NE part (A B) of a ftraight line cannot be in a plane (Z X); and. 1. At the point. B in A B erect in the plane ZX the 2. At the point B in B G erect in the plane ZX the LBC.P.11. B.1. BECAUSE VABG is a L, likewife V GBC, & they meet in the fame point B. 1. The lines A B & B C are in the fame straight line A C. 2. But the line B D is a part of the ftraight line above the plane (Sup.). 3. Confequently, VDBGV GBAGBC, that is, the part 4. Which is impoffible. 5. Therefore, B D cannot be a part of the straight line AB (Arg. 1). And as the fame demonftration may be applied to any other part of BC. 6. It follows, that all the parts of a ftraight line are in the fame plane. Which was to be demonftrated. P.14. B.1. Ax.10.B.1. Ax. 8.B.1, Two PROPOSITION II. THEOREM II. WO ftraight lines which cut one another in (E); are in one plane (Z X) and three straight lines which conftitute a triangle (EAD) are in the fame plane (Z X). Draw G F. BECAUSE Preparation. ECAUSE the part A FGD of the AEA D is not in one plane (Z X) with EFG (Sup.). 1. It follows, that the parts GD, C G of the line CD are in different planes, & the parts AF, FB of the straight line AB, are in different planes, as alfo AFGD & FEG. 2. Which is impoffible. 3. Since then the parts of the two lines & of the ▲ can not be in different planes. 4. They must confequently be in the fame plane. P. 1. B.IL. Which was to be demonstrated. 1. & 1 1. F two planes (RS & PL) cut one another, their common fection is a ftraight line (A B). The section will be two ftraight lines. As A X B for the plane RS; & A Y B for the plane P L, BECAUS ECAUSE the straight lines A X B & AY B have the same extremities A & B. 1. Those two ftraight lines A X B & A Y B include a space A X B Y. 2. Which is impoffible. 3. Confequently, the fection of the planes P L & R S can not be two ftraight lines AXB & A Y B. 4. Therefore their common section, is a straight line A B. Ax.12. B.1. Which was to be demonftrated. IF P PROPOSITION IV. THEOREM IV. F two ftraight lines (AB & C D) interfect each other, and at the point (E) of their interfection a perpendicular (EF) be erected upon those lines (AB & CD): it will be allo perpendicular to the plane (PL) which paffes through thofe lines (A B & CD). THE to thofe lines at the point E. Preparation. Thefis. EF is to the plane P L. 1. Take EC at will, & make E B, ED & AE each equal 2. Join the points A & C, alfo B & D. 3. Thro' the point E in the fame plane P L, draw the straight 4. Draw A F, GF, CF, DF, HF & B F. DEMONSTRATION. HE AAEF, CEF, BEF, & DEF have the fide E F common. The fides AE, CE, BE, & DE equal (Prep. 1) & the adjacent L VAEF, CEF, BEF, & DEF equal (Hyp. 3). 1. Confequently the bafes A F, CF, BF, & DF are equal. In the AEC & DE B, the fides A E, CE, ED & EB are = (Prep. 1.) & the VAEC & DE B also equal. 2. Therefore, ACBD. 3. And VEAC VEBD. The AGAE & E BH have AEG HEB. P. 4. B.1. P.15. B.1. }P. P. 4. B. 1. 4. Confequently, the fides G A & GE are to the fides HB & EH. P.26. B.1. In the AAFC & F D B, the three fides AF, FC & AC of the firft are to the three fides F B, F D & D B of the second (Arg. 1, & 2). 5. Therefore, the three of the AAFC are to the three of 6. Therefore, GF FH. Infine, in the AGFF & FEH, P. 8. B.1. to the the fides G F, GE, & FE are of P. 8. B.1. to the fides FH, EH, & EF 7. Confequently, the three of the the AFE H, each to each, that is But thofe V FEG & FEH are formed by the ftraight line EF falling upon G H (because G E & E H are in the fame ftraight line) (Prep. 3). 8. Therefore, thofe V FEG & FEH are L, & FEL upon GH. (P.13. B.1. But HG is in the fame plane, with the lines A B & CD (Prep. 3). And E F is upon thofe lines (Hyp. 3). 9. Confequently, EF is upon the fame plane P L. D.10. B.I. D. 3. B.II. Which was to be demonstrated. |