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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 two arbitrary points are taken
on the same side of the great circle among the latitudes, and
great circles are inscribed through the resulting points and the
pole, it will be the case that as the circular-arc of the great
circle among the latitudes which is between the initial great
circle and the next one through the poles is to the circular-arc
of the inclined circle between the same circles, so the next circular-arc
of the great circle among the latitudes between the great circles
through the pole and the points taken is to some circular-arc
smaller than the circular-arc of the inclined circle between the
two taken points.
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 on the inclined circle DEG let there be taken two points Z,
H on the same side of the great circle among the latitudes BE.
And through points Z, H and pole A, let great circles AZQ, AHK
be inscribed. I say that it is the case that as circular-arc BQ
is to circular-arc DZ, so is circular-arc QK to some circular-arc
smaller than circular-arc ZH.
For either ZH is commensurable with DZ or it is not.
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Case 1: Let it first be commensurable,
and let DZ, HZ be divided into parts at points L, M, N, and through
L, M, N, and pole A let there be inscribed great circles LX, MO,
NP.
And so, since DL, LM, MZ, ZN, NH
are in succession and are equal to one another, therefore BX,
XO, OQ, QP, PK are successively larger than one another starting
from the largest BX. And so since BX, XO, OQ, QP, PK are successively
larger than one another, but DL, LM, MZ, ZN, NH are in succession
and equal to one another, and the number of BX, XO, OQ is equal
to the number of DL, LM, MZ, while the number of QP, PK is equal
to the number of ZN, NH, therefore BQ to DZ has a larger ratio
than QK to ZH. Therefore if we make it that as BQ is to DZ, so
is QK to some other circular-arc,
it will be to one smaller than ZH. Therefore, as circular-arc
BQ is to circular-arc DZ, so is QK to some circular-arc smaller
than ZH.
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Case 2: Let ZH not be commensurable
with DZ. I say that it is likewise the case that as circular-arc
BQ is to circular-arc DZ, so is circular-arc QK to some circular-arc smaller than circular-arc
ZH.
For if it isn't, either it is to one larger than ZH or to it.
Case 2a:
First, if it is possible, let it be to one larger than
ZH, LZ, as it holds in the case of diagram 2. And since there
are three unequal arcs, LZ, ZH, ZD, let there be taken some circular-arc
ZM smaller than ZL, larger than ZH, and commensurable with ZD,
and through M and pole A let great circle MN be inscribed.
And so, since ZM is commensurable with ZD, it is therefore
the case that as BQ is to DZ, so is QN to some circular-arc
smaller than ZM. But as BQ is to DZ so is QK to ZL. Therefore
as QK is ZL so is QN to a circular-arc
smaller than ZM. And by alternate proportion. Therefore as QK
is to QN, so is ZL to a circular-arc
smaller than ZM. But QK is smaller than QN. Therefore LZ is also
smaller than ZM. But it is also larger, which is impossible. Therefore
it is not the case that as BQ is to DZ so is QK to some circular-arc
larger than circular-arc ZH.
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I say that it is not to it either.
Case 2b: For if it is possible,
let it be that as BQ is to DZ, so is QK to ZH, as it holds in
the third diagram, and let each of DZ, ZH be bisected at points
L, M, and through each of points L, M and pole A, let great circles
LN, MX be inscribed.
And so, since DL, LZ are in succession
and equal to one another, therefore BN, NQ are successively larger
than one another starting from the largest BN. Therefore, BQ is
more than double QN. In fact, we will similarly prove that KQ
is also less than double QX. And so, since BQ is more than double
QN, and KQ is less than double QX, therefore BQ has a larger ratio
to QN than KQ to QX. And by alternate proportion. Therefore BQ
to QK has a larger ratio than NQ to QX. But as BQ is to QK, so
is DZ to ZH. Therefore NQ to QX has a smaller ratio than DZ to
ZH. But as DZ is toZH, so is LZ to ZM. Therefore NQ to XQ has
a smaller ratio than LZ to ZM. And by alternate proportion. Therefore
NQ to LZ has a smaller ratio than QX to ZM. Therefore, if we make
it that as NQ is to LZ so is QX to some other, it will be to one
larger than ZM, which was proved impossible. Therefore, it is
not the case that as BQ is to DZ, so is QK to ZH.
BQ > 2QN
KQ < 2QX
BQ : QN > KQ : QX
BQ : KQ > QN : QX
BQ : DZ = KQ : ZH (hypothesis)
BQ : KQ = DZ : ZH
DZ : ZH > QN : QX
DZ : ZH = 2*LZ : 2*ZM = LZ : ZM > QN : QX
QX : ZM > QN : LZ
QX : x = QN : LZ, where x > ZM, which is impossible by case
2a.
But it was proved that it is not to a larger either. Therefore,
it is to a smaller. Therefore, it is the case that as BQ is to
DZ, so is QK to a circular-arc smaller
than circular-arc ZH.
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