**Archimedes,**__Quadrature of the Parabola__Prop. 21- translated by Henry Mendell (Cal. State U., L.A.)

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Prop. 20

Prop. 22

21. If a triangle is inscribed in a segment which is enclosed by a straight line and a section of a right-angled cone and has the same base as the segment and height the same, and other triangles are inscribed in the remaining segments having the same base as the segments and height the same, the triangle inscribed in the whole segment will be eight-times each of the triangles inscribed in the left over segments.

(diagram 1) Let segment ABG be as described, and let AG be bisected at D, and let BD be drawn parallel to the diameter. Therefore point B is the vertex of the segment. Therefore, triangle ABG has the same base as the segment and height the same. (diagram 2) Again, let AD be bisected at E and let EZ be drawn parallel to the diameter, and let AB be cut at Q. Therefore, point Z is the vertex of the segment AZB. (by similar triangles, AQ = QB) In fact, triangle AZB has the same base as segment [AZB] and height the same. One must show that triangle ABG is eight-times triangle AZB.

(diagram 3) And so, BD is a third again EZ (Prop, 19) but double EQ. (by similar triangles) Therefore, EQ is double QZ. Thus, triangle AEB is also double ZBA. (diagram 4) For AEQ is double AQZ, and QBE double ZQB.(*) Thus, ABG is eight-times AZB. (diagram 5) It will also be proved similarly of the triangle inscribed in segment BHG.

(*) If BD is 4/3*ZE and QE = 1/2*BE, then QE = 2/3*ZE, or QE = 2*ZQ, while they have the same height, with base ZE, QE. Note that triangle ABE = EBD as they have the same height and equal bases..

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