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19. In a segment is enclosed by a straight-line and section of a right-angled cone, the line drawn from the middle of the base will be a third again in length that drawn from the middle of the half.
For let segment ABG be enclosed by a straight-line and the section of a right-angled cone, and let BD be drawn parallel to the diameter from the middle of AG, while EZ is from the middle of AD, and let ZQ also be parallel to AG. And so, since BD has been drawn in a section of a right-angled cone and parallel to the base, and AD, ZQ are parallel to the line touching the section at B, it is clear that BD has the same ratio to BQ in length that AD has to ZQ in power. (Prop. 3) Therefore BD is also four-times BQ in length. And so, it is obvious that BD is a third again EZ in length.
BD : BQ = T(AD) : T(ED) = T(2*ED) : T(ED) = 4 : 1.
Hence, separando, BD : DQ = 4 : 3.
ZE = DQ.