**Theodosius, Sphaerica iii 2 (pp. 118.1-122.3)**©2002- trans. by Henry Mendell, Cal. State U., L.A.

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(preliminary diagram: moving or still) If a straight line is drawn through a circle and takes out a segment no smaller than a semicircle, and a segment of a circle which is no larger than a semicircle is stood on it and is inclined towards the side no larger than a semicircle. while the circular-arc of the segment standing on it is divided into unequal parts, then the straight-line subtending the smaller circular-arc is smaller than all the straight-lines falling from the same point to the circular-arc no smaller than a semicircle.

As in Theorem iii 1, cases 1b-d are not mentioned in the statement of the theorem. For the theorem to be complete, we need to prove that AE < EG (case 1d)

(diagram 1) For let there be drawn a straight line, AG, in circle ABGD and which takes out a segment ABG no smaller than a semicircle, (diagram 2) and let a segment, AEG, of a circle which is no larger than a semicircle and which inclines towards the side no larger than a semicircle, ADG, be stood on it, (diagram 3) and let circular-arc AEG be divided into unequal parts at point E, and let circular-arc GE be larger than circular-arc EA, and let EA be joined. (diagram 4: moving or still) I say that EA is smaller than all the straight-lines falling from point E to the circular-arc ABG.

(diagram 5)
For let a perpendicular be drawn from point E to the plane of
circle ABGE. It will fall between straight-line AG and circular-arc
ADG since segment AEG inclines towards segment ADG. And let it
fall and let it be EZ and let it come together with the plane
of the circle at point Z, and let the center of circle ABGD be
taken. (diagram 6)
The center of circle ABGD falls either on AG (diagram
7) or between straight-line AG and circular-arc ABG since
segment ABG was supposed no smaller than a semicircle.

Case 1a (general diagram or diagram 8: moving or still): AE < any line from E to arc AB.

(=diagram 7) Let it first be between straight-line AG and circular-arc ABG and let it be point H, and (diagram 9) let ZH be joined and extended in the directions of D, B, (diagram 10) and from point E let a straight-line EQ be extended from point E to circular-arc ABG, and let AZ, ZQ be joined.

And since EZ is perpendicular to the plane of ABGD and therefore
it makes right angles with all the straight-lines touching it
which are in the plane of circle ABGD. Therefore, each of AZ,
ZQ, which are in the plane of circle ABGD, touches EZ. Therefore,
each of angles AZE, QZE is also right. And since AZ is smaller
than ZQ, the sqaure of AZ is also smaller than the square of QZ.
Let the square of ZE be added in common. Therefore, the squares
of AZ, ZE, i.e. the square of AE, are smaller than the squares
of AZ, ZE, i.e. the square of AE. Therefore, AE is smaller than
EQ. (diagram 11: moving or still) We will similarly prove that AE is smaller
than all the straight-lines falling from point E to circular-arc
AB. Therefore AE is smaller than all the lines falling from point
E to circular-arc AB.

Case 1b (general diagram or diagram 12: moving or still): if X, Y are on arc AB, and X is nearer to A than Y, then AE < AX < AY.

And
again we will similarly prove that of all the straight-lines drawn
from point E to the circular-arc between points A, B, the line
nearer to it is always smaller than the line further away.

Case 1c (general diagram or diagram 13 moving or still): let EB > all lines from E to arc ABG and if X, Y are on arc ABG, and Y is nearer to B than X, then EB > AY > AX.

(diagram 14) Let EB be joined. I say that EB is the largest of all the straight-lines falling from point E to circular-arc ABG.

(diagram 15)
For since BZ is larger than ZQ, therefore the square of BZ is
larger than the square of ZQ. Let the square of ZE be added in
common. Therefore, the squares of EZ, ZB, i.e. the square of EB,
are larger than the squares of EZ, ZQ, i.e. than the square of
EQ. Therefore, BE is larger than EQ. We will similarly prove that
EB is larger than all the straight-lines from E to circular-arc
ABG. Therefore, of all the straight-lines falling from E to circular-arc
ABG, EB is the largest.

Case 1d (general diagram or diagram 15: moving or still): EG < any line from E to arc BG and if X, Y are on arc BG, and X is nearer to G than Y, then AE < AX < AY.

(diagram 16) Let EG also be joined. I say that EG is smaller than all the straight-lines falling from point E to circular-arc BG between points B, G.

(diagram 17) For let another line EK be drawn, and let ZG, ZK be joined.

And since GZ is smaller than ZK, therefore the square of GZ
is also smaller than the square of ZK. Let the square of ZE be
added in common. Therefore, the squares of GZ, ZE, i.e. the square
of EG, are smaller than the squares of KZ, ZE, i.e. the square
of EK. Therefore, GE is smaller than EK. Similarly we will prove
that EG is smaller than all the straight-lines falling from point
E to circular-arc BKG between points B, G. (diagram
18: moving or still)
And again we will similarly prove that of all the straight-lines
falling from point E to circular-arc BKG between points B, G,
the line nearer to it is smaller than the line further.

Case 1e: general diagram AE < EG

(diagram 23) Arc AE < arc EG, chord AE < chord EG. (diagram
24 moving
or still)
By case 1a, AE is less than any line from
E to arc AB, and by case 1d EG is less than any line from E to
arc GB. Hence, AE is less than all the lines from A to arc ABG.

Case 2 (general diagram or diagram 25: moving or still): AE < any line from E to arc AB.

It will
be similarly proved that even if ABG is a semicircle, AE is smaller
than all the straight-lines falling from point E to circular-arc
ABG.