**Theodosius, Sphaerica iii 3 (pp. 122.4-124.8)**©2002- trans. by Henry Mendell, Cal. State U., L.A.

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If great circles in a sphere intersect one another, but equal circular-arcs are taken out from each of them in succession on each side of the point where they intersect one another, then the straight-lines joining the limits of the circular-arcs on the same sides are equal to one another.

(diagram 1) For let two great circles, AB, GD, intersect one another at point E, (diagram 2) and let equal circular-arcs be taken out from each of them in succession on each side of point E, AE equal to EB, and GE to ED, (diagram 3) and let GA, BD be joined. I say that GA is equal to BD.

(diagram 4) For the circle inscribed with pole E and radius EA will also have come through B. Either it will also have come through G or not.

(diagram 5) Let it first also go through G. Therefore, it will also have come through D, since circular-arc GE is equal to circular--arc ED. (diagram 6) Let it come, and let there be a common section, AB, of AGBD and AEB, and a common section GD of AGBD and GED.

And since a great circle in a sphere, AEB, intersects a circle, AGBD, of those
in the sphere through the poles, it bisects it and at right angles. Therefore
AB is a diameter of circle AGBD. Similarly we will prove that GD is also a diameter
of circle AGBD.
(diagram 7)
Therefore, the four, AZ, ZG, ZB, ZD, are equal to one another.
And so since two, AZ, ZG, are respectively equal to two, DZ, ZB, and an angle,
AZG, is equal to an angle, DZB, therefore base AG is equal to base DB.

(diagram 8) But again, let the circle inscribed with pole E and radius EA not have come through G, but let it fall beyond it. Therefore, it will also have come through B, but will also fall beyond D. (diagram 9) Let it come and let it be AHBQ, and let circle GED be completed at points H, Q, (diagram 10) and let there be the common section AB of circle AHBQ and AEB, and as well common section HQ of circle AHBQ and HEQ.

(diagram 11) Similarly we will again prove that point Z is the center of circle AHBQ, and that each of AEB, HEQ is perpendicular to circle AHBQ. (diagram 12 = general diagram) Let perpendiculars GK, DL be drawn from points G, D to the plane of circle AHBQ, and let AK, LB be joined.

(diagram 13) And since circular-arc EH is equal to circular-arc EQ (for point E is a pole), where GE is equal to ED, therefore remainder GH is equal to remainder DQ. (diagram 14) And so since HEQ is a perpendicular segment of a circle, and equal circular-arcs HG, DQ are taken out, while perpendiculars have been drawn, GK, DL, therefore GK is equal to DL and HK to QL. (diagram 15) But the whole HZ is also equal to the whole ZQ. Therefore remainder KZ is equal to ZL. (diagram 16) But AZ is also equal to BZ. Therefore AK, LB are equal. (diagram 17) And so since AK is equal to LB, with HK equal to QL, the two AK, KG are equal respectively to the two BL, LD. And angle GKL is equal to DLB, since each of them is right. Therefore base AG is equal to base DB.