PROPOSITION XXXIX. THEOR E M. on the fame Side, are in the fame Parallels. upon the fame Base B C, on the fame Side. I say, they are between the same Parallels. For, let A D be drawn. I say A D is parallel to B C. 32 of tbis, . For, if it be not parallel, draw * the Right Line A E thro' the Point A, parallel to BC, and draw EC. + 37 ofibis. Then the Triangle ABC+ is equal to the Triangle EBC; for it is upon the fame Base B C, and between the fame Parallels BC, A E. But the Triangle A B C, 1 From Hyp. is I equal to the Triangle D BC. 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 B C: and by the same way of Reasoning we prove, that no other Line but A D is parallel to BC. Therefore A D is parallel to BC. Wherefore, equal Triangles confiituted upon the same Base, on the same Side, are in the same Parallels ; which was to be demonstrated. PROPOSITION XL. THEOREM. the same Side, are between the same Parallels. LETABC, CDE, be equal Triangles , constituted upon equal Bases B C, CE. I say, they are between the fame Parallels. For let A D be drawn. I say, A D is parallel to BE. * 31 of tbis. For, if it be not, let A F be drawn * through A, parallel to B E, and draw F E. + 38 of rbis. Then the Triangle A B C is equal + to the Triangle FCE; for they are conftituted upon cqual Bases, and between the same Parallels B E, A F. But the Triangle A B C is equal to the Triangle D CE. There fore fore the Triangle D C E shall be equal to the Triangle FCE, the greater to the less, which is impossible. Wherefore A F is not parallel to B E. And in this manner we demonstrate, that no Right Line can be parallel to B E, but A D. Therefore A D is parallel to B E. And so, equal Triangles constituted upon equal Bases, on the fame Side, are between the fame Parallels ; which was to be demonstrated. PROPOSITION XLI. THE ORE M. Base, and are beiween the same Parallels, the EBC, have the same Base, and be between the Now the Triangle A B C is * equal to the Triangle 37 ofibis. EBC; for they are both constituted upon the fame Base B C, and between the same Parallels B C, A E. But the Parallelogram ABCD is + double the Tri- † 34 of ibis. angle A B C, since the Diameter AC bisects it. Wherefore likewise it shall be I double to the Tri- Ax. 6. angle E BC. If, therefore, a Parallelogram and Triangle have both the same Base,, and are between the same Parallels, the Parallelogram will be double the Triangle; which was to be demonstrated. PROPOSITION XLII.. THE OR EM. Triangle, in an Angle equal to a given Right lin'd Angle. LET the given Triangle be A B C, and the Right lined Angle D. It is required to constitute a Parallelogram equal to the given Triangle A B C, in a Rght-lind Angle equal to D. D4 Bireet * 10 of tbis. Birea * B C in E, join A E, and at the Point E, + 23 ofrbis, in the Right Line E C, constitute + an Angle CEF 1 31 of ibis.equal to D. Also draw I AG thro' A, parallel to E C, and through C the Right Line CG, parallel to Now FECG is a Parallelogram: And because *38 of ibis. BE is equal to EC, the Triangle A B E shall be * equal to the Triangle A EC; for they stand upon equal Bases B E, E C, and are between the same Parallels B C, AG. Wherefore the Triangle A B C is double to the Triangle A EC. But the Parallelo+ 41 of 1bis, gram FECG is also + double to the Triangle A EC; for it has the same Base, and is between the same THEOREM. Parallelograms, tbat stand about tbe Diameter, ore equal between themselves. LETAB CD bea Parallelogram, whose Diameter ; standing about the Diameter BD. Now AK, KC, are called the Complements of the n : I say, the Complement A K is equal to the Complement K C. For fince A B C D is a Parallelogram, and B D is the Diameter thereof, the Triangle A B D is * equal 34 of this, 'to the Triangle BD C. Again, because HKFD is a Parallelogram, whose Diameter is DK, the Triangle HD K shall * be equal to the Triangle DFK; and for the same Reason the Triangle KBG is equal to the Triangle K E B. But since the Triangle BEK is equal to the Triangle BGK, and the Triangle HD K to DF K, the Triangle B E K, together with the Triangle HDK, is equal to the Triangle BGK, together with the Triangle D FK. But the whole Triangle A B D is likewise equal to the whole Triangle BDC. BDC. Wherefore the Complement remaining, , are equal betweem themselves ; which was to be demonftrated. PROPOSITION XLIV. PROBLEM. equal to a given Triangle, in a given Right lined Angle. LE ET the Right Line given be A B, the given Tri angle 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 tbis. Now, because the Right Line HF ralis on the Pa- Therefore HL KF is a Parallelogram, whose Diameter is HK; and A G, ME, are Parallelograms about HK; whereof LB, B F are the Complements. Therefore L B is + equal to BF. But BF is also + 43 of this. equal to the Triangle C. Wherefore likewise LB shall be equal to the Triangle C; and because the Angle GBE is I equal to the Angle A BM, and also 1150f bit. equal to the Angle D, the Angle A BM shall be equal to the Angle D. Theretore, to the given Right Line A B is applied a Parallelogram, equal to the given Triangle C, and the Angle A BM, equal to the given Angle D; which was to be done, PRO. 1 19 of teza. Blce? * B C in E, join 33 ore in the Right Line EC, co EC, and through the Ri Now FECG is a Para ccutie to the Triangle A E for it has the same Base, and THEOREM. Paralelograms, ibat stand abou are equal beiween theinselves. LE ETABCD be a Parallelogram, is DB; and let FH, EG, be ftanding about the Diameter BD. N are called the Complements of the n: 1 piement A K is equal to the Complem Forince A B C D is a Parallelogran the Diameter thereof, the Triangle A B 34 og 20. to the Triangle B D C. Again, becaut a Paralle' gram, whose Diameter is DK geHD K thai] * be equal to the Trian and for the same Reason the Triangle KE to the Triangle K E B. But since the Tria is equal to the Triangle B GK, and the HDK 10 DFK, the Triangle B E K, 10 tre Triangle HDK, is equal to the Triang together with the Triangle D FK. But Triangle A B D is likewise equal to the whole 1 |