Bisect * BC in E, join AE, and at the Point E, in * 10 of this. the Right Line EC, constitute + an Angle CEF† 23 of this. equal to D. Also draw [ A G thro’ A, parallel to EC, # 31 of this. and thro' C the Right Line CG parallel to F E. Now FECG is a Parallelogram: And because BE is equal to EC, the Triangle ÅBE shall be * equal * 38 of thisa to the Triangle A EC; for they stand upon equal Bases BE, EC, and are between the fame Parallels BC, AG. Wherefore the Triangle ABC is double to the Triangle AEC. But the Parallelogram FECG is also double to the Triangle AEC; for it has the fame Base, and is between the fame Parallels. Therefore the Parallelogram FECG, is equal to the Triangle A BC, 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. PROPOSITION XLIII. THEOREM. rallelograms that stand about the Diameter, are equal between themselves. LEI ET ABCD be a Parallelogram, whose Diame ter iş DB; and let FH, EĞ, be Parallelograms standing about the Diameter B D. Now AK, KC, are called the Complements of them: I say the Complement AK is equal to the Complement KC. For fince ABCD is a Parallelogram, and BD is the Diameter thereof, the Triangle ABD* is equal * 34 of this 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 K B G is equal to the Triangle KEB. But since the Triangle BÈ K 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 B GK, together with the Triangle DFK. But the whole Triangle A BD. is likewise equal to the whole Triangle BDC D 3 BDC. Wherefore the Complement remaining, AK, will be equal to the remaining Complement KC, Therefore in every Parallelogram the Complements of the Parallelograms, that stand about the Diameter, are equal between themselves ; which was to be done, PROPOSITION XLIV. PROBLEM. equal to a given Triangle, in a given Right-lined Angle. ET the Right Line given be AB, the given Tri angle C, and the given Right-lined Angle D. It is required to the given Right Line AB, to apply a Parallelogram equal to the given Triangle C. In an Angle equal to D, make the Parallelogram * 42 of flis. BEFG equal to * the Triangle C; in the Angle EBG, equal to D. Place BE in a straight Line with AB, and produce FG to H, and thro' A let AH be † 3r of this, drawn t parallel to either GB, or FE, and join HB. Now because the Right Line HF falls on the Pa1 29 of tbi, rallels AH, EF, the Angles AHF, HFE, are I equal to two Right Angles. And fo BHF, HFĖ, are less than two Right Angles; but Right Lines making less than two Right Angles, with a third Line being infinitely produced, will meet * each other. Where fore HB, FE, produced, will meet each other; whick 31 of this. let be in K, thro' which * draw KL parallel to EA, or FH, and produce AH, GB, to the Points L and M. Therefore HLKF is a Parallelogram, whose Diameter is HK; and AG, ME, are Parallelograms about HK; whereof LB, BF, are the Complements, † 43 of this. Therefore LB ist equal to BF. But BF is also equal to the Triangle C. Wherefore likewise LB Thall be equal to the Triangle C; and because the 1 15 of this. Angle GBE is equal to the Angle A BM, and allo equal to the Angle D, the Angle ABM shall be equal to the Angle D.. Therefore to the given Right Line AB is applied a Parallelogram, equal to the given Triangle C, in the Angle A BM, equal to the given Angle D; which was to be done. PRO: * Ax. 1?. PROPOSITION XLV. PROBLEM. lined Figure, in a given Right-lined Angle. and E the Right-lined Angle given. It is required to make a Parallelogram equal to the Rightlined Figure ABCD in an Angle equal to E. Let D B be joined, and make * the Parallelogram * 42 of this, FH equal to the Triangle ADB, in an Angle HKF, equal to the given Angle E. Then to the Right Line GH apply t the Paralle- † 44 of rbis. logram GM, equal to the Triangle DBC, in an Angle GHM, equal to the Angle E. Now because the Angle E is equal to HKF, or GHM, the Angle HK F shall be equal to GHM, add KÁG to both; and the Angles HKF, KHG, are, together, equal to the Angles KHG, GHM. But HKF, KHG, are 1, together, equal to two Right I 29 of the Angles. Wherefore, likewise, the Angles KHG, GHM, shall be equal to two Right Angles : And so at the given Point H in the Right Line GH, two Right Lines KH, HM, not drawn on the fame Side, make the adjacent Angles, both together, equal to two Right Angles ; and consequently KH, HM** 47 of tbis. make one straight Line. And because the Right Line HG falls upon the Parallels KM, FG, the alternate Angles MÁG, HGF, are f equal. And if HGL be added to both, the Angles MHG, HGL, together, are equal to the Angles HGF, HGL, together. But the Angles MHG, HGL, are * together equal * 29 of this to two Right Angles. Wherefore likewise the Angles HGF, HGL, are together equal to two Right Angles; and fo FG, GL, make one straight Line. And since KF is equal and parallel to HG, as likewise HG to ML, KF fhall be equal and parallel † 30 of obis. to ML, and the Right Lines KM, FL, join them. Wherefore KM, FL, are f equal and parallel. There- 1 34 of this, fore KFLM is a Parallelogram. But since the Triangle ABD is equal to the Parallelogram HF, and D 4 the |