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, together 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, together, 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.

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.

7

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 E 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.

But let the excess of A above B be greater than the ex- Book XI. cess of C above D: And because, by the hypothesis, the three B, C, D, are together greater than the fourth A; C and D together are greater than the excess of A above B: Therefore a magnitude may be taken which is less than C and D together, but greater than the excess of A above B. Let this magnitude be E: and because E is greater than the excess of A above B, B together with E is greater than A: And as, in the preceding case, it may be shown that A together with E is greater than B, and that A together with Bis greater than E: Therefore in each of the cases, it has been shown, that any two of the three, A, B, E, are greater than the third.

And because in each of the cases E is greater than the excess of C above D, E together with D is greater than C; and by the hypothesis, C is not less than D; therefore E together with C is greater than D; and, by the construction, C and D together are greater than E: Therefore any two of the three C, D, E, are greater than the third.

PROP. III. THEOREM.

THERE may be innumerable solid angles, all unequal to one another, each of which is contained by the same four plane angles placed in the same order.

Take three plane angles, A, B, C, of which A is not less than either of the other two, and such, that A and B together are less than two right angles; and, by Problem 1, and its corollary, find a fourth angle D such, that any three whatever of the angles A, B, C, D, be greater than the remaining angle, and such, that A and B together be not less than C and D together: And, by Problem 2, find a fifth angle E such, that any two of the angles, A, B, E, be

B

H

greater than the third, and also that any two of the angles

Z

Book XI. C, D, E, be greater than the third: And because A and B together are less than two right angles, the double of A and B together is less than four right angles: But A and B together are greater than the angle E; wherefore the double of A, B, together is greater than the three angles A, B, E, together, which three are consequently less than four right angles; and every two of the same angles A, B, E, are greater than the third; therefore, by prop. 23, 11, a solid angle may be made contained by three plane angles, equal to the angles A, B, E, each to each. Let this be the angle F, contained by the three plane angles GFH, HFK, GFK, which are equal to the angles A, B, E, each to each: And because the angles C, D, together are not greater than the angles A, B, together, therefore the angles C, D, E, are not greater than the angles A, B, E: But these last three are less than four right angles, as has been demonstrated: wherefore also the angles C, D, E, are together less than four right angles, and every two of them are greater than the third; therefore a solid angle may be made, which shall be contained by three plane angles equal to the angles C, 23. 11. D, E, each to eacha: And, by prop. 26, 11, at the point

E

B

M

G

H

F, in the straight line FG, a solid angle may be made equal to that which is contained by the three plane angles that are equal to the angles C, D, E: Let this be made, and let the angle GFK, which is equal to E, he one of the three; and let KFL, GFL, be the other two which are equal to the angles C, D, each to each. Thus there is a solid angle constituted at the point F, contained by the four plane angles GFH, HFK, KFL, GFL, which are equal to the angles A, B, C, D, each to each.

:

Again Find another angle M such, that every two of the three angles A, B, M, be greater than the third, and also every two of the three C, D, M, be greater than the third: And, as in the preceding part, it may be demonstrated, that

the three A, B, M, are less than four right angles, as also Boox XI. that the three C, D, M, are less than four right angles.

Make therefore a solid angle at N contained by the three plane angles ONP, PNQ, ONQ, which are equal to A, B, M, each to each: And by prop. 26, 11, make at the point N, in the straight line ON, a solid angle contained

R

P

by three plane angles, of which one is the angle ONQ equal to M, and the other two are the angles QNR, ONR, which are equal to the angles C, D, each to each. Thus, at the point N, there is a solid angle contained by the four plane. angles ONP, PNQ, QNR, ONR, which are equal to the angles A, B, C, D, each to each. And as the two solid angles at the points P, N, each of which is contained by the above-named four plane angles, are not equal to one another, or, that they can not coincide, will be plain by considering that the angles GFK, ONQ, that is, the angles E, M, are unequal by the construction; and therefore the straight lines GF, FK, cannot coincide with ON, NQ, nor consequently can the solid angles, which therefore are unequal.

And because from the four plane angles A, B, C, D, there can be found innumerable other angles that will serve the same purpose with the angles E and M: it is plain that innumerable other solid angles may be constituted which are each contained by the same four plane angles, and all of them unequal to one another. Q.E. D.

And from this it appears, that Clavius and other authors are mistaken, who assert that those solid angles are equal which are contained by the same number of plane angles that are equal to one another, each to each. Also it is plain, that the 26th prop. of book 11, is by no means sufficiently demonstrated, because the equality of two solid angles, whereof each is contained by three plane angles which are equal to one another, each to each, is only assumed, and not demonstrated.

a 23. 11.

Book XI.

PROP. I. B. XI.

THE words at the end of this, "for a straight line can"not meet a straight line in more than one point," are left out, as an addition by some unskilful hand; for this is to be demonstrated, not assumed.

Mr. Thomas Simpson, in his notes at the end of the second edition of his Elements of Geometry, p. 262, after repeating the words of this note, adds, "Now, can it possibly "show any want of skill in an editor (he means Euclid or "Theon) to refer to an axiom which Euclid himself hath "laid down, book 1, No. 14." he means Barrow's Euclid, for it is the 10th in the Greek, "and not to have de"monstrated what no man can demonstrate?" But all that in this case can follow from that axiom is, that, if two straight lines could meet each other in two points, the parts of them betwixt these points must coincide, and so they would have a segment betwixt these points common to both. Now, as it has not been shown in Euclid, that they cannot have a common segment, this does not prove that they cannot meet in two points, from which their not having a common segment is deduced in the Greek edition: But, on the contrary, because they cannot have a common segment, as is shown in Cor. of 11th Prop. Book 1, of 4to edition, it follows plainly, that they cannot meet in two points, which the remarker says no man can demonstrate.

Mr. Simpson, in the same notes, p. 265, justly observes, that in the corollary of Prop. 11, Book 1, 4to edit. the straight lines AB, BD, BC, are supposed to be all in the same plane, which cannot be assumed in 1st Prop. Book 11. This, soon after the 4to edition was published, I observed and corrected as it is now in this edition: He is mistaken in thinking the 10th axiom he mentions here to be Euclid's; it is none of Euclid's, but is the 10th in Dr. Barrow's edition, who had it from Herigon's Cursus, vol. 1. and in place of it the corollary of 11th Prop. Book 1, was added.

PROP. II. B. XI.

THIS proposition seems to have been changed and vitiated by some editor; for all the figures defined in the 1st Book of the Eléments, and among them triangles, are, by the hypothesis, plane figures; that is, such as are described in a plane wherefore the second part of the enunciation needs no demonstration. Besides, a convex superficies may