Then the whole line CZ is 139,655; and CS, which is a mean proportional between CZ and CO will be 93,410.
At the same place it was stated that GC was found to be 98,779 parts.
To prove this, let PE be drawn in the same figure parallel to DM, and meeting CM at E.In the right-angled triangle CLD the side CL is 99,324 (CM being 100,000), because CL is the sine of the complement of the angle LCM, which is 6 degrees 40 minutes.

And since the angle LCD is 45 degrees 20 minutes, being equal to GCS, the side LD is found to be 100,486: whence deducting ML 11,609 there will remain MD 88,877.
Now as CD (which was 141,289) is to DM 88,877, so will CP 105,032 be to PE 66,070.

But as the rectangle MEH (or rather the difference of the squares on CM and CE) is to the square on MC, so is the square on PE to the square on C_g_; then also as the difference of the squares on DC and CP to the square on CD, so also is the square on PE to the square on _g_C.

But DP, CP, and PE are known; hence also one knows GC, which is 98,779.
_Lemma which has been supposed_.
If a spheroid is touched by a straight line, and also by two or more planes which are parallel to this line, though not parallel to one another, all the points of contact of the line, as well as of the planes, will be in one and the same ellipse made by a plane which passes through the centre of the spheroid.
Let LED be the spheroid touched by the line BM at the point B, and also by the planes parallel to this line at the points O and A.It is required to demonstrate that the points B, O, and A are in one and the same Ellipse made in the spheroid by a plane which passes through its centre.
[Illustration] Through the line BM, and through the points O and A, let there be drawn planes parallel to one another, which, in cutting the spheroid make the ellipses LBD, POP, QAQ; which will all be similar and similarly disposed, and will have their centres K, N, R, in one and the same diameter of the spheroid, which will also be the diameter of the ellipse made by the section of the plane that passes through the centre of the spheroid, and which cuts the planes of the three said Ellipses at right angles: for all this is manifest by proposition 15 of the book of Conoids and Spheroids of Archimedes.

Further, the two latter planes, which are drawn through the points O and A, will also, by cutting the planes which touch the spheroid in these same points, generate straight lines, as OH and AS, which will, as is easy to see, be parallel to BM; and all three, BM, OH, AS, will touch the Ellipses LBD, POP, QAQ in these points, B, O, A; since they are in the planes of these ellipses, and at the same time in the planes which touch the spheroid.