the Proportion to one another that is compounded of their Sides; which was to be demonftrated.

PROPOSITION . XXIV.

THEOREM.

In every Parallelogram, the Parallelograms that are about the Diameter, are fimilar to the Whole, and alfo to one another.

LET ABCD be a Parallelogram, whofe Diameter is A C; and E G, HK, be Parallelograms about the Diameter AC. I fay, the Parallelograms E G, H K, are fimilar to the Whole ABCD, and alfo to each other.

*

29.

For, because EF is drawn parallel to BC, the Side of the Triangle ABC, it fhall be *, as B E is to EA, * 2 of thisà fo is CF to FA. Again, becaufe F G is drawn parallel to C D, the Side of the Triangle A CD, it fhall be as CF is to FA, fo is * DG to G A. But CF is to FA (as has been proved), as BE is to EA. Therefore, as BE is to E A, fo is + DG to GA; + 11. 5. and by compounding, as BA is to A E, fo is $ DAS 18. 5. to A G; and by Alternation, as BA is to A D, fo is AE to AG. Therefore, the Sides of the Parallelograms A B C D, E G, which are about the common Angle B AD are proportional. And because G F is parallel to D C, the Angle AGF is equal to the Angfe A DC, and the Angle G F A equal to the Angle DCA; and the Angle D A C is common to the Triangles A DC, AGF. Wherefore the Triangle ADC will be equiangular to the Triangle A GF. For the fame Reafon, the Triangle AC B is equiangular to the Triangle AF E. Therefore, the whole Parallelogram A BCD is equiangular to the Parallelogram EG; and fo A D is to DC, as AG ist tot 4 of this GF; DC is to CA, as G F is to FA; and AC is to C B, as AF is to FE; and, moreover, CB is to BA, as FE is to E A. Wherefore, fince it has been proved, that DC is to CA, as G F is to FA; and AC is to CB, as AF is to FE; it fhall be, by Equality of Proportion, as DC is to C B, fo is GF to FE. Therefore the Sides that are about the equal

Angles of the Parallelograms ABCD, EG, are proportional; and, accordingly, the Parallelogram ABCD is fimilar to the Parrallelogram E G. For the fame Reason, the Parallelogram A B C D is fimilar to the Parallelogram K H. Therefore, both the Parallelograms EG, HK, are fimilar to the Parallelogram ABC D. But Right-lined Figures that 21 of this are fimilar to the † fame Right-lined Figure, are fimilar to one another. Therefore the Parallelogram EG is fimilar to the Parallelogram HK. And fo, in every Parallelogram, the Parallelograms that are about the Diameter, are fimilar to the Whole, and alfo to one another; which was to be demonftrated.

44. 3.

PROPOSITION XXV.

PROBLEM.

To defcribe a Right-lined Figure fimilar to a Rightlined Figure which shall be given, and equal to another Right-lined Figure given.

ET ABC and D, be two given Right-lined Figures; it is required to defcribe another Figure, fimilar to A B C and equal to D.

On the Side B C of the given Figure A B C *, make the Parallelogram B E equal to the Right-lined Figure A B C; and on the Side C E make the Parallelogram C M equal to the Right-lined Figure D, in the Angle FCE, equal to the Angle C B L. † 14. 1. Then B C, CF, as alfo L E, E M, will be + in two 113 of this ftrait Lines. Find ‡ G H a mean Proportional be

this.

tween B C and CF; and on GH let there be de8 of this fcribed the Right-lined Figure KG H, fimilar, and alike fituate, to the Right-lined Figure A B C.

And then, because B C is to GH, as G H is to CF; and fince, when three Right Lines are proportional, the firft is to the third, as the Figure defcribed Cor. 20 of on the firft is + to a fimilar and alike fituate Figure defcribed on the fecond: it fhall be, as B C is to CF, fo is the Right-lined Figure A B C to the Right-lined Figure KGH. But as BC is to CF, fo is t the Parallelogram BE to the Parallelogram EF. Therefore, as the Right-lined Figure ABC is to the Rightlined Figure KG-H, fo is the Parallelogram B E to

of this.

the

the Parallelogram EF. Wherefore (by Alternation), as the Right-lined Figure A B C is to the Parallelogram B E, fo is the Right-lined Figure KGH to the Parallelogram EF. But the Right-lined Figure ABC is equal to the Parallelogram BE. Therefore the Right-lined Figure K G H is alfo equal to the Parallelogram E F. But the Parallelogram EF is equal to the Right-lined Figure D. Therefore the Rightlined Figure KGH is equal to D. But KGH is fimilar to ABC. Confequently, there is defcribed the Right-lined Figure KGH fimilar to the given Figure ABC, and equal to the given Figure D; which was to be done.

PROPOSITION XXVI.

THEOREM.

If from a Parallelogram be taken away another fimilar to the Whole, and in like manner fituate, having alfo an Angle common with it; then is that Parallelogram about the fame Diameter with the Whole.

L ET the Parallelogram A F be taken away from

the Parallelogram A B CD, fimilar to ABCD, and in like manner fituate, having the Angle DAB common. I fay, the Parallelogram ABCD is about the fame Diameter with the Parallelogram A F..

For, if it be not, let A H C be the Diameter of the Parallelogram BD, and let G F be produced to H; alfo, let H K be drawn parallel to AD, or B C.

24 of this.

Def. 1. of

this.

Hyp.

Then, becaufe the Parailelogram ABCD is about the fame Diameter as the Parallelogram KG, the Parallelogram ABCD fhall be fimilar to the Parallelogram KG; and fo, as DA is to AB, fo is +GA to AK. But because of the Similarity of the Parallelograms ABCD, EG; as DA is to AB, fo is GA to A E. And therefore, as G A is ‡ to ‡ 11. 5. AE, fo is GA to A K. And fince G A has the fame Proportion to AK as to A E, A E is + equal to † 9. 5. AK, the lefs to a greater, or the greater to a lefs; which is abfurd. Therefore, the Parallelogram ABCD is not about the fame Diameter as the Parallelogram A H. And therefore it will be about the

>

N 2

fame

+43.1.

36. I.

fame Diameter with the Parallelogram AF. Therefore, if from a Parallelogram be taken away another fimilar to the Whole, and in like manner fituate, having also an Angle common with it; then is that Parallelogram about the fame Diameter with the Whole; which was to be demonftrated.

PROPOSITION XXVII.

THEOREM.

Of all Parallelograms applied to the fame Right Line, and wanting in Figure by Parallelograms fimilar, and alike fituate to that described on the balf Line, the greatest is that which is applied to the balf Line, and it is fimilar to the Defect.

LET AB be a Right Line, bifected in the Point C; and let the Parallelogram A D be applied to the Right Line A B, wanting in Figure the Parallelogram C E, fimilar and alike fituate to that defcribed on half of the Right Line A B. I fay, AD is the greatest of all Parallelograms applied to the Right Line A B, wanting in Figure by Parallelograms fimilar and alike fituate to CE. For, let the Parallelogram AF be applied to the Right Line A B, wanting in Figure the Parallelogram HK, fimilar and alike fituate to the Parallelogram CE. I fay, the Parallelogram A D is greater than the Parallelogram AF.

For, because the Parallelogram CE is fimilar to the * 26 of this. Parallelogram H K, they ftand * about the fame Diameter. Let D B, their Diameter, be drawn, and the Figure defcribed; then, fince the Parallelogram C F is + equal to F E, let HK, which is common, be added; and the Whole C H is equal to the Whole K E. But CH is equal to CG, because the Right Line AC is equal to CB; therefore CG is equal to KE; add the common Parallelogram CF, and the Whole A F is equal to the Gnomon L N M; and fo C E, that is the Parallelogram A D, is greater than the Parallelogram AF. Therefore, of all the Parallelograms applied to the fame Right Line, and wanting in Figure by Parallelograms fimilar and alike fituate, to that defcribed on the half Line, the greatest is that which is applied to the half Line, and it is fimilar to the Defect; which was to be demonftrated. PRO

PROPOSITION XXVIII.

PROBLEM.

To a Right Line given to apply a Parallelogram equal to a Right-lined Figure given, deficient by a Parallelogram, which is fimilar to another given Parallelogram; but it is necessary that the Right-lined Figure given, to which the Parallelogram to be applied must be equal, be not greater than the Parallelogram which is applied to the half Line, fince the Defects must be fimilar, viz. the Defect of the Parallelogram applied to the half Line, and the Defect of the Parallelogram to be applied.

LE

ET AB be given Right Line, and let the given Right-lined Figure, to which the Parallelogram to be applied to the Right Line A B muft be equal, be C, which must not be greater than the Parallelogram applied to the half Line, the Defects being fimilar; and let D be the Parallelogram, to which the Defect of the Parallelogram to be applied is fimilar. Now it is required to apply a Parallelogram equal to the given Right lined Figure C to the given Right Line A B, deficient by a Parallelogram fimilar to D.

Let A B be bifected in E, and on E B defcribe* the * 18 of bis. Parallelogram E B F G, fimilar and alike fituate to D; and compleat the Parallelogram A G.

Now, A G is either equal to C, or greater than it, because of the Determination. If AG be equal to C, what was propofed will be done; for the Parallelogram A G is applied to the Right Line A B equal to the given Right-lined Figure C, deficient by the Parallelogram EF, fimilar to the Parallelogram D. But, if it be not equal, then H E is greater than C; but E F is equal to HE; therefore E F fhall also be greater than C. Now make † the Parallelogram + 25 of big. K L M N fimilar and alike fituate to D, and equal to the Excefs, by which E F exceeds C. But D is fimilar to EF; wherefore KM fhall also be fimilar to EF. Therefore let the Right Line KL be homologous to G E, and L M to GF: Then because E F

N 3

ទ