**Archimedes,**__On the Equilibria of Planes__Prop. I 11-12- translated by Henry Mendell (Cal. State U., L.A.)

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11. If two triangles are similar to one another and points are similarly placed in relation to the triangles and one point is the center of weight of the triangle in which it is, then the other point is the center of weight of the triangle in which it is.

(diagram 1) Let there be two triangles, ABG, DEZ, and as AG is to DZ so let AB be to DE and BG to EZ, and in the mentioned triangles let there be points positioned similarly, Q, N [viz. triangles ABG, DEZ] and let Q be the center of weight of triangle ABG. I say that N is also the center of weight of triangle DEZ.

(diagram 2) For otherwise, still, if it is possible, let H be the center of weight of triangle DEZ, and let QA, QB, QG, DN, EN, ZN, DH, EH, ZH be joined. And so, since triangle ABG is similar to triangle DEZ, and points Q, H are the centers of their weights, while the centers of weights of similar figures are similarly positioned [so that they make equal angles at the corresponding sides, respectively(*1*)], (diagram 3) therefore angle HDE is equal to angle QAB. (diagram 4) But angle QAB is equal to EDN since points Q, N are similarly positioned]. Angle EDN, therefore, is also equal to EDH, the larger to the smaller, which is impossible. (Assumption 5) Therefore, it is not the case that point N is not the center of weight of triangle DEZ. Therefore it is.

(*1*) This bracketed text is a loose paraphrase of the second part of Assumption 5.

12. If two triangles are similar, but the center of weight of one of them is on a straight-line which is drawn from some angle to the middle of the base, then the center of weight of the remaining triangle will be on the line drawn similarly.

(diagram 1) Let there be two triangles, ABG, DEZ, and as AG is to DZ, so let AB be to DE and BG to ZE, and with AG bisected at H let BH be joined, and let the center of weight, Q, of triangle ABG be on BG. I say that the center of weight of triangle EDZ is also on the similarly drawn straight-line.

(diagram 2) Let DZ be bisected at M and let EM be joined, and as BH is to BQ let ME so be made to EN. (diagram 3) And let AQ, QG, DN, NZ be joined. (diagram 4) Since AH is half of GA and DM is half of DZ, therefore, as BA is to ED, so too is AH to DM. And the sides are proportional about equal angles (namely, BAH and EDM, cf. Euclid, *Elements* VI 6). Therefore, angle AHB is equal to DME, and angle ABQ is equal to DEM. (*1*) (diagram 5) And as AH is to DM, so is BH to EM. But as BH is to BQ so too is ME to EN. Therefore, ex aequali, as AB is to DE, so too is BQ to EN.(*2*)And the sides are proportional about equal angles.(*3*)But if this is the case, angle BAQ is equal to EDN. (diagram 6) Thus a remaining angle, QAG, is also equal to angle NDZ. (diagram 7) For the same reasons, angle BGQ is equal to EZN, (diagram 8) while QGH is equal to NZM. (diagram 9) But ABQ was also proved equal to DEM. Thus a remaining, angle QBG, is equal to NEZ. For all these reasons, in fact, points Q, N are positioned similarly [they make equal angles with the corresponding sides]. And so, since points Q, N are positioned similarly, and Q is the center of weight of triangle ABG, N is, therefore, also the center of weight of DEZ. (Assumption 5)

(*1*) This step is required for the next step in the proof and is referred to later as already proved. Presumably, it dropped out of the text.

(*2*) It is unclear why this is ex aequali and not alternando and mere transitivity of equality, i.e., AB : ED = AH : DM, AH : DM = BH : EM and BH : ME = BQ : EN (since BH : BQ = ME : EN). Hence AB : DE = BQ : EN.

(*3*) This sentence is a repeat of an earlier sentence, each referring to a fairly basic theorem, as noted in the comments, which may raise doubts that at least one of them is a scholium that has crept into the text. So too one might wonder whether the ‘if this’ in the next sentence is somewhat overblown.