Lib. VI. Per 11.15hat is, if it be made as AC is to FI, fo is FI to a third AQ; the Triangle X is to Triangle Z, as AC the first to the third Proportional AQ. See Defin. 10.5.

Because the Triangles X, Z are like, BA will be to (2) Per 4.1.6. L1 (a) as AC is to IF. But by the Construction, as AC is to IF, fo is IF to AQ. Therefore alfo BA is to (b) Per 15. LI, (b) as IF to AQ. Therefore in the Triangles QBA and Z, the Sides about the Angles A, I, (which by the Definition of like Triangles are equal) are reci(c) Per 1.1.6. procal. Therefore QB A and Z are equal (c). But the

1.6.

Triangle X is to QBA, as the Bafe AC to the Bafe (d) Per 1.1.6. AQ (d). Therefore X is to Z, as AC to AQ. Q.E.D.

Fig.38.

(e) By the

fame. (f) Per 6.1.6.

Coroll. Hence is their Error to be corrected, who think that like Figures are in the fame Proportion to one another, that their Sides are. For if of two, not only like Triangles, but also Squares, Pentagons, Hexagons, &c. yea, and Circles alfo, the Sides or Diameters be betwixt themselves as 2 to 1, the Figures or Areas themselves are as 4 to i: If the Sides be betwixt themselves as 3 to 1, the Figures themselves or Areas are as 9 to 1; to wit, in a duplicate Proportion of thofe Sides.

PROP. XX. Theorem.

IKE Polygons (ABCDE, FG HIK) are divided, (1.) into like Triangles (P, S, and Q,T, and R,V) in Number equal: (2.) And proportional to the Wholes: And (3.) the Proportion of the Polygons is duplicate to that of the Sides, (A B, FG) which are betwixt the equal Angles (B, G, and BAE, G FK).

Part I. Because the Polygons are alike, they are mutually (per Defin.1. l. 6.) equiangular, and their Angles equal, BAE to GFK, and B to G, and BCD to GHI, and CDE to HIK, and E to K. Because therefore A B is to BC (e) as FG to GH, and the Angles B and G are equal, the Triangles P, S, (f) are like. In like manner it will be demonftrated that R and V are like. Then because the Wholes BCD, GHI, and the fubducted ones BCA, GH F, are equal, the remaining ones alfo, A CD, FHI, are equal. In the fame manner

I might fhew that A DC, FIH, are equal. Therefore (per Corol. 9. pr. 32. l. 1.) the third CAD is equal to the third HFI. "Where alfo (a) the Triangles Q and T are (a) Per 4.1.6: alike. The firft Part therefore is manifeft.

Part II. Because P and S are alike, the Proportion of

P to S is duplicate to that of (b) CA to HF. But for (b) By the the fame Caufe alfo the Proportion of Q to T is dupli- toregoing. cate to the Proportion of CA to HF. Therefore P is to S as Q to T. In the fame manner I will fhew that a's Qis to T, fo R is to V. Therefore as one Antecedent P is to one Confequent S, fo all the Antecedents P, Q, R, taken together, are to all the Confequents S, T, V, taken together, that is, fo is Polygon to Polygon. Which was the other, Part.

Part III. The Proportion of P to S is duplicate (c) tofc) By the that of A B to FG. But the Proportion of Polygon to foregoing. Polygon is the fame with the Proportion of P to S, as I have already fhew'd. Therefore alfo the Proportion of Polygon to Polygon is duplicate to the Proportion of AB to GF. Which was the third Part.

I.

Corollaries:

AL LL ordinate or regular Figures, as Squares, equilateral Triangles, Pentagons, &c. are betwixt themselves in the duplicate Proportion of the Sides, For all regular Figures are like, as is manifeft from Defin. 1. 6.

2. If in any like Figures whatfoever, the Sides A B, Fig. 38. FG, which are placed betwixt equal Angles, be known, the Proportion of the Figures is alfo known. As for example, Let A B be of two Feet, and FG of fix Feet and as 2 is to 6, fo let 6 be to fome other Number; to wit, 18. The leffer Figure is to the greater, as 2 is to 18, or as I is to 9. Now a third proportional Number is found, if (per Corol. 3. pr. 17. 1.6.) the fecond of the given ones be multiplied by it felf, and the Product divided by the first.

3. From the fame Propofition is drawn the excellent Fig. 39. Method of increafing or diminishing any rectilinear Figure in a given Proportion. As if I would make a Pentagon, whofe Side is A B fivefold of another. Find a Mean proportional BX (d) betwixt the Terms of the (d) Per 13.

K 2

Pro

Proportion given, AB, BC;
tagon like to the given one.
the given one.

upon this frame (a) a PenThis fhall be quintuple of

For by the 20. the Pentagon AB is to BX, which is like to it, as AB the firft is to BC the third Proportional.

Moreover, feeing the Proportion of Circles alfo is duplicate to the Proportion of their Diameter, as will be thew'd, p. 2. l. 12. This Practice belongs likewife to Circles.

[Schol. Seeing the Proportion of the Squares E, K, is duplicate of the Proportion of their Sides O R, SV; from thence the duplicate Proportion of the Sides OR, SV is wont commonly to be express'd by the Proportion of ORq to SVq.]

PROP. XXI. Theorem.

Igures (A, B) which are like to the fame (C) are alfo like betwixt themselves.

Fl

This is manifeft from Defin. 1. 1.6. and from Axiom I../. I.

PROP. XXII. Theorem.

Fig. 40, 41. F four or more right Lines (FI, LQ, and 0 R, SV) be proportional; like Figures, and in like Sort defcribed by them (A, B, and E, K) must also be proportional. And converfely.

Fig. 24.

For

The Demonftration of the firft Part is manifeft. because the Proportions of A to B and E to K are duplicate to the Proportions of FI to LQ, and OR to SV, which are by Hypothefis equal; themselves alfo must be equal.

The fecond Part is manifeft alfo.

[Coroll. If the right Line AB be cut in any manner in &; the Rectangle contain'd under the Parts AC, CB, is a Mean proportional betwixt their Squares. Likewife the Rectangle contain'd under theWhole AB,

and one Part AC or CB is a Mean proportional betwixt the Square of the whole AB, and the Square of the faid Part AC or CB. For (per Coroll. 1. p. 8. 1. 6.) it is manifeft that AC:CF::CF:CB. Therefore AC Square: CF Square:: C F Square: CB Square. That is, AC Square Rectangle ACB:: Rectangle * Per 17.1.6. · AC B: CB Square. Q.E. D.

*

Moreover, (per Coroll. 2. p. 8..1. 6.) BA: AF:: AF: AC. Therefore BAq: AFq:: AFq: ACq. That is, B Aq: BAC Rectangle BAC Rectangle† Per 17.1.6. † ACq. In the fame manner AB q: A B C :: ABC: BCq. Q. E.D.]

E

PRO P. XXIII. Theorem.

"Quiangled Parallelograms (X, Z) have betwixt. themselves a Proportion that is compounded of the Proportions of their Sides (AC to CB, and LC to CF)

That is, if you make CB to be to O, as LC to CF, X is to Z, as AC is to O.

Let IL, SB, meet together in Q. The Parallelo

Fig. 42.

gram X (a) is to the Parallelogram R, as AC is to (a) Per 1.1.6. CB; and R is (b) to Z, as LC is to CF; that is, as (b) By the CB is to O. Therefore ex æquo X is to Z, as AC is fame. to 0. Q.E.D.

Corollaries.

FRom hence, and from 34. 1. 1. it is manifeft,

I. That Triangles which have one Angle (at C) equal, have that Proportion betwixt themfelves, which is compounded of the Proportions of the right Lines AC to CB, and LC to CF. Which Lines contain the equal Angle.

2. That Rectangles, and confequently all Parallelograms whatfoever, have betwixt themfelves that Proportion which is compounded of the Proportions of the Bafe to the Bafe, and the Height to the Height. And in the fame manner we reafon about Triangles.

Fig. 42.

134

Fig. 42.

3. Hence the Proportion of Triangles and Parallelograms may be readily learned. Let X and Z be the Parallelograms, and their Bafes AC, CB, and CL, CF (a) Per 12. be their Heights. Let it be made (a) as the Altitude CL, is to the Altitude CF, fo is one of the Bases CB, to O. The Parallelogram X is to the Parallelogram Z, as AC to O.

1.6.

Fig. 43.

PROP. XXIV. Theorem.

IN every Parallelogram (as SF) the Parallelograms which are about the Diameter (A B), to wit, (CE, OI) are both like to the whole Parallelogram, and to each other.

By 27. 1. the Angles C, S, and L, F, are equal. By the fame, E is equal to I, that is, by the fame, equal to A itfelf; but B is common both to the whole SF, and the Part CL. Therefore the whole SF, and the Part CL, are equiangular. It remains to be fhew'd, that they have the Sides oppofite to the equal Angles proportional.

Because in the Triangles BCE, BSA, CE is parallel to SA, BC (by Corol. 1. pr. 4. 1.6.) will be to CE, as BS to SA: And CE will be to E B (by the fame But because in the Triangles Coroll.) as SA to A B. ELB, AFB alfo, EL is parallel to AF; E B (by the fame Coroll.) will be to EL, as A B to A F. Therefore ex æquo CE is to EL, as SA to AF. Therefore (by Defin. 1. 1. 6.) CL and the whole CF are like. the fame manner, I might fhew OI to be like to the whole SF. Therefore (per 21. 1.6.) CL and O1 are alfo like betwixt themselves. 2. E. D.

PROP. XXV. Problem.

In

O change a given Polygon (A) into another like to a given one (B).

Fig. 46.

Or to make a Polygon equal to a given one and like to another given one (B).

(4)

Upon