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 B E, but A D. Therefore A D 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 demonftrated. PROPOSITION XLI. THE ORE M. If a Parallelogram and a Triangle have the fame LET the Parallelogram ABCD, and the Triangle For join A C. Now the Triangle A B C is equal to the Triangle* 37 of this. EBC; for they are both conftituted upon the fame Bafe BC, and between the fame Parallels BC, A E. But the Parallelogram ABCD is + double the Tri-t 34 of thite angle ABC, fince the Diameter AC bifects it. Wherefore likewise it shall be I double to the Tri-Ax. 6. angle EBC. If, therefore, a Parallelogram and Triangle have both the fame Base, 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 Rightlined Angle, L ET the given Triangle be A B C, and the Rightlined Angle D. It is required to conftitute a Parallelogram equal to the given Triangle ABC, in a Right-lined Angle equal to D. D 4 Bile 10 of this. Bifect* B C in E, join A E, and at the Point E, †23 of this. in the Right Line E C, conftitute + an Angle CEF 31 of this. equal to D. Alfo draw ‡ A G thro' A, parallel to EC, and thro' C the Right Line CG, parallel to F E. Now FECG is a Parallelogram: And because 38 of this. BE is equal to E C, the Triangle A BE fhall be * equal to the Triangle AEC; for they ftand upon equal Bafes B E, E C, and are between the fame Parallels BC, AG. Wherefore the Triangle A B C is double to the Triangle AEC. But the Parallelo41 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 FECG 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 A B C, in an Angle CEF equal to a given Angle D; which was to be done. 1 PROPOSITION XLIII. THEOREM. In every Parallelogram, the Complements of the Parallelograms, that stand about the Diameter, are equal between themselves. LETABCD be a Parallelogram, whofe Diameter is DB; and let FH, EG, be Parallelograms ftanding about the Diameter B D. Now A K, K C, are called the Complements of them: I fay, the Complement A K is equal to the Complement K C. For fince A B C D is a Parallelogram, and B D is 34 of this. the Diameter thereof, the Triangle A B D is equal to the Triangle BD C. Again, because H K FD is a Parallelogram, whofe Diameter is DK, the Triangle HDK fhall be equal to the Triangle D F K; and for the fame Reafon the Triangle K BG is equal to the Triangle KE B. But fince the Triangle BEK is equal to the Triangle BG K, and the Triangle HDK to DFK, the Triangle B EK, together with the Triangle HD K, is equal to the Triangle BG K, together with the Triangle D F K. But the whole Triangle ABD is likewife equal to the whole Triangle BDC. BDC. Wherefore the Complement remaining, AK, will be equal to the remaining Complement KC. Therefore, in every Parallelogram, the Complements of the Parallelograms that ftand about the Diameter, are equal between 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. 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. Triangle C, in the Angle E B G, equal to D. Place BE in a ftrait Line with A B, and produce F G to H, and through A let AH be drawn † parallel to either † 32 of this GB, or FE, and join H B. Now, because the Right Line H F falls on the Parallels AH, EF, 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 Ax. 12. HB, FE, produced, will meet each other; which let be in K, through which * draw KL parallel to E A,* 31 of this. or F H, and produce A H, G B, to the Point L and M. Therefore H L K F is a Parallelogram, whofe Diameter is HK; and AG, ME, are Parallelograms about HK; whereof L B, BF are the Complements. Therefore L B is † equal to B F. But BF is alfo† 43 of this. equal to the Triangle C. Wherefore likewife LB Thall be equal to the Triangle C; and because the An gle GBE 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 ABM, equal to the given Angle D; which was to be done. PRO PROPOSITION XLV. PROBLEM. To make a Parallelogram equal to a given Rightllined Figure, in a given Right-lined Angle. LET ABCD be the given Right-lined Figure, and E the Right-lined Angle given. It is required to make a Parallelogram equal to the Rightlined Figure A BCD, in an Angle equal to E. 24 of this. Let DB be joined, and make the Parallelogram FH equal to the Triangle A D B, in an Angle HKF, equal to the given Angle E, +44 of this * Then to the Right Line GH apply + the Parallelogram G M, equal to the Triangle DBC in an Angle G HM, equal to the Angle E. Now, because the Angle E is equal to HK F, or GHM, the Angle HKF fhall be equal to G H M, add K HG to both; and the Angles H KF, KHG, are, together, equal to the Angles K HG, GHM. 29 of this. But H KF, KHG, are ‡, together, equal to two Right Angles. Wherefore, likewife, the Angles. KHG, GHM, fhall be equal to two Right Angles: And fo, at the given Point H in the Right Line GH, two Right Lines K H, H M, not drawn on the fame. Side, make the adjacent Angles, both together, equal 14 of this. to two Right Angles; and confequently K H, HM *, make one ftrait Line. And because the Right Line HG falls upon the Parallels K M, F G, the alternate Angles M HG, HG F, are ‡ equal. And if HGL be added to both, the Angles MHG, HGL, together, are equal to the Angles H G F, HGL, together, *29 of this. But the Angles M HG, HG L, are together equal to two Right Angles. Wherefore, likewife, the An-. gles HGF, HGL, are together equal to two Right Angles; and fo, FG, GL, make one ftrait Line, And fince K F is equal and parallel to H G, as like+30 of this, wife HG to ML, KF fhall be + equal and parallel to M L, and the Right Lines K M, FL, join them. $33 of this. Wherefore K M, FL, are equal and parallel. There * fore KF L M is a Parallelogram. But fince the Triangle ABD is equal to the Parallelogram HF, and the the Triangle DBC to the Parallelogram GM; therefore the whole Right-lined Figure ABCD will be equal to the whole Parallelogram KF L M. Therefore, the Parallelogram K F L M is made equal to the given Right-lined Figure A B CD, in an Angle FKM, equal to the given Angle E; which was to be done. Coroll. It is manifeft from what has been faid, how to apply a Parallelogram to a given Right Line,_equal to a given Right-lined Figure in a given Right lined Angle. PROPOSITION XLVĮ, PROBLEM. To defcribe a Square upon a given Right Line, LETAB be the Right Line given, upon which it is required to defcribe a Square. * Draw A Cat Right Angles to A B from the Point * 11 of this, A given therein; make + A D equal to A B, and thro'† 3 of this. the Point D draw ‡ DE parallel to A B; alfo thro' 11 of this B draw B E parallel to AD. Then A DEB is a Parallelogram; and fo AB is equal to DE, and AD to B E. But B A is equal to AD. Therefore the four Sides BA, A D, BE, ED, are equal to each other. # 34 of this. And fo the Parallelogram A DEB is equilateral: I fay, it is likewife equiangular. For, because the Right Line A D falls upon the Parallels AB, DE, the Angles BAD, ADE, are + equal to two Right Angles. † 29 of this. But, BAD is a Right Angle. Wherefore ADE is alfo a Right Angle; but the oppofite Sides and oppo fite Angles of Parallelograms are equal. Therefore, ‡ 34 of this. each of the oppofite Angles A B E, BED, are Right Angles; and confequently A D BE is a Rectangle : But it has been proved to be equilateral. Therefore, * it is neceffarily a Square, and is described upon the Def. 30. Right Line AB; which was to be done. Coroll. Hence every Parallelogram that has one Right Angle, is a Rectangle. PRO |