**Archimedes,**__On the Equilibria of Planes__Prop. I 15- translated by Henry Mendell (Cal. State U., L.A.)

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On the Equilibria of Planes, Table of Contents

15. In every trapezium having two sides parallel to one antoher, the center of weight is on the straight-line which joins the bisections of the parallels, which is divided in such a way that the segment of it having as limit the bisection of the smaller of the parallels to the remaining segment has this ratio, what the line equal to double the larger with the smaller together has the double of the smaller with the larger of the parallels.

(diagram 2) Let there be a trapezium, ABGD, having parallels, AD, BG, and (diagram 3) let EZ join the bisections of AD, BG. And so, it is obvious that the center of the trapezium is on EZ. (diagram 4) For if you extend GDH, ZEH, BAH, it is clear that they will come to the same point (*e1*), (diagram 5) and the center of weight of triangle HBG will be on HZ, and similarly the center of weight of triangle AHD is on EH. (I prop. 13) Therefore, the center of weight of the remainder, trapezium ABGD will also be on EZ. (I prop. 8 and assumption 7) (diagram 6) Joining points B, D, let BD be divided in three equals at points K, Q, and through them parallel to BG, let LQM, NKT be drawn (diagram 7) and let DZ, BE, OC be joined. In fact, the center of weight of triangle DBG will be on QM, since QB is a third part of BD (*e2*) [and MQ is drawn parallel to the base through point Q]. But the center of weight of triangle DBG is also on DZ. Thus C is the center of weight of the mentioned triangle. And for these reasons too point O is the center of weight of triangle ABD. Therefore, when the magnitude is composed from both triangles ABD, BDG which is the trapezium, the center of weight will be on straight-line OC. (I prop. 8) (diagram 8) But the center of weight of the mentioned trapezium is also on EZ. Thus the center of weight of trapezium ABGD is point P. (diagram 9) But triangle BDG would have to ABD a ratio which OP has to PC. (I props. 6 and 7) (diagram 10) But as triangle BDG is to ABD so is BG to AD, and as OP is to PC so is RP to PS (by similar triangles). Therefore, as BG is to AD, so too is RP to PS. Thus as two BG with AD is to two AD with BG, so too is two RP with PS to two PS with PR. But two RP with PS together is RSP that is PE. (That is, since BQ = QR, SZ = SR = SP + PR). Also, since QK = KD, SR = RE = SP + PR, so that 2 PR + SP = SE. Therefore, the proposed has been proved. Namely, 2 BG + AD : 2 AD + BG = EP : PZ.

Corollary: If we draw a parallel to BG through P, UF, then 2 BG + AD : 2 AD + BG = AU : UB. Archimedes uses this theorem in *Quadrature of the Parabola*, prop. 10

Lemma: If a : b = c : d, then 2 a + b : 2 b + c = 2 c + d : 2 d + c

proof:

1.
a : b = c : d

2. a + b : b = c + d : d (1 componendo)

3. a + b : 2 b + a = c + d : c + 2 d (2 componendo)

4.
a : a + b = c : c + d (1 componendo)

5. 2 a : b : a + b = 2 c + d : c + d (4 componendo)

7.
2 a : b :2 b + a = 2 c + d : c + 2 d (3, 5 ex aequali)

Commentary of Eutocius, pp. 274.19-26

"For it you extend GDH, ZEH, BAH, it is clear that they will come to the same point," For when BAH, ZEH are extended and meet one another at H and GD, by being extended will fall on the same point. For as HB is to HA so is ZH to HE and BZ to AE and ZG to ED and, of course, GH to DH. (back)

Commentary of Eutocius, pp. 276.1-19:

"In fact, the center of weight of triangle DBG will be on QM, since QB is a third part of BD." For let there be a triangle ABG and let AE, BZ, GD be joined from the angles to the bisections of the sides. (I prop. 14)Therefore, H is the center of weight of triangle ABG. It is also obvious that all the triangles are equal to one another,, and that the lines joined to the bisections of the sides go through H, so that the centers are not more than the single one. For, since AD, DB, BE, EG, GZ, ZA are equal, the triangles which have point H as vertex and whose bases are the mentioned straight-lines will also be equal. Thus AH is double HE. And so, if we draw QK parallel to BG through H, AQ is double QB. Thus, universally, if one side of a triangle is cut so that the part at the vertex is double the part at the base, and through the point taken a parallel to the base is drawn, the center of weight of the triangle will be on the drawn line.(back)

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