**Archimedes,**__On the Equilibria of Planes__Prop. II 4,**with the note of Eutocius**

- translated by Henry Mendell (Cal. State U., L.A.)

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of Ancient Mathematics

On the Equilibria of Planes, Table of Contents

4. The center of the weight of every segment enclosed by a straight-line and the section of a right angled cone is on the diameter of the segment.

(diagram 1) Let there be a segment as stated, ABG; let BD be its diameter. One must prove that the center of the weight of the mentioned segment is on BD.

(diagram 2) For otherwise, let it be E, and let EZ be drawn through it parallel to BD, and let a triangle, ABG having the same base and equal height be inscribed in the segment, and (diagram 3) triangle ABG also have this ratio to area K: the ratio that GZ has to ZD. (diagram 4) But let a rectilinar figure be inscribed familiarly in the segment so that the remaining segments are less than K. (*e1*) (diagram 5) In fact, the center of the weight of the inscribed rectilinear figure is on BD. (II prop. 2) Let it be Q and let QE be joined and extended, and let GL be drawn parallel to BD. It is clear, in fact, that the rectilinear figure inscribed in the segment has a larger ratio to the remaining segments than triangle ABG has to K. But as triangle ABG is to K so is GZ to ZD. Therefore, the inscribed rectilinear figure also has a larger ratio to the left over segments than GZ to ZD, that is, LE to EQ. (diagram 6) And so, let ME have the same ratio, of the rectilinear figure to the segments. And so, since E is the center of the whole segment, and Q of the rectilinear figure inscribed in it, it is clear that the center of the weight of a remainder, the magnitude composed from the left over segments is the point that arises
when QE is extended and some straight line is taken out which has to QE a ratio that the inscribed rectilinear figure has to the left over segments.(I prop. 8) Thus point M would be the center of the weight of the magnitude composed from the left over segments, which is impossible. (diagram 7) For when a line is drawn through M parallel to DB to the same points, all the left over segments will be (on it?). And so it is clear that the center of the weight is on BD.

Commentary of Eutocius, pp. 286.10-15:

On the 4th

“But let a rectilinar figure be inscribed familiarly in the segment so that the remaining segments are less than K.” But this is obvious from what was stated in the tenth book of the *Elements* (X 1?) and the first book of *On the Sphere and Cylinder* (Heiberg suggests I 6). (At most these books show how might prove the claim.)