*

Bifect* BC in E, join AE, and at the Point E, in* 10 of this. the Right Line EC, conftitute † an Angle CEF 23 of this. equal to D. Alfo 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 ABE fhall be * equal 38 of this to the Triangle AEC; for they ftand upon equal Bafes 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 alfo 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 ABC, and has the Angle CEF equal to the Angle D. Wherefore the Parallelogram FECG is conftituted equal to the given Triangle ABC, in an Angle CEF equal to a given Angle D; which was to be done.

PROPOSITION XLIII.

THEOREM.

In every Parallelogram the Complements of the Parallelograms that stand about the Diameter, are equal between themselves.

LET

ET ABCD be a Parallelogram, whose Diameter is DB; and let FH, EG, be Parallelograms ftanding about the Diameter B D. Now AK, KC, are called the Complements of them: I fay 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, becaufe HK FD is a Parallelogram, whofe Diameter is DK, the Triangle HDK fhall be equal to the Triangle DFK; and for the fame Reason the Triangle KBG is equal to the Triangle KE B. But fince 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 BG K, together with the Triangle DFK. But the whole Triangle ABD is likewife equal to the whole Triangle D 3 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 stand about the Diameter, are equal between themfelves; which was to be done.

PROPOSITION XLIV.

PROBLEM.

To apply a Parallelogram to a given Right Line, equal to a given Triangle, in a given Right-lined Angle.

ET the Right Line given be AB, the given Triangle 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 this. BEFG equal to the Triangle C; in the Angle EBG, equal to D. Place BE in a ftraight Line with AB, and produce FG to H, and thro' A let AH be +31 of this, drawn + parallel to either GB, or FE, and join HB. Now because the Right Line HF falls on the Pa29 of this rallels AH, EF, the Angles AHF, HFE, are 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 *Ax. 12, infinitely produced, will meet each other. Wherefore HB, FE, produced, will meet each other; which *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, whofe Diameter is HK; and AG, ME, are Parallelograms about HK; whereof LB, BF, are the Complements, +43 of this. Therefore LB is † equal to BF. But BF is alfo equal to the Triangle C. Wherefore likewife LB fhall be equal to the Triangle C; and because the 15 of this. Angle GBE is equal to the Angle ABM, and alfo equal to the Angle D, the Angle ABM fhall 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 ABM, equal to the given Angle D; which was to be done. PRO:

PROPOSITION XLV.

PROBLEM.

To make a Parallelogram equal to a given Rightlined Figure, in a given Right-lined Angle.

Land ABC Right the angle

ET ABCD be the given Right-lined Figure,

quired to make a Parallelogram equal to the Rightlined Figure ABCD in an Angle equal to E.

*

Let DB 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 + the Paralle- † 44 of this. logram GM, equal to the Triangle DBC, in an Angle G HM, equal to the Angle E.

or

*

41 of this.

Now because the Angle E is equal to HKF, GHM, the Angle HKF fhall be equal to GHM, add KHG to both; and the Angles HKF, KHG, are, together, equal to the Angles KHG, GHM. But HKF, KHG, are ‡, together, equal to two Right ‡ 29 of th 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 KH, HM, not drawn on the fame Side, make the adjacent Angles, both together, equal to two Right Angles; and confequently KH, HM* make one straight Line. And because the Right Line HG falls upon the Parallels KM, FG, the alternate Angles MHG, HGF, are 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 ftraight Line. And fince KF is equal and parallel to HG, as likewife HG to ML, KF fhall be equal and parallel † 30 of this. to ML, and the Right Lines K M, FL, join them. Wherefore KM, FL, are equal and parallel. There-134 of this. fore KFLM is a Parallelogram. But fince the Triangle ABD is equal to the Parallelogram HF, and

D 4

*

the