PROPOSITION XV. THEOREM XIII. F two straight lines (A B & A C) situated in the same plane (A X), and meeting one another (in A), be parallel, to two straight lines (D É & D F) meeting one another, and situated in another plane (D Z); those planes (A X & D Z) will be parallel. Hypothefis. Thesis. A B&AC ftuated in the plane A X The plane A X in which are the lines & meeting each other in A, are plle. to AB@ AC is plle. to the plane DZ DE & EF meeting each other in D, & in which aretbe lines DEED F. fituated in the plane D Z. Preparation. P.1. BIT. 2. Draw G H plle, to DE, & GL plle. to DF. P.31. B. 1. DEMONSTRATION. BECA US ECAUSE the lines GH & GL are plle. to D E & DF (Prep. 2). 1. They will be also plle. to A B & AC. P. 9. B.11. And G L being plle. to A C. 2. The V CAG+ AGL are = 2 L. P.29. B. 1. But YAGL is a L (Prep. 1). 3. Consequently, VCAG is also a L. 4. It may be demonstrated after the same manner that V BAG is a L 5. Therefore, G A is I to the plane A X. 4: B.in. But G A is also I to the plane D Z (Prep. 1). 6. Wherefore, the plane A X is plle. to the plane D Z. Which was to be demonstrated. P. P.14. B.u. PROPOSITION XVI. THEOREM XIV. F two parallel planes (Z X & Y P) be cut by another plane (A B D C), Thesis. 1. The planes 2 X & P Y are plle. The common Sections CD&AB II. Tbey are cut by the plane ABCD. are plle. If not, The lines A B & C D being produced will meet somewhere. Preparation. Pof.2. B. 1. another : (B A F being entirely in the plane P Y, & DCF entirely P. 1. B.11. 2. Which is impoffible (Hyp. 1). Wherefore, AB & C D do not meet one another. 4. Therefore, A B & C D are plle. D.35. B. 1. Which was to be demonstrated. 3 PROPOSITION XVII. THEOREM XV. If two Iraight lines (AC & BD) be cut by parallel planes (XZ, PY&QM): they shall be cut in the same ratio, (that is, A E:EF=BF:FH &c). Hypothesis. Thesis. 1. AC&B D are two straight lines. AE: EGBF;FH. II, Cut by the plle. planes X Z, PY&QM. Preparation Pof.1. B. 1. 3. Draw E I & IF. DEMONSTRATION. & -P.16. B.11. 2. Likewise, E I is plle. to G H. 3. Consequently, AI:1H= BF: FH. P. 2. B. 6. AI:IH = AE: EG. 5. Therefore, AE: EG=BF: F H. P.IT. B. 5. Which was to be demonstrated. 4. And, IF PROPOSITION XVIII. THEOREM XVI. F a straight line (A B) is perpendicular to a plane (Z X): every plane (as Q E) which passes thro' this line (A B) shall be perpendicular to this plane (Z X). Hypothesis. Thesis. A B is I to the plane Z X. Every plane (as Q E) which passes thro the L A B is I to the plane Ż X. Preparation. P. 3. B. I. P.31. B. I. DEMONSTRATION. Beca D. ECAUSE the straight line A B is I to the plane ZX, & DC is plle. to A B (Hyp. & Prep. 3). 1. The line D C is I to the plane Z X: P. 8. B.u. 2. Consequently, CD is also I to the common section É F. 3. B.I. 3. Therefore, the plane E Q in which the lines A B & C D are, is I to the plane Z X. D. 4. B.11. And as the fame demonstration may be applied to any other plane which passes thro’ the I AB, we may conclude, 4. That every plane which passes thro' this line is I to the plane Z X. Which was to be demonstrated. 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 section (AB) shall be perpendicular to the same plane (Z X). Hypothesis. Thesis. 1. The planes CD & E F are I to the plane Z X. The common seation A Bis I ļl. They cut one another in AB, to the plane Z X. DEMONSTRATION. Becau ECAUSE CB, the common section of the plane CD with the P. B.1. 1. There may be erected at the point B in C B a I (P. 11. B. 11.) which will be in the plane CD (Hyp. 1.) P.18. B.II. And because the line F B the common section of the planes FE & X Z is also in the plane X Z. P. 3. B.In. 2. There may be erected at the faine point B & at the same side with the foregoing another I which will fall in the plane F E. P.18. B.I. But from the point B only one I can be raised. 3. Consequently, those I must coincide, that is, those two lines must form but one which is common to the two planes. Which was to be demonstrated. P.13. B.11. 4 |