and fo the Right Line GE is greater than KL, and GF than LM. Make GX equal to KL, and GO equal to LM, and complete the Parallelogram XGOP. Therefore XO is equal and fimilar to

KM, but KM is fimilar to EF, therefore XO is **21 of thise fimilar to EF, and fo XO is + about the fame Dia-t 26 of this. meter with FE: Let GPB be their Diameter, and

the Figure be described.

Then fince EF is equal to C and KM together, and XO is equal to KM, the Gnomon Y remaining, is equal to the remaining Figure C; and because OR is equal to XS, let SR, which is common, be added; then the whole OB is equal to the whole XB, but XB is equal to TE, fince the Side AE is equal to the Side E B. Wherefore TE is equal to OB. Add XS, which is common, and then the whole TS is equal to the whole Gnomon ; but the Gnomon ro has been proved equal to C; and fo TS fhall be equal to C; and fo the Parallelogram TS is applied to the Right Line AB, equal to the given Right-lined Figure C, and deficient by a Parallelogram SR, fimilar to the Parallelogram D, because SR is fimilar to FE; which was to be done.

PROPOSITION XXIX.

THE ORE M.

To a Right Line given, to apply a Parallelogram equal to a Right-lined Figure given, exceeding by a Parallelogram, which shall be fimilar to another given Parallelogram.

ET AB be a given Right Line, and let C be the given Right-lined Figure to which that to be applied to AB must be equal. Likewise let D be the Parallelogram to which the exceeding Parallelogram is to be fimilar; it is required to apply a Parallelogram to the Right Line AB, equal to the given Rightlined Figure C, exceeding by a Parallelogram fimilar to D.

*

Bifect AB in E, and let the Parallelogram EL be defcribed upon the Right Line EB, fimilar and alike* 18. 1. fituate to D; and that † the Parallelogram G H equal † 25 of this,

to EL and C together, but fimilar to D, and alike fituate. Therefore GH is fimilar to EL; let KH be a Side homologous to FL, and KG to FE. Then because the Parallelogram GH is greater than the Parallelogram EL, the Right Line KH will be greater than FL, and KG greater than FE. Let FL, FE, be produced, and let FLM be equal to KH, FEN equal to KG, and complete the Parallelogram MN. Therefore MN is equal and fimilar to GH; but GH 1 of this. is fimilar to EL, and fo MN fhall be ‡ fimilar to * 26 of this. EL; and accordingly EL is about the fame Diameter with MN. Let FX be their Diameter, and defcribe the Figure.

* 46. 1.

Then fince GH is equal to EL and C together, as likewife to MN; therefore MN fhall be equal to EL and C. Let EL, which is common, be taken away, then the Gnomon Yo remaining, is equal to C; and fince AE is equal to EB, the Parallelogram AN will be alfo equal to the Parallelogram EP, that is, to LO; and if EX, which is common, be added, then the whole Parallelogram AX is equal to the Gnomon 14*, but the Gnomon Y is equal to C. Therefore AX fhall be alfo equal to C. Wherefore the Parallelogram AX is applied to the given Right Line AB, equal to the given Right-lined Figure C, and exceeding by the Parallelogram PO, fimilar to the Parallelogram D; which was to be done.

PROPOSITION XXX.

PROBLEM.

To cut a given terminate Right Line according to extreme and mean Ratio.

L to cut the fame according to extreme and quean
ET AB be a given terminate Line; it is required

Ratio.

Defcribe* BC the Square of AB, and apply the Parallelogram CD to AC, equal to the Square BC, † 29 of this, exceeding + by the Figure AD fimilar to BC; but BC is a Square, therefore AD fhall also be a Square.

Now because BC is equal to CD, take away CE which is common; then BF remaining fhall be equal

to

to AD remaining; but BF is equiangular to AD; therefore the Sides that are about the equal Angles

are reciprocally proportional; and fo as FE is to ‡ 14 of this. ED, fo is AE to EB, but FE is equal to AC, that * 34. 1. is, to AB, and ED to AE. Wherefore as BA is to

L

AE, fo is AE to EB, but AB is greater than AE; therefore AE ist greater than EB; and fo the Right † 14. 5. Line AB is cut according to extreme and mean Ratio in the Point E; and AE is the greater Segment thereof; which was to be done.

Otherwife thus: Let AB be the Right Line given; it is required to cut the fame into extreme and mean Ratio.

Divide AB fo in C, that the Rectangle contained ‡ 11. 2. under AB, BC, be equal to the Square of AC.

*

Then because the Rectangle under AB, BC, is equal to the Square of AC, it fhall be as BA is* 17 of this. to AC, fo is AC to CB; and fo the Right Line AB is cut into mean and extreme Ratio; which was to be done.

PROPOSITION XXXI.

THEOREM.

Any Figure defcribed upon the Side of a Right-angled Triangle fubtending the Right Angle, is equal to the Figures defcribed upon the Sides containing the Right Angle, being fimilar and alike fituate to the former Figure.

ET ABC be a rectangular Triangle, having the Right Angle BAC. I fay the Figure defcribed on BC, is equal to the two Figures together defcribed on BA, AC, which are fimilar and alike fituate to the Figure defcribed on B C.

For draw the Perpendicular AD.

Then because the Right Line AD is drawn in the Right-angled Triangle ACB, from the Right Angle A, perpendicular to the Bafe BC; the Triangles ABD, ADC, which are about the Perpendicular AD, will be ** 8 of this. fimilar to the whole Triangle ABC, and alfo to each other. Then because the Triangle ABC is fimilar to the Triangle ABD, it fhall be as CB is to BA, so

+Cor. 20. of this.

24. 5.

# 29. I.

is BA to BD; and fince when three Right Lines are proportional, the first shall be † to the third, as a Figure described on the firft, to a fimilar and alike fituate Figure defcribed on the fecond. Wherefore as CB is to BD, fo is a Figure described on CB to a fimilar and alike fituate Figure defcribed on BA. For the fame Reafon as BC is to CD, fo is a Figure defcribed on BC to one described on CA. Wherefore alfo, as BC is to BD and DC together, fo is the Figure described on BC, to thofe two together that are, defcribed fimilar and alike fituate on BA, AC; but BC is equal to BD and DC together: Therefore the Figure described on BC is equal to those together described on BA, AC; fimilar and alike fituate to that on BC. Wherefore, any Figure defcribed upon the Side of a Right-angled Triangle fubtending the Right Angle, is equal to the Figures defcribed upon the Sides containing the Right Angle, being fimilar and alike fituate to the former Figure; which was to be de

monftrated.

PROPOSITION XXXII.

THEOREM.

If two Triangles having two Sides proportional to two Sides, be fo compounded, or set together at one Angle, that their homologous Sides be parallel, then the other Sides of thefe Triangles will be in one strait Line.

ET there be two Triangles ABC, DCE, having two Sides BA, AC, of the one, proportional to two Sides CD, DE, of the other, viz. Let BA be to AC, as CD is to DE; alfo let AB be parallel to DC, and AC to DE. I fay BC, CE, are both in one ftrait Line.

*

For because AB is parallel to DC, and the Right Line AC falls on them, the alternate Angles BAC, ACD, will be equal to each other. And by the fame Reason, the Angle CDE is equal to the Angle ACD; wherefore the Angle BAC is equal to the Angle CDE. Then because ABC, DCE, are two Triangles, having one Angle A equal to one Angle Angle

D, and the Sides about the equal Angles proportional, viz. BA to AC, as CD to DE, the Triangle ABC will be equiangular to the Triangle DCE; where- * 6 of his. fore the Angle ABC is equal to the Angle DCE; but the Angle ACD has been proved to be equal to the Angle BAC; therefore the whole Angle A CE is equal to the two Angles ABC, BAC; and if ACB, which is common, be added, then the Angles ACE, ACB, are equal to the Angles BAC, ACB, CBA; but the Angles BAC, ACB, CBA, are equal to two Right Angles. Therefore the Angles ACE, ACB, will alfo be equal to two Right Angles, and fo at the Point C in the Right Line AC, two Right Lines BC, CE, tending contrary ways, makes the adjacent Angles ACE, ACB, equal to two Right Angles; therefore BC fhall be † in the fame Right † 14. 1. Line with CE. Wherefore, if two Triangles having two Sides proportional to two Sides, be fo compounded, or fet together at one Angle, that their homologous Sides be parallel, then the other Sides of thefe Triangles will be in one ftrait Line; which was to be demonftrated.

PROPOSITION XXXIII.

THEOREM.

In equal Circles the Angles have the fame Proportion with their Circumferences on which they ftand, whether the Angles be at the Centers, or at the Circumferences; and fo likewife are the Sectors, as being at the Centers.

ET ABC, DEF, be equal Circles, and let the

e at

and the Angles BAC, EDF, at their Circumferences. I fay, as the Circumference BC is to the Circumference EF, fo is the Angle BGC to the Angle EHF; and fo is the Angle BAC to the Angle EDF; and fo is the Sector BGC to the Sector EHF.

For affume any Number of continuous Circumferences CK, KL, each equal to BC; and alfo any Number FM, MN, each equal to EF, and join GK, GL, HM, HN.