**Theodosius, Sphaerica iii 5 (pp. 128.23-132.16)**©2002- trans. by Henry Mendell, Cal. State U., L.A.

Return to Vignettes of Ancient Mathematics

If the pole of parallel lines [latitudes] is on the circular-arc of a great circle, and two great circles cut this [circle] at right angles, where one is one of the latitudes, while the other is inclined to the latitudes, and equal circular-arcs are taken from the inclined circle in succession on the same side of the great circle among the latitudes, and parallel circles are described through the points which arise, they take out between them unequal circular-arcs from the initial great circle and the one nearer the great circle among the latitudes is always larger than the one further.

(diagram 1) For let point A on the circular-arc of great circle ABG be the pole of latitudes, and let two great circles intersect this circle at right angles, BZG and DZE, where BZG is one of the latitudes, while DZE is inclined to the latitudes, (diagram 2) and from the inclined circle DZE let equal circular-arcs KQ, QH be taken in succession on the same side of the great circle among the latitudes, BZG, (diagram 3) and through points K, Q, H let latitude circles OKP, NQX, LHM be inscribed. (diagram 4) I say that circles OKP, NQX, LHM take out unequal arcs of the initial great circle ABG and that the one nearer to BZG is larger than one further. I say that circular-arc ON is larger than circular-arc NL.

(diagram 5)
For let great circle AQR be inscribed through points A, Q.

(diagram 6) And since point A is the pole of circle OKP, circular-arc, ANO is equal to circular-arc AQR. Again since point A is the pole of circle NQX, circular-arc ALN is equal to circular-arc ASQ. Therefore, remainder NO is equal to remainder QR. (diagram 7) Similarly we will prove that circular-arc NL is also equal to SQ. Therefore, NO is equal to QR, and LN to SQ. (diagram 8) And since a great circle, AQR, in a sphere intersects one of the circles in the sphere through its poles, OKP, it bisects it and at right angles. Therefore, circle AQR is perpendicular to circle OKP. (diagram 9 or close-up diagram 10) In fact, a segment of a circle, RQ, stands perpendicular on the diameter from R of some circle, OKP, and the conditions that go with this (see Theorem iii 1), and a circular-arc RQ is taken which is smaller than half the segment standing on OKP. Therefore (by Theorem iii 1, case 2), the straight-line joining QR is smaller than the line joining QK. And the circles are equal since they are great circles. Therefore, circular-arc QR is smaller than circular-arc QK. (close-up diagram 11) Similarly we will prove that QS is also smaller than QH. [when we say similarly, a segment SQ of a circle stands perpendicular on the diameter from point S on the a circle, namely LHM, and the conditions that go with this, and a circular-arc SQ is taken which is less than half the segment standing on LSM]. (close-up diagram 12) And KQ is equal to QH. Therefore each of KQ, QH is larger than each of RQ, QS. (diagram 13) And since BZG is parallel to LHM, and BZG falls on circles HQK, AQR within a common section, that is at the center of the sphere, and therefore circle LHM falls outside circles HQK, AQR in a common section outside the surface of the sphere, where point Q is.(this is not proved by Theodosius but the proof of this lemma is easy) (diagram 14 or diagram 15) And so since two great circles HQK, SQR in a sphere, intersect one another, and from one of them, HQK, equal circular-arcs, KQ, QH, have been taken in succession on each side of point Q, and through points, H, K parallel planes, LHM, OKP, were extended, where LHM falls on the common section of planes HQK, SQR outside the surface of the sphere, where point Q is, but one of the equal circular-arcs, KQ, QH, is larger than each of RQ, QS, therefore circular-arc RQ is larger than circular-arc QS (by Theorem iii 4). But RQ is equal to ON, and QS to NL. Therefore, circular-arc ON is larger than circular-arc NL.

Lemma: Let BZG be parallel to LHM, and let BZG fall on a common section of HQK (oblique to BZG), AQR (perpendicular to BZG). Then it falls on the common section outside the sphere where point Q is.

Since HQK and AQR are great circles, the common section goes through the center C. Let it be CQX. Since Q is between H and K, it is also between R and S, which lies on the common section of circles LHSM and ASQR. Otherwise, circle NQX would intersect LHSM. Let this be YS. It is now obvious that YS meets CQX outside the sphere, since the problem reduces to the intersection of two lines through a circle.