A E IF D B PROPOSITION XV. THEOREM VIII. F two ftraight lines (AB, DE,) cut one another in (C), the vertical or oppofite angles (ECA, DCB, & ACD, BCE,) are equal. Hypothefis. AB, DE, are ftraight lines which cut one another in the point C. DEMONSTRATION. Thefis. I. V ECAV DCB, BECAUSE the ftraight line AC falls upon the straight line DE (Hyp.). 1. The fum of the two adjacent VECA + ACD is to two L. Again, fince the ftraight line DC falls upon the ftraight line AB (Hyp.). 2. The fum of the adjacent V ACD + DCB is alfo to two L. 3. Confequently, the VECA + ACD are to V ACD + DCB. Taking away therefore from thofe equal fums (Arg. 3.) the common ACD. P. 13. B. 1, P. 13. B. 1, Ax. 1. 4. The remaining V ECA, DCB, which are vertically oppofite, are equal. Ax. 3. Which was to be demonftrated. I, In the fame manner it will be proved: 5. That V ACD is to V BCE, which is vertically oppofite to it. Which was to be demonftrated. II. COROLLARY I. FROM this it is manifeft, that if two straight linescut one another, the angles they make at the point where they cut, are together equal to four right angles. COROLLARY II. AND confequently, that all the angles made by any number of lines meeting in one point, are together equal to four right angles, IF PROPOSITION XVI. E THEOREM IX. F one fide as (AB) of a triangle (ACB) be produced, the exterior angle (CBF) is greater than either of the interior oppofite angles (ACB, CAB,). Hypothefis. T Preparation. Thefis. The exterior CBF > the interior oppofite V ACB or CAB. 1. Divide CB into two equal parts at the point D. (Fig. 1.) 2. From the point A to the point D, draw the line AD, & pro- 3. Make DE DÅ. 4. From the point B to the point E, draw the ftraight line BE. DEMONSTRATION. HE ftraight lines AE, BC, (Fig. 1.) interfect each other at the point D. (Prep. 2.). 1. Confequently, the oppofite vertical VCDA, BDE, are to one another. P. 15. B. 1. Wherefore fince in the ▲ ACD, DEB, the fide CD is to the fide DB (Prep. 1.), AD = DE (Prep. 3.), & V contained CDA is = to contained BDE (Arg. 1.). 2. It follows, that the remaining V of the one are equal to the remaining Vof the other, each to each of thofe to which the equal fides are oppofite. P. 4. B. 1. But the ACD, DEE, are oppofite to the equal fides AD,DE, (Prep. 3.). 3. Therefore V ACD is to V DBE. But CBF being the whole, & VDBE its part. 4. It follows, that V CEF > \ DBE. 5. Wherefore the exterior V CBF is alfo > the interior V ACB. In the fame manner, dividing the fide AB into two equal parts in the point D (Fig. 2.) it will be proved. 6. That the exterior VABƒ is the interior V CAB. But this ABf is vertically oppofite to VCBF. 7. Wherefore V ABƒ = ▼ CBF. Ax. S. C. N. P. 15. B. 1. Which was to be demonstrated. 8. Confequently, the exterior V CBF is > the interior V CAB. B A C D PROPOSITION XVII. THEOREM X. ANY NY two angles as (ABC, ACB,) of a triangle (BAC), are less than two right angles. Hypothefis. ABC is a A. Thefis. The ABC+ACB are < two L. Preparation. Produce the fide BC (upon which the two V ABC, ACB, are BECAUSE DEMONSTRATION. ECAUSE V ACD is an exterior V of the ▲ BAC. 1. It is its interior oppofite one ABC. Since therefore V ACD is >V ABC; if the V ACB be added to each. 3. Confequently, thofe V ACD + ACB are to two L. Pof. 2. P. 16. B. i. Ax. 4. P. 13. B. 1. 4. It follows, that the V ABC + ACB are < two L. C. N. Which was to be demonftrated. A C m D B PROPOSITION XVIII. THEOREM. XI. IN every triangle (ACB); the greater fide is oppofite to the greater angle. P. 3. B. 1. 2. From the point C to the point D, draw the ftraight line CD. Pof. 1. DEMONSTRATION. BECAUSE the fide AD is to the fide AC (Prep. 1.). 1. The ▲ ACD is an ifofceles A. = D. 25. B. 1. 2. Confequently, the Vm & n at the base CD are to one another. P. 5. B. 1. But m being an exterior V of ▲ DCB. 3. It follows, that it is the interior oppofite V D3C. But miston (Arg. 2.) P. 16. B. 1. 4. Therefore n is also > ▼ DBC. C. N. And if to Vn be added V p. 5. Much more will n+p or V ACB, oppofite to the greater fide AB, be > DBC, or ABC, oppofite to the leffer fide AC. C. N. Which was to be demonftrated. A 33 PROPOSITION XIX. THEOREM XII. IN every triangle (BAC), the greater angle, has the greater fide oppofite to it. BECAUSE C. N. to the fide CB (Sup. 1.). D. 25. B. 1. P. 5. B. 1. ECAUSE the fide AB is 1. The A BAC is an ifofceles A. 2. Confequently, the VC & A at the base, are to one another. But thofe VC & A are not 3. Therefore neither are the fides AB, CB to one another (Hyp.). to one another. 1. It follows, that VC oppofite to the leffer fide AB, is <VA oppofite to the greater fide CB. But C is not <VA (Hyp.). 2. Confequently, the fide AB cannot be The fide AB being therefore neither nor the fide CB (Cafe 2.). the fide CB. to the fide CB (Cafe 1.); 3. It follows, that this fide AB is > the fide CB. Which was to be demonftrated. P. 18. B. I. C. N. E |