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BKC. Also, let MHL be a parallel
BV since DBC is a right angle, GK is v perpendicular to BČ, and HI to ML D' (So. Ge. I. 9). Now, by similar triangles, VK: VI = RC : IL ; but BK = MI, therefore VK : VI = BK · KC: MI · IL = GK2 : HI2. Whence GVN is a parabola, of which VK is a diameter, and GK its semiordinate.
Let DFO be a section of the cone ABDC, different from any that has been mentioned, meeting the base in the line DO. Let DO be cut at right angles by the diameter BC, and let ABC, the plane of the axis, and BC, intersect the surface, and the plane FDO, in the straight lines AB, AFC, and FK, and let AHG be a tangent plane through AB. Then FK is not parallel to AB; otherwise, as DK is parallel to the tangent BG, the plane FDO would be parallel to the tangent plane AGB (So. Ge. I. 13), contrary to the hypothesis ; let them meet, Vt v therefore, in the point E, which B - ----will be in the opposite surface, because, by hypothesis, the plane FDO does not meet the other on every side. Also, let LMI be a parallel section, through any point N in FK, intersecting ABC in the diameter (IV. 1, Cor. 4) LI, and DKF, in the straight line MN. Ìhen LI is parallel to BC, and MN to DK (So. Ge.
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Edinburgh : Printed by W. & R. Chambers,