**Archimedes,**__Quadrature of the Parabola__Prop. 18- translated by Henry Mendell (Cal. State U., L.A.)

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Prop. 17

Prop. 19

Definitions before prop. 18:

I call **base** the straight line of segments enclosed by a straight-line and a curved line, and **height** the largest perpendicular drawn from the curve line to the base of the segment, and **vertex** the point from which the largest perpendicular is drawn.

18. If a straight line is drawn from the middle of the base in a segment which is enclosed by a straight line and a section of a right-angled cone, the point will be a vertex of the segment at which the line drawn parallel to the diameter cuts the section of the cone.

(diagram 1) For let there be a segment, ABG, enclosed by a straight line and section of a right-angled cone, and from the middle of AG let DB be drawn parallel to the diameter. And so in the section of the right-angled cone BD has been drawn parallel to the diameter, and AD, DG are equal, it is clear that AG and the line touching the section of the cone at B are parallel. (diagram 2) And so, it is obvious that the line drawn from B is the largest of those perpendiculars drawn from the section to to AG. And so, point B is the vertex of the segment.

This assumes that there is only one vertex to the section, something which we may want proved from fthe properties of cones.

The converse is easy to prove:

If a tangent is drawn at the vertex of a segment then the tangent is parallel to the base and a line drawn from the vertex that is the diameter of the segment or parallel to it will bisect the base.

(general diagram for first part = diagram 2)

(general diagram for second part = diagram 5)

(diagram 1) Let B be the vertex of segment ABG. I say that a tangent to the section at B will be parallel to base AG and that a line drawn from B to the base AG parallel to the diameter will bisect AG.

(diagram 2) Suppose base AG is is not bisected by the parallel to the diameter from B, and let the parallel to the diameter be BE. (diagram 3) Then, bisect AG at D and draw DZ to the section. By Prop. 18, Z is the vertex of the section. Hence, there are two vertices of the segment, which is impossible (as noted above, we may want to prove this from the properties of cones). Hence, B and Z are the same point. But BD and BE are parallel, which is also impossible.

(diagram 4) Suppose now that the tangent, HQ, at B is not parallel to AG. (diagram 5) Again, bisect AG at D and draw BD. By Prop. 1, the tangent, KL, at B is parallel to base AG, and B is the vertex, as was just proved. Hence, there are two tangents at B, which is impossible (cf. the proof of Prop. 2).