than the square of AL, that is of KG: Add to each the square Book VI. of KE; therefore the squareb of AK is not greater than a

b 6. 2. the squares of EK, KG; that is, than the square of

C EG; and consequently the

E straight line AK or GL

D is not greater than GE. Now, if GE be equal

K to GL, the circle EHF touches AB in L, and therefore the square of H


M. AL is equal to the rect

c 36. 3. A

B angle EA, AH, that is, to the given rectangle C, D; and that which was Q

PO Ο. required is done : But if EG, GL be unequal, EG must be the greater: And therefore the circle EHF cuts the straight line AB: Let it cut it in the points M, N, and upon NB describe the square NBOP, and complete the rectangle ANPQ: Because LM is equal tod LN, and it has been proved that AL is equal to LB; d 3. 3. therefore AM is equal to NB, and the rectangle AN, NB equal to the rectangle NA, AM, that is, to the rectangle e Cor.36.3. EA, AH, or the rectangle C, D: But the rectangle AN, NB is the rectangle AP, because PN is equal to NB: Therefore the rectangle AP is equal to the rectangle C, D; and the rectangle AP equal to the given rectangle C, D has been applied to the given straight line AB, deficient by the square BP. Which was to be done.

4. To apply a rectangle to a given straight line that shall be equal to a given rectangle, exceeding by a square.

Let AB be the given straight line, and the rectangle C, D the given rectangle; it is required to apply a rectangle to AB equal to C, D, ex

E ceeding by a square. Draw AE, BF at right

D angles to AB, on the contrary sides of it, and make A E equal


OP to C, and BF equal to D: Join EF, and bisect it in G; and

B from the centre G, at the dis M

N tance GE, describe a circle meeting AE again in H; joint


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Book VI. HF, and draw GL parallel to AE; let the circle meet AB

produced in M, N, and upon BN describe the square NBOP, and complete the rectangle ANPQ; because the angle EHF in a semicircle is equal to the right angle EAB, AB, and HF are parallels, and therefore A# and BF are equal, and the rectangle EA, AH equal to the rectangle EA, BF, that is, to the rectangle C, D. And because ML is equal to LN,

and AL to LB, therefore MA is equal to BN, and the recta 35. 3. angle AN, NB to MA, AN, that is a to the rectangle EA,

AH, or the rectangle C, D: Therefore the rectangle AN, NB, that is, AP, is equal to the rectangle C, D; and to the given straight line AB the rectangle AP has been applied equal to the given rectangle C, D, exceeding by the square BP. Which was to be done.

Willebrordus Snellius was the first, as far as I know, who gave these constructions of the 3d and 4th Problems in his 6 Apollonius Batavus:" And afterwards the learned Dr Halley gave them in the Scholium of the 18th Prop. of the 8th Book of Apollonius's Conics restored by him.

The 3d Problem is otberwise enunciated thus: To cut a given straight line AB in the point N, so as to make the rectangle AN, NB, equal to a given space : Or, which is the same thing, Having given AB the sum of the sides of a rectangle, and the magnitude of it being likewise given, to find its sides.

And the 4th Problem is the same with this, To find a point N in the given straight line AB produced, so as to make the rectangle AN, NB equal to a given space : Or, which is the same thing, Having given AB the difference of the sides of a rectangle, and the magnitude of it, to find the sides.


In the demonstration of this, the inversion of proportional is twice neglected, and is now added, that the conclusion may be legitimately made by help of 24th Prop. of B. 5, as Clavius had done.


The enunciation of the preceding 26th Prop. is not general enough; because not only two similar parallelograms that have an angle common to both, are about the same diameter; but likewise two similar parallelograms that have vertically opposite angles, have their diameters in the same straight line : But there seems to have been another, and that a direct demonstration of these cases, to which this 32d Proposition Book VI. was needful: And the 32d may be otherwise, and something more briefly demonstrated, as follows.


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If two triangles which have two sides of the one, &c.

Let GAF, HFC be two triangles which have two sides AG, GF, proportional to the two sides FH, HC, viz. AG to FG, as FH to HC; and let AG be parallel to FH, and GF to HC; AF and FC are in a straight line.

Draw CK parallel a to FH, and let it meet GF produced a 31. 1. in K. Because AG, KC are each of them parallel to FH, they are

G parallel to one another, and there



b 30. 1. fore the alternate angles AGF,

F FKC are equal : And AG is tó E

H GF, as (FH to HC, that is c) CK

c 34. l. to KF; wherefore the triangles AGF, CKF are equiangular",


d 6. 6. and the angle AFG equal to the

К. angle CFK: But GFK is a straight line; therefore AF and FC are in a straight line

e 14. 1. The 26th Prop. is demonstrated from the 32d as follows:

If two similar and similarly placed parallelograms have an angle common to both, or vertically opposite angles; their diameters are in the same straight line.

First, Let the parallelograms ABCD, AEFG have the angle BAD common to both, and be similar, and similarly placed, ABCD, A EFG are about the same diameter.

Produce EF, GF, to H, K, and join FA, FC; then because the parallelograms ABCD, AEFG are similar, DA is to AB, as GA to AE: Wherefore the remainder DG is a to a Cor.19.5. the remainder EB, as GA to AE: But DG is equal to FH, EB to HC, and AE to GF: Therefore as FH to HC, so is AG to GF; and FH, HC are parallel to AG, GF; and the triangles AGF, FHC are joined at one angle in the point F; wherefore AF, FC are in the same straight line b.

b 32, 6. Next, Let the parallelograms KFHC, GFEA, which are similar, and similarly placed, have their angles KFH, GFE, vertically opposite; their diameters AF, FC are in the same straight line.

Because AG, GF are parallel to FH, HC; and AG is to GF, as FH to HC; therefore AF, FC are in the same straight line b.

Book VI.


The words, “ because they are at the centre," are left out, as the addition of some unskilful hand.

In the Greek, as also in the Latin translation, the words ä pruxe, “any whatever,” are left out in the demonstration of both parts of the proposition, and are now added as quite necessary; and, in the demonstration of the second part, where the triangle BGC is proved to be equal to CGK, the illative particle ägs in the Greek text ought to be omitted.

The second part of the proposition is an addition of Theon's, as he tells us in his commentary on Ptolemy's Morean Lurražus,

p. 50.

PROP. B. C. D. B. VI.

These three propositions are added, because they are frequently made use of by geometers.


Book XI. The similitude of plane figures is defined from the equality of

their angles, and the proportionality of the sides about the equal angles; for from the proportionality of the sides only, or only from the equality of the angles, the similitude of the figures does not follow, except in the case when the figures are triangles: The similar position of the sides which contain the figures, to one another, depending partly upon each of these : And for the same reason, those are similar solid figures which have all their solid angles equal, each to each, and are contained by the same number of similar plane figures: For there are some solid figures contained by similar plane figures, of the same number, and even of the same magnitude, that are neither similar nor equal, as shall be demonstrated after the notes on the 10th Definition : Upon this account it was necessary to amend the definition of similar solid figures, and to place the definition of a solid angle before it: and from this and the 10th Definition, it is sufficiently plain how much the Elements have been spoiled by unskilful editors.

DEF. X. B. XI.

SINCE the meaning of the word “equal" is known and established before it comes to be used in this definition ; therefore the proposition which is the 10th Definition of this Book XI. book is a theorem, the truth or falsehood of which ought to be demonstrated, not assumed; so that Theon, or some other editor, has ignorantly turned a theorem, which ought to be demonstrated, into this 10th Definition: That figures are similar, ought to be proved from the definition of similar figures; that they are equal ought to be demonstrated from the axiom, “ Magnitudes that wholly coincide, are equal to one another;" or from Prop. A of Book 5, or the 9th Prop. or the 14th of the same book, from one of which the equality of all kind of figures must ultimately be deduced. In the preceding books, Euclid has given no definition of equal figures, and it is certain he did not give this: For what is called the first def. of the third book is really a theorem, in which those circles are said to be equal, that have the straight lines from their centres to the circumferences equal, which is plain, from the definition of a circle; and therefore has by some editor been improperly placed among the definitions. The equality of figures ought not to be defined, but demonstrated : Therefore, though it were true, that solid figures contained by the same number of similar, and equal plane figures are equal to one another, yet he would justly deserve to be blamed who would make a definition of this proposition, which ought to be demonstrated. But if this proposition be not true, must it not be confessed that geometers have, for these thirteen hundred years, been mistaken in this elementary matter? and this should teach us modesty, and to acknowledge how little, through the weakness of our minds, we are able to prevent mistakes, even in the principles of sciences, which are justly reckoned amongst the most certain ; for that the proposition is not universally true, can be shown by many examples : The following is sufficient.

Let there be any plane rectilineal figure, as the triangle ABC, and from a point D within it draw a the straight line a 12. 11. DE at right angles to the plane ABC; in DE take DE, DF equal to one another, upon the opposite sides of the plane, and let G be any point in EF; join DA, DB, DC; EA, EB, EC; FA, FB, FC; GA, GB, GC: Because the straight line EDF is at right angles to the plane ABC, it makes right angles with DA, DB, DC which it meets in that plane; and in the triangles EDB, FDB, ED and DB are equal to FD and DB, each to each, and they contain right angles; therefore the base EB is equal o to the base FB; in the same manner, b 4. I. EA is equal to FA, and EC to FC: and in the triangles EBA, FBA, EB, BA, are equal to FB, BA; and the base

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