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On the Equilibria of Planes, Table of Contents

II Prop. 4
II Prop. 6

5. If a rectilinear-figure is familiarly inscribed in a segment enclosed by a straight-line and the section of a right angled cone, the center of the weight of the whole segment is nearer to the vertex of the segment than the center of the inscribed rectilinear-figure.

Note: the proof is in two parts. The first part sets up the case where the familiarly inscribed figure is a triangle, the second part sets up the next familarly inscribed figure, which happens to be a pentagon. However, the argument of the second case tells you how to proceed to the next familiarly inscribed figure (a 9-gon), and to the 17-gon and so forth.

Part 1

(general diagram)

Diagram for first part of II 5(diagram 1) Let segment ABG be as stated, and its diameter be DB, and let first a triangle, ABG, be inscribed familiarly in it, (diagram 2) and let BD be cut at E so that BE is double ED. And so point E is the center of the weight of triangle ABG. (I prop. 14 corollary) (diagram 3) And let each of AB, BG be bisected at Z, H, and let ZK, LH be drawn through Z, H and parallel to BD. (diagram 4) Therefore, the center of the weight of segment AKB will be on ZK and the center of the weight of segment BGL on HL. Let them be Q, I (I prop. 14), and let QI be joined. (diagram 5) (By now, ZH needs also to be drawn) And since QZHI is a parallelogram,(*e1*) and ZN is equal to NH, therefore CQ is also equal to CI. (diagram 5) Thus the center of the weight of the magnitude composed from both segments AKB, BLG is in the middle of QI [since they are equal segments], that is point C. (I prop. 4) (diagram 6) Since the center of the weight of triangle ABG is point E, and C of the magnitude composed from AKB, BLG, it clear, accordingly, that the center of the weight of a whole, segment ABG, is on CE, that is between points C, E. . (I prop. 8 corollary) Thus the center of the whole segment would be nearer to the vertex of the segment than that of the triangle inscribed familiarly.


Part 2

(general diagram)

Figure for second part of II 5(diagram 1) Again, let there be inscribed familiarly in the segment a rectilinear pentagon, AKBLG, and let there be a diameter of the whole segment, BD, and each diameter for each of segments KZ, LH (diagram 2) [and since a rectilinear figure has been inscribed familiarly in segment AKB, the center of the weight of the whole segment is nearer to the vertex than that of the recilinear figure].And so let the center of the weight of the segment be Q, of the triangle be I, and again let the center of the weight of segment BLG be M and that of the triangle be N. (II prop. 5 first part) (diagram 3) In fact, C will be the center of the weight of the magnitude composed from both segments AKB, BLG, and T of the magnitude composed from both triangles AKB, BLG.(*e2*) (diagram 4) And so, again, since E is the center of the weight of triangle ABG, and C of that from both segments AKB, BLG (I prop. 4), it is clear that the center of the weight of a [the] whole, segment ABG, is on CE cut off so that the ratio which triangle ABG has to both segments together, AKB, BLG, is the same ratio that the segment of it having its end-point as C to the smaller segment. (I Prop. 5 corollary) (diagram 5) But the center of the weight of pentagon AKBLG is on straight-line ET cut off so that the ratio that triangle ABG has to AKB, BLG triangles, is the same ratio that the segment of it having its end-point as T to the remainder. (I Prop. 8) (diagram 6) And so, since triangle ABG has a larger ratio to triangles KAB, LBG than to the segments,(*e3*) it is clear, accordingly, that the center of the weight of segment ABG is nearer to vertex B than that of the inscribed rectilinear figure. And the same argument occurs in the case of all rectilinear figures inscribed familiarly in the segments.

(*e1*) Eutocius, 286.16-286.24:

On the 5th.

Diagram for first part of II 5“And since QZHI is a parallelogram, ...” For they are diameters of equal segments. And as they are equally distant from axis BD they are similarly divided by centers Q, I; as KQ is to QZ, LI is to IH. And alternando. For this reason QZ is also equal to IH. But they are also parallel. For all diameters of the parabola are parallel. Therefore QZHI is a parallelogram.


(*e2*) Eutocius, 286.25-288.9:

On the second part of the 5th.

Figure for second part of II 5“In fact, C will be the center of the weight of the magnitude composed from both segments AKB, BLG, and T of the [magnitude composed] from both triangles AKB, BLG.” For it has been prove in the preceeding (first part of prop. 5) that QM, which joins the centers of the segments is bisected by BD at C, as it is parallel to ZH, and NI is bisected at T. Thus C is the center of weight of the magnitude composed from segments AKB, BLG and T of the magnitude composed from triangles AKB, BLG.


(*e3*)288.9-.288.21:

Figure for second part of II 5(Diagram 6 for II 5b) “And so, since triangle BAG has a larger ratio to triangles AKB, BLG than to the segments” and so forth. (trivial differences in order of the letters)

(diagram 1 or (still diagram 2) For since it has been proved that E is the center of the weight of triangle ABG, and T of triangles ABK, BLG, it is obvious that the center of the weight of rectilinear figure AKBLG is on TE cut at R in inverse ratio of that which ABG has to triangles AKB, BLG. And since triangle ABG has a larger ratio to triangles KAB, BLG than to the segments (for the segments are larger than the triangles), it is clear that if we cut ET in the ratio which the triangle has to the segments, the point which is the center of the whole segment because of the inverse ratio will fall higher than R.

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