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

PROPOSITION XXIV.

THEOREM.

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

LE

ET ABCD be a Parallelogram, whose Diameter is AC; and EG, HK, be Parallelograms about the Diameter AC. I fay the Parallelograms EG, HK, are fimilar to the whole ABCD, and alfo to each other.

*

For because EF is drawn parallel to B C, the Side #2 of this. of the Triangle ABC, it shall be * as BE to EA, fo is CF to FA. Again, because F G is drawn parallel to CD, the Side of the Triangle ACD, it shall be as CF to FA, fo is DG to GA. But CF is to FA, (as has been proved) as BE is to EA. Therefore, as BE is to EA, fo is † DG to GA; and by compounding, as BA is to AE, fo is ‡ DA to AG; and by Alternation, as BA is to AD, fo is EA to AG. Therefore the Sides of the Parallelograms ABCD, EG, which are about the common Angle BAD, are proportional. And because G F is parallel to DC, the Angle AGF is equal to the Angle ADC, and the Angle GFA equal to the Angle DCA; and the Angle DAC is common to the two Triangles ADC, AGF. Wherefore the Triangle ADC will be equiangular to the Triangle AGF. For the fame Reason, the Triangle ACB is equiangular to the Triangle AFE. Therefore the whole Parallelogram ABCD is equiangular to the Parallelogram EG; † 4 of this and fo AD is to DC as AG is + to G F. But DC is to CA as GF is to FA; and AC is to CB, as AF is to FE; and moreover, CB is to BA as FE is to EA. Wherefore, fince it has been proved, that DC is to CA, as GF is to FA; and AC is to CB, as AF is to FE; it fhall be, by Equality of Proportion, as DC is to CB, fo is GF to FE. Therefore the Sides that are about the equal Angles of the Parallelograms

ABCD,

ABCD, EG, are proportional; and accordingly the Parallelogram ABCD is fimilar to the Parallelogram EG. For the fame Reason the Parallelogram ABCD is fimilar to the Parallelogram KH. Therefore, both the Parallelograms EG, HK, are fimilar to the Parallelogram ABCD. But Right-lined Figures that are fimilar to the fame Right-lined Figure, are fimi- *21 of this lar 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 anether; which was to be demonftrated.

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.

L'

ET ABC be a given Right-lined Figure, to which
it is required to describe another fimilar and equal

to D.

*

On the Side BC of the given Figure ABC, *make* 44.4. the Parallelogram BE equal to the Right-lined Figure ABC; and on the Side CE make the Parallelogram CM equal to the Right-lined Figure D, in the Angle FCE, equal to the Angle CBL Then BC, CF, as alfo LE, EM, will be † in two strait Lines. † 14. 1. Find GH a mean Proportional between BC, CF, † 13 of this. and on G H let there be defcribed the Right-lined 18 of this. Figure KGH fimilar and alike fituate to the Rightlined Figure ABC.

And then because BC is to GH, as GH is to CF, and fince when three Right Lines are proportional, the firft is to the third as the Figure defcribed on the first

*

ist to a fimilar and alike fituate Figure defcribed on † Cor. 29. the fecond, it fhall be as BC is to CF, fo is the Right- of this. lined Figure ABC to the Right-lined Figure KGH. But as BC is to CF, fo is the Parallelogram BE to † 1 of this. the Parallelogram EF. Therefore, as the Right-lined Figure ABC is to the Right-lined Figure KGH, fo is the Parallelogram BE to the Parallelogram EF.

Where

Wherefore, (by Alternation) as the Right-lined Figure ABC is to the Parallelogram BE, fo is the Rightlined Figure KGH to the Parallelogram EF. But the Right-lined Figure ABC is equal to the Parallelo gram BE. Therefore the Right-lined Figure KGH is alfo equal to the Parallelogram EF. But the Pa→ rallelogram EF is equal to the Right-lined Figure D. Therefore the Right-lined Figure KGH is equal to D. But KGH is fimilar to ABC. Confequently there is described 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, baving alfo an Angle common with it, then is that Parallelogram about the fame Diameter with the whole.

ET the Parallelogram AF be taken away from

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the Parallelogram ABCD 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 AF.

For if it be not, let AHC be the Diameter of the Parallelogram BD, and let GF be produced to H; alfo let HK be drawn parallel to AD, or B C.

Then because the Parallelogram ABCD is about the fame Diameter as the Parallelogram KG, the Pa24 of this. rallelogram ABCD fhall be * fimilar to the ParalletDef. t. of logram KG; and fo as DA is to AB, fo is † GA to

this.

111. 5.

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AK. But because of the Similarity of the Parallelograms ABCD, EG, as DA is to AB, fo is GA to AE. And therefore as GA is t to AE, fo is GA to AK. And fince GA has the fame Proportion to AK as to AE, AE is + equal to AK, the lefs to a greater, which is abfurd. Therefore the Parallelogram ABCD is not about the fame Diameter as the Parallelogram AH. And therefore it will be about the fame Diameter with the Parallelogram AF. Therefore, if from a Parallelogram be taken away another fimilar

to

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; which was to be demonftrated.

PROPOSITION XXVII.

THE ORE M.

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

ET AB be a Right Line, bisected in the Point C, and let the Parallelogram AD be applied to the Right Line AB, wanting in Figure the Parallelogram CE, fimilar and alike fituate to that described on half of the Right Line AB. I fay, AD is the greatest of all Parallelograms applied to the Right Line AB, wanting in Figure by Parallelograms fimilar and alike fituate to CE. For let the Parallelogram AF be applied to the Right Line AB, wanting in Figure the Parallelogram HK, fimilar and alike fituate to the Parallelogram CE. I fay, the Parallelogram AD is greater than the Parallelogram A F.

26 of this,

For because the Parallelogram CE is fimilar to the Parallelogram HK, they ftand* about the fame Diameter, let DB their Diameter be drawn, and the Figure defcribed. Then fince the Parallelogram CF is t t 43. 1, equal to F E, let HK, which is common, be added; and the whole CH is equal to the whole KE. But CH is equal to CG, because the Right Line AC is ‡ 36. 3, equal to CB. Therefore the whole AF is equal to the Gnomon LNM; and fo CE, that is, the Parallelogram AD is greater than the Parallelogram AF. Therefore, of all Parallelograms applied to the fame Right Line, and wanting in Figure by Parallelograms fimilar and alike fituate, defcribed on the half Line, the greatest is that which is applied to the half Line, being fimilar to the Defect; which was to be demon

ftrated.

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# 18 of this.

PROPOSITION XXVIII.

PROBLEM.

To a Right Line given to apply a Parallelogram equal to a Right Line Figure given, deficient by a Parallelogram, which is fimilar to another given Parallelogram; but it is necessary that the Rightlined Figure given, to which the Parallelogram to be applied must be equal, be not greater than the Parallelogram which is applied to the balf 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.

ET AB be a given Right Line, and let the given Right-lined Figure, to which the Parallelogram to be applied to the Right Line A B must be equal, be C, which muft 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 AB, deficient by a Parallelogram fimilar to D.

Let AB be bifected in E, and on EB describe * the Parallelogram EBFG, fimilar and alike fituate to D, and complete the Parallelogram AG.

Now AG 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 AG is applied to the Right Line AB, 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 HE is greater than C; but EF is equal to HE. Therefore EF fhall alfo be +23 of this. greater than C. Now make the Parallelogram

KLMN fimilar and alike fituate to D, and equal to the Excefs, by which EF 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 GE, and LM to GF. Then because EF is equal to C and KM together, EF will be greater than KM;

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