**Archimedes,**__On the Equilibria of Planes__Prop. I 9-10- translated by Henry Mendell (Cal. State U., L.A.)

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10. The center of weight of every parallelogram is the point at which the diameters meet.

10. Alternative proof of proposition 10

9. The center of weight of every parallelogram is on the straight-line that joins the bisections of the opposite sides of the parallelogram.

(diagram 1)Let there be a parallelogram, ABGD, and EZ on the bisection of AB, GD. I say that the center of weight of parallelogram ABGD will be on EZ. (diagram 2) For otherwise, still, if it is possible, let it be Q, and let QI be drawn parallel to AB. (diagram 3) With EB in fact being bisected repeatedly the remaining will at some point be smaller than IQ. And let each of AE, EB be divided into equals to EK, and from the points in the divisions, let there be drawn parallels to EZ. (diagram 4) The whole parallelogram will in fact be divided into parallelograms equal and similar to KZ. And so, when the equal and similar parallelograms are fitted onto one another the centers of their weights will also fall on one another. (Assumption 4) There will be, in fact, certain magnitudes, parallelograms equal to KZ and even in number, and their centers of weight positioned on a straight line, and the middle ones are equal as are all those on each side of the middle one themselves are equal and the straight-lines between the centers are equal. Since the magnitude, therefore, is composed from all of them, the center of weight will be on the straight-line joining the centers of weight of the middle areas. (Corollary 1 to Props. 4 & 5) But that is not so. For Q is outside the middle parallelograms. And so it is clear that the center of weight of parallelogram ABGD is on straight-line EZ.

10. The center of weight of every parallelogram is the point at which the diameters meet.

(diagram 1) Let there be a parallelogram ABGD and EZ bisecting AB, GD in it, and KL bisecting AG, BD. (diagram 2) The center of weight of parallelogram ABGD, in fact, is on EZ. For this was shown. (I Prop. 9) (diagram 3) And for the same reasons it is also on KL. (diagram 4) Therefore Q is the center of weight. (diagram 5) But the diameters of the parallelogram meet at Q. Thus what was proposed has been proved.

In another way. (diagram 4 = general diagram)

(diagram 1) It is also possible to prove the same thing in another way. Let there be a parallelogram ABGD and it diameter DB. Therefore, triangles ABD, BDG are equal and similar to one another. (diagram 2) Thus, when the triangles are fitted to one another, the centers of their weights will also fall on one another. (Assumption 4) (diagram 4) In fact, let the center of weight of triangle ABD be point E, and let DB be bisected at Q, and let EQ be joined and extended, and let ZQ, equal to QE, but cut off. (diagram 3) In fact, when triangle ABD is fitted to triangle BDG and side AB placed on ZQ, straight-line QE will also fit onto ZQ, and point E will fall on Z. (Assumption 4) But it also fits on the center of weight of triangle BDG. (diagram 4) And so since point E is the center of weight of triangle ABD, and Z of DBG, it is clear that the center of weight of the magnitude composed from both triangles is the middle of straight-line EZ, which is point Q. (I Prop. 4)

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