q. E. D. PROP. XXII. THEOR. Ir every two or 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.* Let ABC, DEF, GHK be three plane angles, whereof every two are greater than the third, and are contained by the equal straight lines AB, BC, DE, EF, GH, HK ; if their extremities be joined by the straight lines AC, DF, GK, a triangle may be made of three straight lines equal to AC, DF, GK; that is, every two of them together greater than the third. If the angles at B, E, H are equal; AC, DF, GK are also equal (4. 1.), and any two of them greater than the third: but if the angles be not all equal, let the angle ABC be not less than either of the two at E, H; therefore the straight line AC is not less than either of the other two DF, GK (4. Cor. 24. 1.); and it is plain that AC, together with either of the other two, must be greater than the third: also, DF, with GK, are greater than AC: for at the point B in the straight line AB make (23. 1.) the angle ABL * See Note. equal to the angle GHK, 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, and the angle ABL to the angle GHK, the base AL is equal to the base GK; and because the angles at E, H are greater than the angle ABC, of which the angle at H is equal to ABL; therefore the remaining angle at E is greater than the angle LBC; and because the two sides LB, BC are equal to the two DE, EF, and that the angle DEF is greater than the angle LBC, the base DF is greater (24. 1.) than the base LC: and it has been proved that GK is equal to AL; therefore DF and GK are greater than AL and LC; but AL and LC are greater (20. 1.) than AC: much more then 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 (22. 1.) the sides of which shall be equal to AC, DF, GK. Q. E. D. PROP. XXIII. PROB. 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.* Let the three given plane angles be ABC, DEF, GHK, any two of which are greater than the third, and all of them together less than four right angles. It is required to make a solid angle contained by three plane angles equal to ABC, DEF, GHK, each to each. From the straight lines containing the angles, cut off AB, BC, DE, EF, GH, HK, all equal to one another; and join AC, DF, GK; then a triangle may be made (22. 11.) of three straight lines equal to AC, DF, GK. Let this be the triangle LMN (22. 1.) so that AC be equal to LM, DF to MN, and GK to LN; and about the triangle LMN describe (5. 4.) a circle, and find its centre X, which will either be within the triangle, or in one of its sides, or without it. L R First; let the centre X be within the triangle, and join LX, MX, NX: AB is 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 ABC is equal to the angle LXM (8. 1.). For the same reason, the angle DEF is equal to the angle MXN, and the angle GHK to the angle NXL; therefore the three angles ABC, DEF, GHK are equal to the three angles LXM, MXN, NXL: but the three angles LXM, MXN, NXL are equal to four right angles (2. Cor. 15. 1.): therefore also the three angles ABC, DEF, GHK, are equal to four right angles; but by the hypothesis, they are less than four right angles; which is absurd; therefore AB is M 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, X N the side of it on which is the centre X, describe the triangle LOM, the sides LO, OM of which are equal to AB, BC; and because the base LM is equal to the base AC, the angle LOM is equal to the angle ABC (8. 1.): and AB, that is LO, by the hypothesis, is less than LX; wherefore LO, OM fall within the triangle LXM; for if they fell upon its sides, or without it, they would be equal to, or greater than LX, XM (21. 1.): therefore the angle LOM, that is the angle ABC, is greater than the angle LXM (21. 1.): in the same manner it may be proved that the angle DEF is greater than the angle Next, let the centre X of the circle L Ꭱ X N Ꭱ fall in one of the sides of the triangle, viz. in MN, and join XL: in this case also AB is greater than LX. If not, AB is either equal to LX, or less than it; first, let it be equal to XL: therefore AB and BC, that is, DE and EF, are equal to MX and XL, that is, to MN: but by the construction, MN is equal to DF; therefore DE, EF are equal to DF, which is impossible (20. 1.): wherefore AB is not equal to LX; nor is it less; for then, much more, an absurdity would follow therefore AB is greater than LX. But let the centre X of the circle fall without the triangle LMN, and join LX, MX, NX. In this case likewise AB is greater than LX: if not, it is either M L N X equal to or less than LX: first, let it be equal; it may be proved in the same manner, as in the first case, that the angle ABC is equal to the angle MXL, and GHK to LXN; therefore the whole angle MXN is equal to the two angles ABC, GHK; but ABC and GHK are together greater than the angle DEF; therefore also the angle MXN is greater than DEF. And because DE, EF are equal to MX, XN, and the base DF to the base MN, the angle MXN is equal (8. 1.) to the angle DEF: and it has been proved that it is greater than DEF, which is absurd. Therefore AB is not equal to LX. Nor yet is it less; for then, as has been proved in the first case, the angle ABC is greater than the angle MXL, and the angle GHK greater than the angle LXN. At the point B in the straight line CB make the angle CBP equal to the angle GHK, and make BP equal to HK, and join CP, AP. And because CB is equal to GH; CB, BP are equal to GH, HK, each to each, and they contain equal angles; wherefore the base CP is equal to the base GK, that is, to LN. And in the isosceles triangles ABC, MXL, because the angle ABC is greater than the angle MXL, therefore the angle MLX at the base is is greater (32. 1.) than the angle ACB at the base. For the same reason, because the angle GHK, or CBP, is greater than the angle LXN, the angle XLN, is greater than the angle CBP. Therefore the whole angle MLN is greater than the whole angle ACP. And because ML, LN are equal to AC, CP, each to each, but the angle |