† 37 of this. PROPOSITION XXXIX. THEOREM. • Equal Triangles constituted upon the fame Base, LET ABC, DBC, be equal Triangles, confti- 4 *32 of this. For, if it be not parallel, draw * the Right Line A E thro' the Point A, parallel to B C, and draw EC. Then the Triangle ABC + is equal to the Triangle EBC; for it is upon the fame Bafe B C, and between the fame Parallels BC, AE. But the Triangle A B С, From Hyp, is equal to the Triangle DBC. Therefore the Triangle DBC is also equal to the Triangle E BC, a greater to a lefs, which is impossible. Wherefore A E is not parallel to BC: and by the fame Way of Reafoning we prove, that no other Line but A Dis parallel to B C. Therefore AD is parallel to B C. Wherefore, equal Triangles conftituted upon the fame Base, on the fame Side, are in the fame Parallels; which was to be demonstrated. PROPOSITION XL. THEOREM. Equal Triangles constituted upon equal Bases, on the fame Side, are between the same Parallels. LETABC, CDE, be equal Triangles, conftituted upon equal Bases BC, CE. I fay, they are between the fame Parallels. For let AD be drawn. I fay, A D is parallel to BE. For, if it be not, let AF be drawn * through A, parallel to B E, and draw F E. * 31 of this. + 38 of this. Then the Triangle A BC is equal + to the Triangle FCE; for they are conftituted upon equal Bases, and between the fame Parallels BE, AF. But the Triangle A B C is equal to the Triangle DCE. There fore fore the Triangle DCE shall be equal to the Triangle FCE, the greater to the less, which is impoffible. Wherefore AF is not parallel to BE. And in this manner we demonftrate, that no Right Line can be parallel to B E, but AD. Therefore AD is parallel to BE. And so, equal Triangles constituted upon equal Bafes, on the fame Side, are between the fame Parallels; which was to be demonftrated. PROPOSITION XLI. THEOREM. If a Parallelogram and a Triangle have the same LET the Parallelogram ABCD, and the Triangle For join A C. 37 of ibis. Now the Triangle A B C is * equal to the Triangle EBC; for they are both conftituted upon the fame Bafe B C, and between the same Parallels B С, АЕ. But the Parallelogram ABCD is † double the Tri-† 34 of tbis. angle ABC, fince the Diameter AC bisects it. Wherefore likewise it shall be ‡ double to the Tri- Ax. 6. angle EBC. If, therefore, a Parallelogram and Triangle have both the fame Bafe, and are between the fame Parallels, the Parallelogram will be double the Triangle; which was to be demonftrated. PROPOSITION XLII.. To constitute a Parallelogram equal to a given LET the given Triangle be ABC, and the Rightlined Angle D. It is required to conftitute a Parallelogram equal to the given Triangle A B C, in a Rght-lin'd Angle equal to D. D 4 Bifect * 10 of this. Bisect * B C in E, join A E, and at the Point E, † 23 ofrbis, in the Right Line E C, constitute + an Angle CEF I 38 of this. equal to D. Also draw I AG thro' A, parallel to * 38 of this. E C, and through C the Right Line CG, parallel to FE. Now FECG is a Parallelogram: And because BE is equal to EC, the Triangle A BE shall be * equal to the Triangle AEC; for they stand upon equal Bases B E, EC, and are between the same Parallels BC, AG. Wherefore the Triangle ABC is double to the Triangle AEC. But the Parallelo+ 41 of ibis, gram FECG is alfo + double to the Triangle A EC; for it has the same Base, and is between the fame Parallels. Therefore the Parallelogram FECG is equal to the Triangle ABC, and has the Angle CEF equal to the Angle D. Wherefore, the Parallelogram FECG is constituted equal to the given Triangle ABC, in an Angle CEF equal to a given Angle D; which was to be done. * 34 of this. PROPOSITION XLIII. THEOREM. In every Parallelogram, the Complements of the Parallelograms, that stand about the Diameter, are equal between themselves. LETABCD be a Parallelogram, whose Diameter is DB; and let F H, EG, be Parallelograms standing about the Diameter BD. Now AK, к с, are called the Complements of then: I fay, the Complement AK is equal to the Complement K C. For fince A B C D is a Parallelogram, and BD is the Diameter thereof, the Triangle A B D is * equal to the Triangle BDC. Again, because HKFD is a Parallelogram, whose Diameter is DK, the Triangle HDK shall * be equal to the Triangle DFK; and for the fame Reason the Triangle KBG is equal to the Triangle KEB. But since the Triangle BEK is equal to the Triangle BGK, and the Triangle HDK to DFK, the Triangle BEK, together with the Triangle HDK, is equal to the Triangle BGK, together with the Triangle DFK. But the whole Triangle A B D is likewise equal to the whole Triangle BDC. BDC. Wherefore the Complement remaining, PROPOSITION XLIV. To apply a Parallelogram to a given Right Line, LET the Right Line given be A B, the given Triangle C, and the given Right-lined Angle D. It is required to the given Right Line A B, to apply a Parallelogram equal to the given Triangle C, in an Angle equal to D. Make the Parallelogram BEFG equal to the * * 42 of this. Now, because the Right Line HF falis on the Pa- Therefore HLKF is a Parallelogram, whose Diameter is HK; and AG, ME, are Parallelograms about HK; whereof LB, BF are the Complements. Therefore LB is † equal to BF. But BF is also + 43 of this. equal to the Triangle C. Wherefore likewife LB shall be equal to the Triangle C; and because the Angle GBE is ‡ equal to the Angle A BM, and also 1 15 of this. equal to the Angle D, the Angle ABM shall be equal to the Angle D. Therefore, to the given Right Line A B is applied a Parallelogram, equal to the given Triangle C, and the Angle ABM, equal to the given Angle D; which was to be done. PRO 1 4 *10 of test. Bicct * B C in E, join † 23 of rhu, in the Right Line EC, се 31 of this equal to D. Also draw 42 EC, and through C the Ri FE. 38 of this. Now FECG is a Para BE is equal to EC, the T equal to the Triangle AE( equal Bafes B E, EC, and al rallels BC, AG. Wherefor double to the Triangle A E +4 of gram FECG is alfo + double for it has the fame Base, and Parallels. Therefore the Para equal to the Triangle ABC CEF equal to the Angle D. ra.elegram FECG is conftitut Triangle ABC, in an Angle Cl Angle D; which was to be done. 34 of aber PROPOSITIO THEOREM In every Parallelogram, the Con is DB; and let FH, EG, be For fince ABCD is a Parallelogran |