PROPOSITION XXXVI. THEORE M. Parallelograms conftituted upon equal Bafes, and between the fame Parallels, are equal between themfelves. Le conftituted upon the equal and ET the Parallelograms ABCD, EFGH, be conftituted upon the equal Bafes BC, FG, and between the fame Parallels AH, BG. I fay, the Parallelogram ABCD is equal to the Parallelogram EFGH. * 35 of this. For join BE, CH. Then becaufe BC is equal * Hyp. to F G, and FG to EH; BC will be likewife 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 fame Way, are † equal, and pa- † 33 of this. fallel: Wherefore EBCH is a Parallelogram, and is tequal to the Parallelogram ABCD; for it has the fame Bafe BC, and is conftituted between the fame Parallels BC, AD. For the fame Reafon, the Parallelogram EFGH is equal to the fame Parallelo gram EBCH. Therefore the Parallelogram ABCD fhall be equal to the Parallelogram EFGH. And fo Parallelograms conftituted upon equal Bafes, and between the fame Parallels, are equal between themfelves; which was to be demonftrated. PROPOSITION XXXVII. THEORE M. Triangles conftituted upon the fame Bafe, and between the fame Parallels are equal between themselves. ET the Triangles ABC, DBC, be conftituted upon the fame Bafe BC, and between the fame Parallels AD, BC. I fay, the Triangle ABC, is equal to the Triangle DBC. For produce AD both ways to the Points E and F; and through B draw * BÉ parallel to CA; and * 31 of thit. through C, CF, parallel to BD. Wherefore both EBCA, DBCF, are Parallelo * 35 of this. grams; and the Parallelogram EBCA is equal to the Parallelogram DBCF; for they ftand upon the fame Base BC, and between the fame Parallels B C, † 34 of this. EF. But the Triangle ABC is † one half of the Parallelogram E BCA, because the Diameter A B bifects it; and the Triangle DBC is one half of the Parallelogram DBCF, for the Diameter DC bifects it. But Things that are the Halves of equal Things, are equal between themselves. Therefore the Triangle ABC, is equal to the Triangle DBC. Wherefore, Triangles conftituted upon the fame Bafe, and between the fame Parallels, are equal between themfelves; which was to be demonstrated. Ax. 7. 31 of this. PROPOSITION XXXVIII. THEOREM. Triangles conftituted upon equal Bafes, and between Lupon For produce AD both Ways to the Points G, H: thro' B draw BG parallel to CA; and thro' E, EH, parallel to DC. Wherefore both GBCA, DCEH, are Parallelograms, and the Parallelogram GBCA is † equal † 36 of this to the Parallelogram DCEH: For they ftand upon equal Bafes, BC, CE, and between the fame Paral134 of this lels BE, GH. But the Triangle ABC is one half of the Parallelogram GBCA, for the Diameter AB bifects it; and the Triangle DCE ‡ is one half of the Parallelogram DCEH, for the Diameter DE bifects it. But Things that are the Halves of equal Things, are equal between themfelves. Therefore the Triangle ABC, is equal to the Triangle DCE. Wherefore Triangles conftituted upon equal Bafes, and between the fame Parallels, are equal between themfelves; which was to be demonftrated. * PRO Equal Triangles conftituted upon the fame Bafe, on the fame Side, are in the fame Parallels. ET ABC, DB C, be equal Triangles, conftiupon the fame Base B C, on the fame Side. I fay they are between the fame Parallels. For let AD be drawn. I fay AD is parallel to BC. 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 this.. t EBC; for it is upon the fame Bafe BC, and between the fame Parallels BC, AE. But the Triangle ABC is equal to the Triangle DBC. Therefore the From Hyp. Triangle DBC is alfo equal to the Triangle EBC, a lefs to a greater, which is impoffible. Wherefore AE is not parallel to BC: And by the fame Way of Reasoning we prove, that no other Line but AD is parallel to BC. Therefore AD is parallel to BC. Wherefore equal Triangles conftituted upon the fame Bafe, on the fame Side, are in the fame Parallels; which was to be demonftrated. PROPOSITION XL. THEOREM. Equal Triangles conftituted upon equal Bases, on the fame Side, are between the fame Parallels. LET ABC, CDE, be equal Triangles, confti tuted upon equal Bafes BC, CE. I fay they are between the fame Parallels. For let AD be drawn. I fay AD is parallel to BE. For if it be not, let AF be drawn * thro' A, pa* 31 of this: rallel to BE, and draw F E. Then the Triangle ABC is equal to the Triangle † 38 of this: FCE; for they are conftituted upon equal Bafes, and between the fame Parallels BE, AF. But the Triangle ABC is equal to the Triangle DCE. There D 2 fore *37 of this. fore the Triangle DCE fhall be equal to the Triangle FCE, the greater to the lefs, which is impoffible. Wherefore AF is not parallel to BE. And in this Manner we demonftrate, that no Right Line can be parallel to BE, but AD. Therefore AD is parallel to BE. And fo equal Triangles conftituted upon equal Bafes, on the fame Side, are between the fame Parallels; which was to be demonstrated. PROPOSITION XLI. PROBLEM. If a Parallelogram and a Triangle have the fame LET the Parallelogram ABCD, and the Trian gle EBC, have the fame Base, and be between the fame Parallels, BC, AE. I fay the Parallelogram ABCD is double the Triangle E B C. For join A C. * Now the Triangle ABC is equal to the Triangle EBC; for they are both conftituted upon the fame Bafe BC, and between the fame Parallels B C, AE. +34 of this. But the Parallelogram ABCD is † double the Triangle ABC, fince the Diameter AC bifects it. Wherefore likewise it shall be double to the Triangle EBC. If, therefore, a Parallelogram and Triangle have both the fame Bafe, and are between the fame Parallels, the Parallelogram will be double to the Triangle; which was to be demonftrated. PROPOSITION XLII. THEOREM. To conftitute a Parallelogram equal to a given Triangle, in an Angle equal to a given Right-lined Angle. ET the given Triangle be ABC, and the Rightlined Angle given D. It is required to conftitute a Parallelogram equal to the given Triangle ABC, in a Right-lined Angle equal to D. Bifect |