PROPOSITION XXXVI. THEOREM. between the same Parallels, are equal between ET the Parallelograms ABCD, EFGH, be between the fame Parallels AH, BG. I say, the Parallelogram ABCD is equal to the Parallelogram EFGH. For join BE, CH. Then because B C is * equal * Hypo to FG, and FG to EH; BC will be likewise equal to EH; and they are parallel, and BE, CH, joins them. But two Right Lines joining Right Lines which are equal and parallel the same Way, are + equal, and pa- † 33 of this. tallel: Wherefore EBCH is a Parallelogram, and is f equal to the Parallelogram ABCD; for it has the # 35 of this, fame Base B C, and is constituted between the same Parallels BC, AD. For the same Reason, the Parallelogram EFGH is equal to the fame Parallelogram EBCH. Therefore the Parallelogram ABCD Thall be equal to the Parallelogram EFGH. And so Parallelograms constituted upon equal Bases, and between the same Parallels, are equal between themselves; which was to be demonstrated. PROPOSITION XXXVII. THEORE M. upon the same Base BC, and between the same Parallels AD, BC. I say, the Triangle AB C, is equal to the Triangle D B’ć. For produce AĎ both ways to the Points E and F; and through B draw * B É parallel to CA; and * 31 of thit. through C, CF, parallel to BD. D Where Wherefore both EBCA, DBCF, are Parallelo• 35 of this. grams; and the Parallelogram EBCA is * equal to the Parallelogram DBCF; for they stand upon the fame Base B C, and between the same Parallels BC, + 34 of this EF. But the Triangle A B C is tone half of the Pa rallelogram EBCA, because the Diameter AB bifects it; and the Triangle DBC is one half of the Parallelogram DBCF, for the Diameter DC bisects it. But Things that are the Halves of equal Things, | Ax. 7. are fequal between themselves. Therefore the Triangle ABC, is equal to the Triangle DBC. Wherefore, Triangles constituted upon the fame Bafe, and between the same Parallels, are equal between themfelves; which was to be demonstrated. PROPOSITION XXXVIII. THEOREM. Triangles constituted upon equal Bases, and between the same Parallels, are equal between themselves. ET the Triangles ABC, DCE, be constituted Came. Parallels BE, AD. I say the Triangle ABC is equal to the Triangle DCE. For produce AD both Ways to the Points G, H: 31 of this. thro' B draw * BG parallel to CA; and thro' E; EH, parallel to DC. Wherefore both GBCA, DCEH, are Paralle lograms, and the Parallelogram GBCA is † equal † 36 of this, to the Parallelogram DCEH: For they stand upon equal Bases, BC, CE, and between the fame Paral1 34 of bis leis BE, GH. But the Triangle ABC is t one half of the Parallelogram GBCA, for the Diameter AB bifects it; and the Triangle DCE I is one half of the Parallelogram DCEH, for the Diameter DE bisects it. But Things that are the Halves of equal Things, are * equal between themselves. Therefore the Triangle AB C, is equal to the Triangle DCE. Wherefore Triangles constituted upon equal Bafes, and between the same Parallels, are equal between thenselves; which was to be demonstrated. PRO THEOREM. on the Same Side, are in the same Parallels. tuted upon the fame Base B C, on the fame Side. For if it be not parallel, draw* the Right Line AE * 31 of this. thro’ the Point A, parallel to BC, and draw E C. Then the Triangle ABC, + is equal to the Triangle † 37 of ibis. EBC; for it is upon the same Bale BC, and between the fame Parallels BC, AE. But the Triangle ABC is equal to the Triangle D BC. Therefore the I From Hyp. Triangle DBC is also equal to the Triangle E B C, a less to a greater, which is impossible. Wherefore AE is not parallel to BC: And by the same way of Reasoning we prove, that no other Line but AD is parallel to BC. Therefore AD iš parallel to B C. Wherefore equal Triangles constituted upon the same Base, on the fame Side, are in the same Parallels ; which was to be demonstrated. PROPOSITION XL. THEOREM. the Same Side, are between the fame Parallels. ET ABC, CDE, be equal Triangles, consti tuted upon equal Bases B C, CE. I say they áre between the same Parallels. For let AD be drawn. I say AD is parallel to B E. For if it be not, let A F be drawn * thro' A, pa*** 31 of this: rallel to BE, and draw F E. Then the Triangle ABC is f'equal to the Triangle + 38 of this. FCE; for they are constituted upon equal Bases, and between the fame Parallels BE, AF. But the Triángle ABC is equal to the Triangle D CE: There fore D2 fore the Triangle DCE shall be equal to the Triangle F CE, the greater to the less, which is impofiible. Wherefore AF is not parallel to BE. And in this Manner we demonstrate, that no Right Line can be parallel to BE, but AD. Therefore AD is parallel to BE. And fo equal Triangles constituted upon equal Bases, on the same Side, are between the same Parallels'; which was to be demonstrated. PROPOSITION XLI. PROBLEM Base, and are between the same Parallels, the ET the Parallelogram ABCD, and the Trian gle EBC, have the same Base, and be between the fame Parallels, BC, AE. I say the Parallelogram A B C D is double the Triangle EBC. For join AC * 37 of this. Now the Triangle ABC is * equal to the Triangle EBC; for they are both constituted upon the same Bafe BC, and between the same Parallels BC, AE. + 34 of this. But the Parallelogram ABCD is + double the Tri angle ABC, since the Diameter AC bifects it. Wherefore likewise it shall be double to the Triangle EBC. If, therefore, a Parallelogram and Triangle have both ihe same Base, and are between the fame Parallels, the Parallelogram will be double to the Triangle; which was to be demonstrated. PROPOSITION XLII. THEOREM angle, in an Angle equal to a given Right-lined lined Angle given D. It is required to constitute a Parallelogram equal to the given Triangle AB C, in a Right-lined Angle equal to D. Bisect |