Draw the right Line gh any how cutting a the Sides of the Angle, and make b AG=ag; upon AG raife a Triangle equal-fided to the other gab, fo that AH be equal to ah, and GH gb; then fhall you have the Angle Ada. Q. E. F. PROP. XXIV. If two Triangles (ABC, abc) have two Sides (AB, AC) of the one, each equal to two Sides (ab, ac) of the other, and have the Angle A greater than a 23. I. b 3.1. d byp. e conftr. 8 5. I. h 9 ax. the Angle bac, contain'd under the equal right Lines; the Base (BC) of the one shall also be greater than the Bafe (bc) of the other. a Make the Angle bag equal to A, and the Side agac, and join bg, cg. Cafe 1. When bg falls above bc, then because AB ab, and AC ag, and the Angle A a; therefore isf BC=bg. = = = But fince ac ag, therefore is the g Angle acg hagc. And fo the Angle acg hbgc, and confequently the Angle begh bgc. Therefore bg (BC) Q. E. D. bc. Cafe Cafe 2. When the Bafe bg coincides with bc, then it is plain that bg (BC) be. a 9 ax. Cafe 3. But when bg falls below bc, because ag+gb bac+cb; if from both fides youb 21. 1. take away the equal Lines ag, ac, there will remain bg (BC) bc. Q. E. D. с PROP. XXV. c 5 ax. (ab, ac) of the o ther; and if the Bafe BC be greater than the Bafe bc; then shall the Angle A contained under the equal Lines be greater than the Angle a. For if you fay that the Angle Aa, then fhall the Bafe BC.be = dbc; which is contrary d 4. I. to the Hypothefis. If you fay that the Angle Aa, then fhall BC be bc; which is like-e 24. 1. wife contrary to the Hypothefis. Q. E. D. PRO P. XXVI. If two Triangles (BAC, bac) have two Angles B, C, equal to two Angles (b, acb) and one Side of the one equal to one Side of the other, either that Side AA which lies between the equal Angles, or that which is fubtended by one of the equal Angles: each remaining Side a 3. 1. b fuppof. c hyp. d 4. I. e hyp. f 9. ax. g hyp. k 16. I. Side of the one shall be equal to the remaining Side of the other; and the remaining Angle of the one shall be equal to the remaining Angle of the other. Hyp. 1. Let BC= be; I fay, BA = ba, and AC ac, and the Angle Abac. For if you fay that ba BA, make a bd BA, and draw cd. Because AB bd, and BC= bc, and the Angle Bb; therefore fhall the Angle bed bed Cc; which is abfurd: therefore AB ab. In like manner, AC alfo the Angle Ad bac. Q. E. D. ac; and fo Hyp. 2. Let AB = ab; I tay, BC= bc, and AC ac, and the Angle A bac. For if you fay that be BC, make be BC, and join ae. Becaufe AB & ab, and BCh be, and the Angle B8b: therefore fhall the Angle bea bed Cic; which is abfurd. Therefore BC be; and confequently, as at firft, AC = ac, and the Angle A bac. k QE. D. If a right Line (EF) fal ling upon two -G right Lines(AB, CD) makest alternate Angles (AEF, DFE) equal the one to the other; then are the right Lines (AB, CD) parallel. 1 If AB, CD be faid not to be parallel, pro116. 1. duce them till they meet, fuppofe in G: Then the outward Angle AEF will be greater than the inward Angle DFE, to which it was equal by the Hypothefis: Which is abfurd. PROP. If a right Line (EF) falling upon -B two right Lines AB, CD, makes the outward Angle D AGE, equal to (CHG)the inward and oppofite Angle on the fame fide; or makes the inward Angles AGH, CHG, on the fame fide equal to two right Angles; then are the right Lines (AB, CD) parallel. = Hyp. 1. Becaufe by the Hypothefis the Ang. AGE CHG, therefore the alternate Angles BGH, CHG are equal: and confequently ABa 15. 1. and CD are parallel. b 111 b 27. I. Hyp. 2. Becaufe by the Hypothefis the Ang. AGH + CHG two right Ang. AGHC 13. 1. +BGH; therefore is 4 CHG BGH: and d 3 ax. confequently AB, CD are e parallel. Q. E. D. e 27. I. PROP. E с = XXIX. If a right Line (EF) falls upon two ParalBlels (AB, CD) the al A G ternate Angles DHG, the inward and oppofite Angle on the fame fide; as alfo the inward Angles AGH, CHG on the fame fide will be equal to two right Angles. It is evident that AGH+CHG = two right Angles; for otherwife AB, CD would not be f parallel, which is contrary, to the Hypothefis.f 13 ax. But alfo the Ang. DHG+CHGs right Angles: Therefore DHG BGE. QE. D. twog 13. 1. AGHh 3 ax. Co-i 15. 1. a 29.I. b 3 ax. B A CORO L. Hence every Parallelogram (AC) that has one right Angle A, is a Rectangle. For A+B= two Dright Angles: therefore fince A is a right Angle, B must be balfo a right Angle. By the fame Argument C and D are right Angles. Let the Line GI cut the three given Lines any how ; then because AB, EF are parallel, the Angle c 29.1. AGI will be EHI: alfo because CD and EF are parallel, the Angle EHI will be <= DIG: Therefore the Angle AGI DIG. Whence AB and CD are parallel. Q.E.D. d I ax. e 27. I. e PRO P. XXXI. = Line BC given. A draw a right Line AD to any given Point of the given right Line; with which at the Point A |