Book XI. bafe AC, the angle LOM is equal to the angle ABC *: And AB, that is, LO, by the hypothefis, is lefs than LX; wherefore LO, OM fall within the triangle LXM; for, if they fell upon its fides, or without it, they d 8. I. f 21. I. f 20. 1. would be equal to, or greater than M L R N the fame angles ABC, DEF, GHK are lefs than four right angles; which is abfurd: Therefore AB is not lefs than LX, and it has been proved that it is not equal to LX; wherefore AB is greater than LX. L R Next, Let the center X of the circle fall in one of the fides of the triangle, viz. in MN, and join XL: In this cafe alfo AB is greater than LX. If not, AB is either equal to LX or lefs than it: Firft, let it be equal to LX: Therefore AB and BC, that is, DE and EF, are equal to MX and XL, that is, to MN: But, by the conftruction, MN is equal to DF; therefore DE, EF are equal to DF, which is impoffible+: Wherefore AB is not equal to LX; nor is it lefs; for then, much more, an abfurdity would follow: Therefore AB is greater than LX. M N X But, let the center X of the circle fall without the triangle LMN, and join LX, MX, NÄ. In this cafe likewife AB is greater than LX: If not, it is either equal to, or less than LX: First, let it be equal; it may be proved in the fame manner, as in the firft cafe, 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 alfo the angle MXN is greater than DEF. And because DE, EF d EF are equal to MX, XN, and the base DF to the bafe Book XI. MN, the angle MXN is equal to the angle DEF: And it has been proved that it is greater than DEF, which is abfurd. d 8. I, Therefore AB is not equal to LX. Nor yet is it lefs; for then, as has been proved in the firft cafe, the angle ABC is greater than the angle MXL, and the angle GHK greater than the angle LXN. At the point B in the ftraight 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; or CBP, is greater than the angle h g R Nh 24. I. X ABP. And ABP is equal to the two angles ABC, CBP, that is, to the two angles ABC, GHK; therefore the angle DEF is greater than the two angles ABC, GHK; but it is also lefs than thefe; which is impoffible. Therefore AB is not less than k 25.1. LX; 220 Book XI. LX; and it has been proved that it is not equal to it; therefore AB is greater than LX. a 12. II. b 3. def. 11. is 47. I. d 8. I. a R A From the point X erect XR at right angles to the plane of the circle LMN. And because it has been proved in all the cafes, that AB is greater than LX, find a fquare equal to the excess of the fquare of AB above the fquare of LX, and make RX equal to its fide, and join RL, RM, RN. Because RX is perpendicular to the plane of the circle LMN, it perpendicular to each of the ftraight lines LX, MX, NX. And because LX is equal to MX, and XR common, and at right angles M to each of them, the base RL is equal to the bafe RM. For the same reafon, RN is equal to each of the two RL, RM. Therefore the three ftraight lines RL, RM, RN are all equal. And because the fquare of c N XR is equal to the excefs of the square of AB above the square of LX; therefore the fquare of AB is equal to the fquares of LX, XR. But the fquare of RL is equal to the fame fquares, because LXR is a right angle. Therefore the square of AB is equal to the fquare of RL, and the straight line AB to RL. But each of the straight lines BC, DE, EF, GH, HK is equal to AB, and each of the two RM, RN is equal to RL. Wherefore AB, BC, DE, EF, GH, HK are each of them equal to each of the ftraight lines RL, RM, RN. And because RL, RM, are equal to AB, BC, and the base LM to the base AC; the angle LRM is equal to the angle ABC. For the fame reason, the angle MRN is equal to the angle DEF, and NRL to GHK. Therefore there is made a folid angle at R, which is contained by three plane angles LRM, MRN, NRL, which are equal to the three given plane angles ABC, DEF, GHK, each to each. Which was to be done, PROP Book XI. PROP. A. THEOR. Feach of two folid angles be contained by three plane See N. angles equal to one another, each to each; the planes in which the equal angles are, have the fame inclination to one another. Let there be two folid angles at the points A, B; and let the angle at A be contained by the three plane angles CAD, CAE, EAD; and the angle at B by the three plane angles FBG, FBH, HBG; of which the angle CAD is equal to the angle FBG, and CAE to FBH, and EAD to HBG: The planes in which the equal angles are, have the fame inclination to one another. B In the ftraight line AC take any point K, and in the plane CAD from K draw the ftraight line KD at right angles to AC, and in the planeCAE the ftraight line KL at right angles to the fame AC: Therefore the angle DKL is the inclination of the plane CAD to the plane A K L M N a 6. def. 11. DF G BM equal to AK, and from the point M draw, in the planes FBG, FBH, the ftraight lines MG, MN at right angles to BF; therefore the angle GMN is the inclination of the plane FBG to the plane FBH: Join LD, NG; and because in the triangles KAD, MBG, the angles KAD, MBG are equal, as alfo the right angles AKD, BMG, and that the fides AK, BM, adjacent to the equal angles, are equal to one another, therefore KD is equal b to b 26. 1. MG, and AD to BG: For the fame reafon, in the triangles. KAL, MBN, KL is equal to MN, and AL to BN: And in the triangles LAD, NBG, LA, AD are equal to NB, BG, and they contain equal angles; therefore the bafe LD is equal to the bafe NG. Laftly, in the triangles KLD, MNG, c 4.1. the fides DK, KL are equal to GM, MN, and the bafe LD to the bafe NG; therefore the angle DKL is equal to the angle d 8. 1. GMN But the angle DKL is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclina a Book XI. tion of the plane FBG to the plane FBH, which planes have therefore the fame inclination to one another: And in the 7. def. 11. fame manner it may be demonftrated, that the other planes in which the equal angles are, have the fame inclination to one another. Therefore, if two folid angles, &c. Q. E. D. See N. a A. II. I' PROP. B. THEOR. F two folid angles be contained, each by three plane angles which are equal to one another, each to each, and alike fituated; thefe folid angles are equal to one another. Let there be two folid angles at A and B, of which the folid angle at A is contained by the three plane angles CAD, CAE, EAD; and that at B, by the three plane angles FBG, FBH, HBG; of which CAD is equal to FBG; CAE to FBH; and EAD to HBG: The folid angle at A is equal to the folid angle at B. Let the folid angle at A be applied to the folid angle at B: and, firft, the plane angle CAD being applied to the plane angle FBG, fo as the point A may coincide with the point B, and the ftraight line AC with BF; then AD coincides with BG, because the angle CAD is equal to the angle FBG: And because the inclination of the plane CAE to the plane CAD is equal to the inclina 2 A B E H F D G tion of the plane FBH to the plane FBG, the plane CAE C coincides with the plane FBH, because the planes CAD, FBG coincide with one another: And because the straight lines AC, BF coincide, and that the angle CAE is equal to the angle FBH; therefore AE coincides with BH, and AD coincides with BG; wherefore the plane EAD coincides with the plane HBG: Therefore the folid angle A coincides with the folid angle B, and confequently they are e b 8. A. 6. qualb to one another. Q. Ë. D. PROP |