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First; AD common EFG are abajoin

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Book V. First, let the parallelograms ABCD, AEFG, have the

angle BAD common to both, and be similar, and similarly placed : ABCD, AEFG 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: Where- A . Cor. 19. 5. fore the remainder DG isa to 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, TL
GF; and the triangles AGF, a

FHC are joined at one angle in the point F; wherefore 32. 6. AF, FC are in the same straight lineb

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 that AG is to GF, as FH to HC; therefore AF, FC are in the same straight lineb.

PROP. XXXIII. B. VI.

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

In the Greek, as also in the Latin translation, the words 0, etuXS, “ 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 apa, 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 Meyaan Eurtagis, p. 50.

PROP. B. Ć. D. B. VI. !! These three propositions are added, because they are frea quently made use of hy geometers. Bori ";!4,372,19

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i DEF. IX. and XI. B. 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 por 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 so 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, 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 coin« cide 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 kinds 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 the centres to the circumferences equal, which is plain from the definition of a circle; and therefore has by some editor

Book XI. been improperly placed among the definitions. The equx

lity 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 • 12. 11. ABC, and from a point D within it draw a the straight

line 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, DĚ, 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 4. 1. contain right angles; therefore the base EB is equal to

the base FB; in the
same manner EA is
equal to FA, and EC

M it 01983 dinud to FC: And in the tri.

* podloge 20 angles EBA, FBA, EB, BA, are equal to FB, BA, and the base EA is equal to the base

FA; wherefore the anc 8. 1. gle EBA is equal e to

the angle FBA, and the
triangle EBA equal b B4
to the triangle FBA,
and the other angles

equal to the other 14. 6. angles; therefore these 31. Deci triangles are similard: In the same manner the triangle

EBC is similar to the triangle FBC, and the triangle EAE to FAC; therefore there are two solid figures, each of which is : contained by six triangles, one of them by three Book XI. triangles the common vertex of which is the point G, and their bases the straight lines AB, BC, CA, and by three other triangles, the common vertex of which is the point E, and their bases the same lines AB, BC, CA: The other solid is contained by the same three triangles the common vertex of which is G, and their bases AB, BC, CA: and by three other triangles of which the common vertex is the point F, and their bases the same straight lines AB, BC, CA: Now the three triangles GAB, GBC, GCA, are common to both solids, and the three others EAB, EBC, ECA, of the first solid, have been shown equal and similar to the three others, FAB, FBC, FCA, of the other solid, each to each : therefore these two solids are contained by the same number of equal and similar planes : But that they are not equal is manifest, because the first of them is contained in the other; Therefore it is not universally true that solids are equal which are contained by the same number of equal and similar planes.

Cor. From this it appears that two unequal solid angles may be contained by the same number of equal plane an

gles.

For the solid angle at B, which is contained by the four plane angles EBA, EBC, GBA, GBC, is not equal to the solid angle at the same point B which is contained by the four plane angles FBA, FBC, GBA, GBC; for this last contains the other : And each of them is contained by four plane angles, which are equal to one another, each to each, or are the self-same, as has been proved: And indeed there may be innumerable solid angles all unequal to one another, which are each of them contained by plane angles that are equal to one another, each to each : It is likewise manifest, that the before-mentioned solids are not similar, since their solid angles are not all equal.

And that there may be innumerable solid angles all unequal to one another, which are each of them contained by the same plane angles disposed in the same order, will be .. plain from the three following propositions.

in PROP. I. PROBLEM. THREE magnitudes, A, B, C, being given, to find a fourth such, that every three shall be greater than the re; maining one. Home

-Let D be the fourth : therefore D must be less than A,

Book XI. B, C, together : Of the three A, B, C, let A be that which

is not less than either of the two B and C: And first, let B
and C together be not less than A; therefore B, C, D, to-
gether are greater than A: And because A is not less than
B; A, C, D, together are greater than B: In the like
manner A, B, D, together are greater than C; Wherefore
in the case in which B and C together are not less than A,
any magnitude D which is less than A, B, C, together, will
answer the problem.
· But if B and C together be less than A ; then, because
it is required that B, C, D, together be greater than A,
from each of these taking away, B, C, the remaining one D
must be greater than the excess of A above B and C: Take
therefore any magnitude D which is less than A, B, C, to-
gether, but greater than the excess of A above B and C:
Then B, C, D, together are greater than A; and because
A is greater than either B or C, much more will A and D,
together with either of the two B, C, be greater than the
other : And, by the construction, A, B, C are together
greater than D.

Cor. If besides it be required, that A and B together
shall not be less than C and D together; the excess of A
and B together above C must not be less than D, that is,
D must not be greater than that excess.

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· PROP. II. · PROBLEM.
· four magnitudes A, B, C, D, being given of which A
and B together are not less than C and D together, and such
that any three of them whatever are greater than the fourth ;
it is required to find a fifth magnitude E such, that any
two of the three. A, B, E, shall be greater than the third,
and also that any two of the three C, D, E, shall be greater
than the third. Let A be not less than B, and C not less
than D.

First, Let the excess of C above D be not less than the
excess of A above B: It is plain that a magnitude E can be
taken which is less than the sum of C and D, but greater
than the excess of C above D; let it be taken; then E is
greater likewise than the excess of A above B; wherefore
E and B together are greater than A; and A is not less than
B; therefore A and B together are greater than B : And,
by the hypothesis, A and B together are not less than C
and D together, and C and D together are greater than E;
therefore likewise A and B are greater than E.

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