to the parallelogram CF is compounded of the ratios of their fides.

Therefore equiangular parallelogram's are to one another in a ratio compounded of the ratios of their fides. Which was to be demonftrated.

k Some may perhaps like the following demonftration of this propofition better than Euclid's.

It will be [by 1. 6] as B C to cè, fo is the parallelogram AC to the parallelogram cн. And as BC is to CG, fo is the rectangle under BC and CD to the rectangle under CG and CD. Wherefore [by equality] as the rectangle under BC and CD is to the rectangle under C G and CD, fo is the parallelogram AC to the parallelogram cн.

Therefore [by alternation] as the rectangle under B C and CD, is to the parallelogram a c, fo is the rectangle under co and CD, to the parallelogram c H. For the fame reason, the rectangle under CE and CG is to the parallelogram CF, as the rectangle under CD and CG is to the parallelogram CH. Wherefore [by equality] as the rectangle under BC and CD, is to the parallelogram A c, fo is the rectangle under CE and co to the parallelogram CF. Therefore [by alternation] as the rectangle under B C and C D is to the rectangle under ce and c G, fo is the parallelogram Ac, to the parallelogram cr. W. W. D.

If there be two triangles having one angle of the one equal to one angle of the other: thofe triangles will be to one another in the ratios compounded of the ratio of the fides containing the equal angles.

This will appear evident by supposing the angles B, D, E, C of the equiangular parallelograms ABCD, ECGF to be joined by right lines. For then becaufe [by 34. 1.] the triangles BDC, ECG are each the one half of the faid parallelograms; fince halves are as the wholes, and fince the vertical angles of thefe triangles at c are equal to one another. Therefore [by this prop.] the two triangles at B D C, E CD will be to one another in a ratio compounded of the ratios of their fides: and fo [by def. 5. 6.] one triangle will be to another, as the rectangle under the fides containing an angle of the one, is to the rectangle under the fides containing an equal angle of the other.

This theorem is alfo eafily demonftrated. If the diagonals of two trapeziums interfect one another in a given angle, thofe trapeziums will be to one another in a ratio compounded of the ratios of the diagonals.

PROP. XXIV. THEOR.

The two parallelograms that ftand about the diameter of every parallelogram, are fimilar to one another, and to the whole parallelogram '.

Let ABCD be a parallelogram, whose diameter is a c, and let EG, HK be parallelograms ftanding about the diaB meter Ac: I fay the parallelo

G

A E

C

H

grams EG, HK are fimilar to one another, and to the whole parallelogram A BC D.

For because EF is drawn parallel to one fide BC of the triangle ABC; [by 2. 6.] it will be D K as BE to EA, fo is CF to Fa. Again, because FG is drawn parallel to the fide CD of the triangle A CD, as CF is to FA, fo will DG be to GA. But it has been proved that as CF is to FA, fo is BE to EA: Therefore [by 11. 5.] as BE is to E A, fo is DG to GA: and by compounding [by 18. 5.] as BA is to AE, fo is DA to AG; and inversely [by 16. 5.] as BA is to -AD, fo is AE to AG: Therefore the fides of the parallelograms ABCD, EG about the common angle BA D are proportional. And because GF is parallel to DC, the angle AGF [by 29. 1.] is equal to the angle ADC. But the angle GFA is equal to the angle DCA, and the angle DAC is common to each of the triangles ADC, AGF: Therefore the triangle ADC will be equiangular to the triangle AGF. By the fame reafon the triangle ABC will be equiangular to the triangle AFE: Therefore the whole parallelogram ABCD is equiangular to the parallelogram EG: Wherefore [by 4. 6.] as A D is to DC, fo is AG to GF. But as DC is to CA, fo is GF to FA, and as AC is to CB, fo is AF to FE; and also as CB is to BA, fo is FE to EA: Wherefore because it has been proved that DC is to CA, as GF is to F A, and as AC is to CB, fo is AF to FE; it will be, by equality, [by 22. 5.] as D C is to CB, fo is GF to FE: Therefore the fides of the parallelograms ABCD, EG which are about the equal angles, are proportional; and accordingly [by 1. def. 6.] the parallelogram A B C D is fimilar to the parallelogram E G. By

the

the fame reafon the parallelogram ABCD is alfo fimilar to the parallelogram KH: Therefore the parallelograms EG, HK are each of them fimilar to the parallelogram ABCD. But [by 21. 6.] thofe right lined figures which are fimilar to the fame right lined figure, are fimilar to one another': Therefore the parallelogram E G is fimilar to the parallelogram HK.

Wherefore the parallelograms that ftand about the diameter of every parallelogram are fimilar to one another, and to the whole parallelogram. Which was to be de monftrated.

Some have obferved here that the parallelograms about the diaméter are fimilar figures, having their fides directly proportional, and the complements are equal parallelograms, having their fides reciprocally proportional to one another: Alfo either of the compliments is a mean proportional between the parallelograms about, the diameter. And these are to one another in the duplicate ratio of their homologous fides..

PRO P. XXV. PROBL.

To defcribe a right lined figure fimilar to one given right lined figure, and equal to another.

Let ABC be the given right lined figure to which the figure is to be made fimilar, and D the right lined figure to which it is to be made equal: it is required to make a right lined figure fimilar to A BC, and equal to D.

For [by 44. and 45. 1.] to the right line BC apply the parallelogram BE equal to the triangle ABC, and to the right line CE apply the parallelogram c M equal to D, and having the angle FCE, equal to the angle CBL: Then [by 14. 1.] BC, CF are both in one right line, as alfo LE, EM. Now [by 13. 6.]

find a

nd a mean propor

tional G H between

BC, CE, and [by B

18. 6.] upon GH

defcribe the right

lined figure KGH

L

C

D

K

Book VI. fimilar to the right lined figure A B C, and alike fi

tuate.

Then because it is as B C to GH, fo is GH to CF, and if three right lines be proportional, as the firft is to the third, fo [by 2. cor. 20. 6.] is the right lined figure defcribed upon the first to a fimilar right lined figure alike fituated defcribed upon the fecond: Therefore it will be as B C is to CF, fo is the triangle ABC, to the triangle KGH. But [by 1. 6.] as BC is to CF, fo is the paralelogram BE to the parallelogram EF: And therefore as the triangle ABC is to the triangle KCH, fo is the parallelogram BE to the parallelogram E F. Wherefore alternately [by 16. 5.] as the triangle ABC is to the parallelogram BE, fo is the triangle KGH to the parallelogram EF. But [by conftr.] the triangle ABC is equal to the parallelogram BE: Therefore the triangle KG H also is equal to the parallelogram EF. But the parallelogram E F is equal to the right lined figure D. Therefore the triangle KGH is equal to the right lined figure D. But [by conftr.] KGH is fimilar to the triangle a BC.

Therefore the right lined figure KGH is defcribed fimilar to the given right lined figure ABC, and equal to the given right lined figure D. Which was to be done.

PROP. XXVI. THEOR..

If a parallelogram be taken away from a parallelogram fimilar to the whole, alike fituate, and both baving one common angle: I say they will also both have the fame common diameter.

For from the parallelogram ABCD let the parallelogram A EFG be taken, fimilar to ABC D, alike fituate, and both having the fame common angle DAB; I fay both the parallelograms ABCD, AEFG will have the fame common diameter.

For if not, let if poffible, AHC be the diameter of them, and through a draw HK parallel to A D or BC.

Then because the parallelograms ARCD, KG have both the fame diameter: [by 24. 6.] the parallelogram

ABCD

G

F

H

D

ABCD will be fimilar to the parallelogram KG: A Wherefore [by 1. def. 6.] as DA is to AB, fo is GA K to A K. But because of E the fimilar parallelograms ABCD, EG, as DA is to AB, fo is GA to AE and therefore [by 11. 5.] as GA is to A E, fo is GA to AK: and so G A has the fame ratio to a K, or AE: Wherefore [by 9. 5.] A E is equal to AK, a less equal to a greater; which is abfurd. Therefore the parallelograms A B CD, KG have not both the fame common diameter. Wherefore the parallelograms ABCD, AE F G have both the fame common dia

meter.

B

C

If therefore a parallelogram be taken away from a par allelogram, fimilar to the whole, alike fituate, and both having one common angle: they will alfo both have one common diameter. Which was to be demonftrated.

PROP. XXVII.

THEOR.

The greatest of all parallelograms applied to the fame right line deficient by parallelograms, fimilar and alike fituate to that applied to half the line, is that applied to the half line, being fimilar to the deficient parallelograms applied to the other half line m

Let the right line be A B, which is bifected in c; and to AB let the parallelogram AD be applied lefs than the parallelogram AE by the parallelogram CE fimilar and alike fituate to the parallelogram defcribed upon the line CB, viz. the one half of the right line AB: I fay AD is the greatest of all the parallelograms applied to the right line AB, and deficient by parallelograms, fimilar and alike fituate to CE. For apply the parallelogram AF to the right line AB, deficient in its figure by the parallelogram KH, fimilar to CE, and alike fituate. I fay the parallelogram A D is greater than the parallelogram AF.

T 3

For