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.

G

Let there be any plane rectilineal figure, as the triangle ABC, and from a point D within it draw (12. 11.) 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 plain 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 (4. 1.) to the base FB; in the same manner 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 EA, is equal to the base FA; wherefore the angle EBA is equal (S. 1.) to the angle FBA, and the triangle EBA equal (4. 1.) to the triangle FBA, and the other angles equal to the other angles; therefore these triangles are similar (4. 6. 1 def.): in the

same

B

E

D

C

F

manner the triangle EBC is similar to the triangle

FBC, and the triangle EAC to FAC; therefore there are two solid figures, each of which is contained by six triangles, one of them by three 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 angles.

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.

PROP. I. PROBLEM.

Three magnitudes, A, B, C being given, to find a fourth such, that every three shall be greater than the remaining one.

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

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 E 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.

But let the excess of A above B be greater than the excess 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 B is 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, than any two of the angles A, B, E be greater than the third,

and also that any two of the angles 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, D, E, each to each (23. 11.):

and by prop. 26. 11. at the point 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, be 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: