**Theodosius, Sphaerica iii 4 (pp. 124.19-128.21)**©2002- trans. by Henry Mendell, Cal. State U., L.A.

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Since the figure is somewhat complex, two view points are given, from the left and from right of arc GED.

General strategy: the theorem will prove that arc GE > arc
ED. The strategy will be to show that because

1. straight-line HN = straight-line MZ (since HL = LZ, while triangle
BLN = triangle ALM).

2. angle DHZ is obtuse

3. angle GNH = angle DHL

4. Therefore, angle GNH is acute.

5. Therefore (by 2, 4), angle GNH < angle DMZ.

6. Therefore (by 1, 5) arc GH < arc DZ

7. But, arc HGE = arc EDZ

8. Hence, EG = EH - GH > EZ - DZ = ED.

(general diagram: right view, left view)

If two great circles in a sphere intersect one another, and from one of them equal circular-arcs are taken in succession on each side of the point where they intersect, and through the points which arise parallel planes are extended, where one falls onto the common section of the planes outside the surface of the sphere, where the mentioned point is, and one of the equal circular-arcs is larger than each of the arcs taken out by the first planes at the same point, then the circular-arc between the point and the plane that does not fall onto common section is larger than the circular-arc of the same circle between the point and plane that does fall on it.

(diagram 1)
For let two great circles, AEB, GED, intersect one another at
point E, (diagram
2) and from one of them equal circular-arcs, AE, EB, are taken
in succession on each side of point E,

(diagram 3, i)
solid right view, ii)
truncated right view,
iii) right view with plane, iv)
clear right view with plane, v)truncated
left view, vi)
left view of some planes, vii)
left view with other planes)

and through points A, B let parallel planes be drawn AD, GB where
AD falls on the common section of AEB, GED outside the surface
of the sphere at pont E, (diagram 4:
left view, right view) and
let one of the equal circular-arcs AE, EB be larger than each
of GE, ED. I say that circular-arc GE is larger than circular-arc
ED.

(general diagram: right view, left view)

(diagram 5: right
view, left view,
plain right view,
plain left view)
For the circle inscribed with pole E and distance EA will have
come through B and will fall beyond points G, D since each of
AE, EB is larger than each of GE, ED. Let it come and let it be
as AHBZ, and let the circles be completed, let circle AD meet
circle AHBZ at point Q, and circle BG at point K,(diagram
6: right
view, left view)
and let AB and HZ be common sections of AHBZ with AEB and HEZ,
(diagram 7: right
view, left view)
and let AQ be the common section of circle ADQ and AHBZ, and let
KB be the common section of KGB and AHBZ,(diagram
8: right
view, left view)
and let DM be be the common section of HEZ and ADQ, and let GN
be the common section of KGB and HEZ. (diagram
9: right
view, left view)
And since plane AD meets the common section of planes HEZ, AEB,
that is EL, it is outside the surface of the sphere at point E,
let it meet it at X, and let EL be extended to X.(diagram
10: right
view, left view)
Therefore, point X is on plane ADQ, but also on HEZ, but points
D, M are on both planes DQ, HEZ. (diagram
11: right
view, left view)
Therefore MD meets it outside the surface of the sphere where
point E is. In fact, it meets it at X.

(general diagram: right view, left view)

(diagram
12: right view, left
view) And since a great circle in a sphere, AEB, intersects through the
poles one of these circles in the sphere, AHBZ, it bisects it and at right angles.
Therefore, AB is a diameter of circle AHBZ. Similarly we will prove that HZ
is also a diameter of circle AHBZ. Therefore, L is its center. (diagram
13: right view, left
view) And since two parallel planes KGB, ADQ are cut by a plane, AHBZ, therefore
their common sections are parallel. Therefore KB is parallel to AQ. (diagram
14: right view, left
view) Again, since two parallel planes, KGB, ADQ are cut by a plane, HEZ,
their common sections are therefore parallel. Therefore GN is parallel to DM.
(diagram 15: right
view, left view) And since
each of planes AEB, HEZ are at right angles to plane AHBZ, therefore their common
section is also at perpendicular to plane AHBZ. But their common section is
EL. Therefore, EL is also perpendicular to plane AHBZ. Thus it will make right
angles with all the straight-lines touching it in the plane of circle AHBZ.
But each of AB, HZ, which are in the plane of circle AHBZ, touches EL. Therefore,
EL is perpendicular to each of AB, HZ.

(diagram 15 (again): right view, left view) And since angle XLN is an outside angle of triangle XLM, it is larger than the opposite and interior angle XML. But XLN is right. Therefore XML is acute. Therefore, XMZ is obtuse.

(diagram 16: right view, left view) And since GN is parallel to DM, and HZ falls across them, therefore angle GNH is equal to angle XML. But XML is acute. Therefore angle GNH is also acute.

(diagram 17: right view, left view) And AM is parallel to NB, and two lines have been drawn through them, AB, MN, and AL is equal to LB. Therefore NL is also equal to LM. But the whole HL is also equal to the whole LZ. Therefore remainder HN is equal to remainder MZ.

(diagram 19: right
view, left view) And so
since HEZ is a segment of a circle, and equal straight-lines
HN, MZ have been taken out, and parallels GN, DM have been drawn, with GNH being
acute and DMZ being obtuse, therefore circular-arc HG is smaller than circular-arc
DZ. And so, since the whole circular-arc HE is equal to the whole circular-arc
ZE, where HG is smaller than DZ, therefore the remainder, circular-arc GE, is
larger than the remainder, circular-arc ED, Q.E.D.

Only two propositions in Theodosius end in 'Q.E.D.', this one and the introductory i 2. Perhaps, he is making a statement about the complexity of this lemma.