4 PROPOSITION XXXIX. THEOREM. Equal Triangles conftituted upon the fame Base, on the fame Side, are in the fame Parallels. LET ABC, DBC, be equal Triangles, conftituted upon the fame Bafe BC, on the fame Side. I fay, they are between the fame Parallels. For, let AD be drawn. I fay AD is parallel to B C. 32 of this. For, if it be not parallel, draw the Right Line AE thro' the Point A, parallel to B C, and draw EC. t37 of this. 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, A E. But the Triangle A B C, From Hyp, is equal to the Triangle D B C. Therefore the Triangle DBC is alfo equal to the Triangle E B C, a greater to a lefs, which is impoffible. Wherefore A E is not parallel to BC: and by the fame Way of Reafoning we prove, that no other Line but AD is parallel to BC. Therefore A D is parallel to BC. Wherefore, equal Triangles conflituted upon the fame Bafe, on the fame Side, are in the fame Parallels; which was to be demonftrated. 31 of this. + 38 of this. PROPOSITION XL. THEOREM. Equal Triangles conftituted upon equal Bafes, on the fame Side, are between the fame Parallels. LETABC, CDE, be equal Triangles, conftituted upon equal Bafes 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 A F be drawn * through A, parallel to B E, and draw F E. Then the Triangle ABC is equal † to the Triangle FCE; for they are conftituted upon equal Bafes, and between the fame Parallels B E, A F. But the Triangle A B C is equal to the Triangle D C E. There fore fore the Triangle D C E fhall be equal to the Triangle FCE, the greater to the lefs, which is impoffible. Wherefore AF is not parallel to B E. And in this manner we demonftrate, that no Right Line can be parallel to B E, but A D. Therefore A D is parallel to B E. And fo, equal Triangles conftituted upon equal Bafes, on the fame Side, are between the fame Parallels; which was to be demonftrated. PROPOSITION XLI. THE ORE M. If a Parallelogram and a Triangle have the fame ET the Parallelogram A BCD, and the Triangle For join A C. Now the Triangle ABC is equal to the Triangle 37 of tbis. EBC; for they are both conftituted upon the fame Base B C, and between the fame Parallels B C, A E. But the Parallelogram A B C D is + double the Tri-+ 34 of this. angle ABC, fince the Diameter AC bifects 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.. THEOREM. To conftitute a Parallelogram equal to a given Triangle, in an Angle equal to a given Rightlin'd Angle. ET the given Triangle be A B C, and the Rightlined Augle D. It is required to conftitute a Parallelogram equal to the given Triangle A B C, in a R ght-lin❜d Angle equal to D. D 4 Bifect * 10 of this. Bifect B C in E, join A E, and at the Point E, +23 of bis, in the Right Line E C, conftitute + an Angle CEF 38 of this equal to D. Alfo draw A G thro' A, parallel to EC, and through C the Right Line CG, parallel to FE. 38 of this. Now FECG is a Parallelogram: And becaufe BE is equal to EC, the Triangle A B E fhall be * equal to the Triangle A EC; for they stand upon equal Bafes B E, EC, and are between the fame Parallels B C, AG. Wherefore the Triangle A B C is double to the Triangle A E C. But the Parallelo+41 of this, gram FECG is alfo + double to the Triangle A EC; for it has the fame Bafe, and is between the fame Parallels. Therefore the Parallelogram F E C G is equal to the Triangle A B C, and has the Angle CEF equal to the Angle D. Wherefore, the Parallelogram FECG is conflituted equal to the given Triangle ABC, in an Angle C E F 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 ftand about the Diameter, are equal between themselves. LE ETABCD be a Parallelogram, whose Diameter is DB; and let F H, E G, be Parallelograms ftanding about the Diameter B D. Now A K, KC, are called the Complements of the n: I fay, the Complement A K 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 B D C. Again, because HKFD is a Parallelogram, whofe Diameter is D K, the Triangle HDK fhall be equal to the Triangle DF K; and for the fame Reafon the Triangle K BG is equal to the Triangle K E B. But fince the Triangle BEK is equal to the Triangle B G K, and the Triangle HDK to DF K, the Triangle B E K, together with the Triangle HDK, is equal to the Triangle BGK, together with the Triangle DFK. But the whole Triangle A B D is likewife equal to the whole Triangle BDC. BDC. Wherefore the Complement remaining, A K, will be equal to the remaining Complement KC. Therefore, in every Parallelogram, the Complements of the Parallelograms that stand about the Diameter are equal betweem themselves; which was to be demonftrated. PROPOSITION XLIV. PROBLEM. To apply a Parallelogram to a given Right Line, equal to a given Triangle, in a given Rightlined Angle. ET 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 BE F G equal to the ** 42 of this. Triangle C, in the Angle E BG, equal to D. Place BE in a ftrait Line with A B, and produce F G to H, and through A let A H be drawn † parallel to either + 32 of this. G B, or F E, and join H B. Now, because the Right Line HF faiis on the Pa rallels A H, E F, the Angles AHF, HFE, are ‡‡ 29 of this. equal to two Right Angles. And fo BHF, HFE, are lefs than two Right Angles; but Right Lines making lefs than two Right Angles, with a third Line, being infinitely produced, will meet each other. Wherefore * Az. 12. HB, FE, produced, will meet each other; which let be in K, through which draw || K L parallel to EA, || 13 of bis. or F H, and produce AH, G B, to the Points L and M. Therefore HLK F is a Parallelogram, whose Diameter is H K; and A G, M E, are Parallelograms about HK; whereof LB, B F are the Complements. Therefore L B is equal to B F. But BF is alfo t43 of this. equal to the Triangle C. Wherefore likewife LB fhall be equal to the Triangle C; and because the An gle G BE is equal to the Angle A B M, and alfo † 15 of this, equal to the Angle D, the Angle ABM fhall 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 A B M, equal to the given Angle D; which was to be done. PRO Now FECG is a Para 33 of ibi. BE is equal to EC, the T 3. † 41 of 1 gram FECG is alfo + double PROPOSITIO LE THEOREM. In every Parallelogram, the Con 42 For fince A B C D is a Parallelograr the Diameter thereof, the Triangle A B • 34 of the to the Triangle B D C. Again, becau! a Parallel gram, whole Diameter is D K gle HDK fhall be equal to the Trian and for the fame Reafon the Triangle Kh to the Triangle KE B. But fince the Tria is equal to the Triangle B G K, and the HDK to DFK, the Triangle B EK, to the Triangle HDK, is equal to the Triang together with the Triangle DFK. But Triangle A B D is likewise equal to the whole |