PROPOSITION XVIII. THEOREM XVI. IF a ftraight line (A B) is perpendicular to a plane (Z X): every plane (as QE) which paffes thro' this line (A B) fhall be perpendicular to this plane (Z X). AB is Hypothefis. Thefis. Every plane (as QE) which passes thro' the LAB is to the plane Z X. Preparation. 1. Let a plane QE pafs thro' A B, which will cut the plane Z X in EF. 2. Take in this ftraight line E F, a point D at will. BECA DEMONSTRATION. ECAUSE the ftraight line A B is to the plane ZX, & DC is plle. to A B (Hyp. & Prep. 3). 1. The line DC is to the plane Z X. 2. Confequently, CD is alfo 1 to the common section E F. P. 3. B. 1, P.31. B. 1. P. 8. B.11. D. 3. B.11. 3. Therefore, the plane E Q in which the lines A B & C D are, is to the plane Z X. D. 4. B.11. And as the fame demonftration may be applied to any other plane which paffes thro' the LA B, we may conclude, 4. That every plane which paffes thro' this line is to the plane Z X. Which was to be demonstrated. IF PROPOSITION XIX. THEOREM XVII. F two planes (CD & E F) cutting one another be each of them perpendicular to a third plane (Z X); their common fection (A B) fhall be perpendicular to the fame plane (Z X). Hypothefis. I. The planes CD & E F are 1 to the plane Z X. DEMONSTRATION. Thefis. The common fection A B is to the plane Z X. BECAUSE CB, the common fection of the plane CD with the plane X Z is alfo in the plane X Z.. 1. There may be erected at the point B in CB a ▲ (P. 11. B. 11.) which will be in the plane C D (Hyp. 1.) And because the line F B the common fection of the planes F E 2. There may be erected at the fame point B & at the fame fide with P. 3. B.11. P.18. B.11. P. 3. B.11. P.18. B.11. 3. Confequently, thofe muft coincide, that is, those two lines must form but one which is common to the two planes. P.13. B.11. But thofe planes have only the line A B in common (Hyp. 2.) Which was to be demonstrated. C A b B E D PROPOSITION XX. THEOREM XVIII. IF three plane angles (CA B, BAD & DA C) form a folid angle A: any two of those angles (as BAD & CA B) are greater than the third (CAD). BECAU ECAUSE the VCA B, d&c + b are equal. 1. It follows that VCAB+d will be c+b. CASE II. When of the three angles C A B, d&c + b two as CAB & d are equal, & the third c + b is less than either of them. BECAUSE ECAUSE VCAB is >vc+b. i. V CAB+Vd will be much >c+b. Ax.4. B. 1. Ax.4. B. X. Which was to be demonftrated. THE CASE III. When the three angles are unequal, & b + c is > CAB or d. Preparation. 1. At the point A in AC make VbVCAB in the plane 2. Make A E=AB. "P23. B.1. P. 3. B.1. D. Pof.1. B.A. 3. From the point C draw thro' E the ftraight line C E D. HE ABCA & CA E have the fides A B & AE equal (Prep.2). The fide C A common & b=VCAB (Prep. 1). 1. Confequently, the fide B C is to the fide CE. But in the ACBD the fides CB+ B D are > C D. Therefore, if from CB+BD be taken away the part C B, & from CD a part to C E. 2. The remainder B D will be > ED. In the ABA D & E AD, the fides A B & A E are = (Prep. 2). & A D common. But the bafe B D is > the bafe ED (Arg. 2). 3. Therefore, Vd is > c. If therefore, CA B be added one fide, & its equal ▷ b on the P.25. B.1. other. 4. VCAB+d will be >Vb+cor CAD. Ax.4. B.1. Which was to be demonftrated. A B Ꭰ PROPOSITION XXI. THEOREM XIX. ALL L the plane angles (B A C, CAD & DA B) which form a folid angle (A); are lefs than four right angles. The Hypothefis. BAC, CAD & DAB form a folid, VA The plane Thefis. Preparation. 1. In the fides BA, AC, & AD take the three points B, C, D. 3. three folid V ; CBA, ABD plane VBCA, Let a plane BCD pafs thro' thofe lines, which will form DEMONSTRATION, ECAUSE the folid V D, is formed by the plane VC DA, BECAU VACBA CD are > VBCD. 1. The 2. Likewife, VABD+ABC are > VDBC. 3. And 4. Hence, the fix plane VCDA+ADB+ABD+ABC+ACB +ACD are the three plane VBDC+DBC+BCD. Pof.1. B. 1. D.11. B.11. P.20. B.11. But thofe three plane VBDC+DBC+BCD are 2 L. P.32. B.1. 5. Therefore, the fix plane VCDA+ADB+ABD+ &c. are > 2 L (Arg. 4.) But the nine of the ABCA, CAD & D A B viz. the fix already mentioned (Arg. 5.) & the three remaining BAC, CAD & DAB are together to 6 L. If therefore the fix VCDA+ADB+ABD+ABC+ACB 6. The remaining plane VBA C+CAD+DA B will be < 4 L. P.32. B:1. |