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of Ancient Mathematics
(preliminary diagram: moving or still)
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 without being in succession, but on the
same side of the great circle among the latitudes, and great circles
are inscribed through the resulting points and the pole, then
they will take out between them unequal circular-arcs from the
great circle among the latitudes, and the one nearer to beginning
of the great circle is always larger than one farther.
For let the pole of the latitudes on a great circular-arc of
the circle ABG be point A, and let two great circles DEG, BE cut
circle ABG at right angles, where BE is one of the latitudes,
while DEG is inclined to the latitudes. And from inclined circle
DEG let equal circular-arcs ZH, QK be taken, which are not in
succession, but on the same side of great circle BE, one of the
latitudes, and through points Z, H, Q, K and the pole A let great
circles be inscribed, AZL, AHM, AQN, AKX. I say that circular-arc
LM is larger than circular-arc NX.
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Case 1: (general
diagram 1)
(diagram 1.1) For
HQ either is commensurable with ZH, QK or is not.
(diagram 1.2)
Let HQ first be commensurable with ZH, QK, and let ZH, HQ, QK
be divided into parts at points O, P, R, S. (diagram
1.3 = general diagram 1) And through points O, P, R, S and
pole A, let there be inscribed great circles OT, PU, RF, SC.
And so, since circular-arcs ZO, OH, HP, PR, RQ, QS, SK are
in succession and equal to one another, LT, TM, MU, UF, FN, NC,
CX in succession are larger than one another beginning from the
largest, LT. (cf. Theodosius, Sphaerica
iii 6) And so, since LT is larger than NC, and TM than
CX, therefore the whole LM is larger than the whole NX.
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Case 2a: (general
diagram 2)
(diagram 2.0) Let
HQ not be commensurable with ZH, QK. I say that circular-arc LM
is likewise larger than circular-arc NX.
For if LM is not larger than circular-arc NX, either it is
less than it or it is equal.
(diagram 2.1)
First, if it is possible, let LM be smaller than NX, as it holds
in the second diagram.
(diagram 2.2)
Let NO lie equal to LM, and through pole A and O let great circle
PO be inscribed, (diagram
2.3) and since there are three unequal circular-arcs of the
same kind, KQ, QP, HQ, let a circular-arc QR be taken larger than
QP, but smaller than QK, and commensurable with HQ (see
scholion with lemma to iii 9), (diagram
2.4 = general diagram 2) and let SH lie equal to QR (since ZH = QK and QR < QK, SH < ZH),
and through points S, R and pole A let great circles ST, RU be
inscribed.
And so since SH is equal to QR, and HQ is commensurable with
each of SH, QR, therefore TM is larger than NU (By
the commensurable case). But LM is larger than TM. Therefore,
ML is much larger than NO. But it is also equal, which is impossible.
Therefore, LM is not smaller than NX.
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Case 2b: (general
diagram 3)
(diagram 3.1) I
say that LM is not equal to NX either.
For if it is possible, let it hold as in the case of diagram
3, (diagram 3.2)
and let ZH, QK be bisected at points O, P, (general
diagram 3.3 = general diagram 3) and through points O, P and
through pole A, let great circles OR, PS be inscribed.
And so since ZO, OH are equal and in succession to one another,
therefore LR, RM in succession with one another are larger than one another, starting with the largest
LR. (cf. Theodosius, Sphaerica iii 6) Therefore
LR is larger than RM. Therefore LM is more than twice MR. Again,
since QP, PK are in succession with one another, NS, SX in succession
with one another are larger than one another,
starting with the largest NS. Therefore NS is larger than SX,
so that XN is less than twice NS. And so since LM is equal to
NX, where LM is more than double MR. XN is less than twice NS,
therefore RM is less than NS when OH, QP are supposed equal, which
is impossible (by case 2a). Therefore
LM is not equal to NX.
2 RM < LM
LM = NX
NX < 2NS
Hence, RM < NS. Note that the impossibility comes from the
fact that OH = 1/2 ZH = 1/2 QK = QP and case 2a.
It was proved that it is not smaller either. Therefore circular-arc
LM is larger than circular-arc NX.
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