L R 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 MXN, and the angle GHK greater than the angle NXL. Therefore the three angles ABC, DEF, GHK are greater than the three angles LXM, MXN, NXL; that is than four right angles; but the same angles ABC, DEF, GHK are less than M four right angles; which is absurd : therefore AB is not less than LX, and it has been proved that it is not equal to LX; wherefore AB is greater than LX. L N R Next, let the centre X of the circle 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 M 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. N X 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 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. R K And because CB is equal to GH; CB, BP are equal to GH, MN is equal to DF; therefore also I N X base AP, the angle DEF is greater (25. 1.) than the angle 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 less than these ; Ee which is impossible. Therefore AB is not less than LX, and it has been proved that it is not equal to it; therefore AB is greater than LX. M X R N From the point X erect (12. 11.) XR at right angles to the plane of the circle LMN. And because it has been proved in all the cases, that AB is greater than LX, find a square equal to the excess of the square of AB above the square of LX, and make RX equal to its side; and join RL, RM, RN. Because RX is perpendicular to the plane of the circle LMN, it is (3. def. 11.) perpendicular to each of the straight lines LX, MX, NX. And because LX is equal to MX, and XR common, and at right angles to each of them, the base RL is equal to the base RM. For the same reason, RN is equal to each of the two RL, RM. Therefore the three straight lines RL, RM, RN are all equal. And because the square of XR is equal to the excess of the square of AB above the square of LX; therefore the square of AB is equal to the squares of LX, XR. But the square of RL is equal (47. 1.) to the same squares, because LXR is a right angle. Therefore the square of AB is equal to the square 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 straight 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 (8. 1.) to the angle ABC. For the same reason, the angle MRN is equal to the angle DEF, and NRL to GHK. Therefore there is made a solid 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. A. THEOR. IF each of two solid angles be contained by three plane angles equal to one another, each to each; the planes in which the equal angles are, have the same inclination to one another.* Let there be two solid 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 same inclination to one another. A B In the straight line AC take any point K, and in the plane CAD from K draw the straight line KD at right angles to AC, and in the plane CAE the straight line KL at right angles to the same AC: therefore the angle DKL is the inclination (9. def. 11.) of the plane CAD to the plane CAE. In BF take BM equal to AK, and from the point A A E M N G H M draw, in the planes FBG, FBH, the straight lines MG, MN at right angles to BF; therefore the angle GMN is the inclination (6. def. 11.) 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 also the right angles AKD, BMG, and that the sides AK, BM, adjacent to the equal angles, are equal to one another; therefore KD is equal (26. 1.) to MG, and AD to BG: for the same reason, 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 base LD is equal (4. 1.) to the base NG. Lastly, in the triangles KLD, MNG, the sides DK, KL are equal to GM, MN, and the base LD to the base NG: therefore the angle DKL is equal (8. 1.) to the angle GMN: but the angle DKL is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclination of the * See Note. plane FBG to the plane FBH, which planes have therefore the same inclination (7. def. 11.) to one another: and in the same manner it may be demonstrated, that the other planes in which the equal angles are, have the same inclination to one another. Therefore, if two solid angles, &c. Q. E. D. PROP. B. THEOR. 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.* Let there be two solid angles at A and B, of which the solid 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 solid angle at A is equal to the solid angle at B. B Let the solid angle at A be applied to the solid angle at B; and, first, the plane angle CAD being applied to the plane angle FBG, so as the point A may coincide with the point B, and the straight 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 planeCAD is equal (A. 11.) to the inclination of the plane FBH to the plane FBG, the plane CAE coincides with the plane FBH, because the planes CAD, AA E G 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 solid angle A coincides with the solid angle B, and consequently they are equal (A. 8. 1.) to one another. Q. E. D. • See Note. |