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of Ancient Mathematics
(general diagram)
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.
(general diagram)
(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.
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