Again, seeing by supposition A Z. Z B:: C D F E. C I A P E are proportionals; A B. A Z:: C D F E + C I A P E. C D F E will also, by compounding, be proportionals. And seeing A L is the half of A B, A L. A Z:: C D F E + C I A P E. 2 C D F E will also be proportionals. But the proportion of C D F E + C I A P E to 2 C D F E is compounded, as was but now shown, of the proportions of moment to moment, &c., and therefore the proportion of A L to A Z is compounded of the proportion of the moment of the complete figure C D F E to the moment of the deficient figure C I A P E, and of the proportion of the weight of the deficient figure C I A P E to the weight of the complete figure C D F E; but the proportion of A L to A Z is compounded of the proportions of A L to B Z and of B Z to A Z. Now the proportion of B Z to A Z is the proportion of the weights reciprocally taken, that is to say, of the weight C I A P E to the weight C D F E. Therefore the remaining proportion of A L to B Z, that is, of L B to B Z, is the proportion of the moment of the weight C D F E to the moment of the weight C I A P E. But the proportion of A L to B Z is compounded of the proportions of A L to A Z and of A Z to Z B; of which proportions that of A Z to Z B is the proportion of the weight C D F E to the weight C I A P E. Wherefore (by [art. 5] of this chapter) the remaining proportion of A L to A Z is the proportion of the distances of the points Z and L from the centre of the scale, which is A. And, therefore, (by [art. 6]) the weight C I A P E shall hang from O in the strait line O Z. So that O Z is one diameter of equiponderation of the weight C I A P E. But the strait line A B is the other diameter of equiponderation of the same weight C I A P E. Wherefore (by the 7th definition) the point Z is the centre of the same equiponderation; which point, by construction, divides the axis so, that the part A Z, which is the part next the vertex, is to the other part Z B, as the complete figure C D F E is to the deficient figure C I A P E; which is that which was to be demonstrated.
Coroll. I. The centre of equiponderation of any of those plane three-sided figures, which are compared with their complete figures in the table of [art. 3], chap. XVII, is to be found in the same table, by taking the denominator of the fraction for the part of the axis cut off next the vertex, and the numerator for the other part next the base. For example, if it be required to find the centre of equiponderation of the second three-sided figure of four means, there is in the concourse of the second column with the row of three-sided figures of four means this fraction 5⁄7, which signifies that that figure is to its parallelogram or complete figure as 5⁄7 to unity, that is, as 5⁄7 to 7⁄7, or as 5 to 7; and, therefore the centre of equiponderation of that figure divides the axis, so that the part next the vertex is to the other part as 7 to 5.
Coroll. II. The centre of equiponderation of any of the solids of those figures, which are contained in the table of [art. 7] of the same chap. XVII, is exhibited in the same table. For example, if the centre of equiponderation of a cone be sought for, the cone will be found to be 1⁄3 of its cylinder; and, therefore, the centre of its equiponderation will so divide the axis, that the part next the vertex to the other part will be as 3 to 1. Also the solid of a three-sided figure of one mean, that is, a parabolical solid, seeing it is 2⁄4, that is ½ of its cylinder, will have its centre of equiponderation in that point, which divides the axis, so that the part towards the vertex be double to the part towards the base.
The diameter of equiponderation of the complement of the half of any of the said deficient figures, divides that line which is drawn through the vertex parallel to the base, so that the part next the vertex is to the other part as the complete figure to the complement.
10. The diameter of equiponderation of the complement of the half of any of those figures which are contained in the table of [art. 3], chap. XVII, divides that line which is drawn through the vertex parallel and equal to the base, so that the part next the vertex will be to the other part, as the complete figure to the complement.
For let A I C B (in the same [fig. 5]) be the half of a parabola, or of any other of those three-sided figures which are in the table of [art. 3], chap. XVII, whose axis is A B, and base B C, having A D drawn from the vertex, equal and parallel to the base B C, and whose complete figure is the parallelogram A B C D. Let I Q be drawn at any distance from the side C D, but parallel to it; and let A D be the altitude of the complement A I C D, and Q I a line ordinately applied in it. Wherefore the altitude A L in the deficient figure A I C B is equal to Q I the line ordinately applied in its complement; and contrarily, L I the line ordinately applied in the figure A I C B is equal to the altitude A Q in its complement; and so in all the rest of the ordinate lines and altitudes the mutation is such, that that line, which is ordinately applied in the figure, is the altitude of its complement. And, therefore, the proportion of the altitudes decreasing to that of the ordinate lines decreasing, being multiplicate according to any number in the deficient figure, is submultiplicate according to the same number in its complement. For example, if A I C B be a parabola, seeing the proportion of A B to A L is duplicate to that of B C to L I, the proportion of AD to A Q in the complement A I C D, which is the same with that of B C to L I, will be subduplicate to that of C D to Q I, which is the same with that of A B to A L; and consequently, in a parabola, the complement will be to the parallelogram as 1 to 3; in a three-sided figure of two means, as 1 to 4; in a three-sided figure of three means, as 1 to 5, &c. But all the ordinate lines together in A I C D are its moment; and all the ordinate lines in A I C B are its moment. Wherefore the moments of the complements of the halves of deficient figures in the table of [art. 3] of chap. XVII, being compared, are as the deficient figures themselves; and, therefore, the diameter of equiponderation will divide the strait line A D in such proportion, that the part next the vertex be to the other part, as the complete figure A B C D is to the complement A I C D.
Coroll. The diameter of equiponderation of these halves may be found by the table of [art. 3] of chap. XVII, in this manner. Let there be propounded any deficient figure, namely, the second three-sided figure of two means. This figure is to the complete figure as 3⁄5 to 1, that is 3 to 5. Wherefore the complement to the same complete figure is as 2 to 5; and, therefore, the diameter of equiponderation of this complement will cut the strait line drawn from the vertex parallel to the base, so that the part next the vertex will be to the other part as 5 to 2. And, in like manner, any other of the said three-sided figures being propounded, if the numerator of its fraction found out in the table be taken from the denominator, the strait line drawn from the vertex is to be divided, so that the part next the vertex be to the other part, as the denominator is to the remainder which that subtraction leaves.
The centre of equiponderation of the half of any of the deficient figures in the first row of the table of art. 3, chapter xvii, may be found out by the numbers of the second row.
11. The centre of equiponderation of the half of any of those crooked-lined figures, which are in the first row of the table of [art. 3] of chap. XVII, is in that strait line which, being parallel to the axis, divides the base according to the numbers of the fraction next below it in the second row, so that the numerator be answerable to that part which is towards the axis.
For example, let the first figure of three means be taken, whose half is A B C D (in [fig. 6]), and let the rectangle A B E D be completed. The complement therefore will be B C D E. And seeing A B E D is to the figure A B C D (by the table) as 5 to 4, the same A B E D will be to the complement B C D E as 5 to 1. Wherefore, if F G be drawn parallel to the base D A, cutting the axis so that A G be to G B as 4 to 5, the centre of equiponderation of the figure A B C D will, by the precedent article, be somewhere in the same F G. Again, seeing, by the same article, the complete figure A B E D, is to the complement B C D E as 5 to 1, therefore if B E and A D be divided in I and H as 5 to 1 the centre of equiponderation of the complement B C D E will be somewhere in the strait line which connects H and I. Let now the strait line L K be drawn through M the centre of the complete figure, parallel to the base; and the strait line N O through the same centre M, perpendicular to it; and let the strait lines L K and F G cut the strait line H I in P and Q. Let P R be taken quadruple to P Q; and let R M be drawn and produced to F G in S. R M therefore will be to M S as 4 to 1, that is, as the figure A B C D to its complement B C D E. Wherefore, seeing M is the centre of the complete figure A B E D, and the distances of R and S from the centre M be in proportion reciprocal to that of the weight of the complement B C D E to the weight of the figure A B C D, R and S will either be the centres of equiponderation of their own figures, or those centres will be in some other points of the diameters of equiponderation H I and F G. But this last is impossible. For no other strait line can be drawn through the point M terminating in the strait lines H I and F G, and retaining the proportion of M R to M S, that is, of the figure A B C D to its complement B C D E. The centre, therefore, of equiponderation of the figure A B C D is in the point S. Now, seeing P M hath the same proportion to Q S which R P hath to R Q, Q S will be 5 of those parts of which P M is four, that is, of which I N is four. But I N or P M is 2 of those parts of which E B or F G is 6; and, therefore, if it be as 4 to 5, so 2 to a fourth, that fourth will be 2½. Wherefore Q S is 2½ of those parts of which F G is 6. But F Q is 1; and, therefore, F S is 3½. Wherefore the remaining part G S is 2½. So that F G is so divided in S, that the part towards the axis is in proportion to the other part, as 2½ to 3½, that is as 5 to 7; which answereth to the fraction 5⁄7 in the second row, next under the fraction 4⁄5 in the first row. Wherefore drawing S T parallel to the axis, the base will be divided in like manner.