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AD (Cor. 2 to Def.); the base BCD, in the diameters BEC, DEF; and the parallel section GOL, in the straight lines GOH, KOL. Then BC is parallel to GH, and FD to KL.

Hence, by similar triangles, ED: OK (= AE: AO) = EC: OH. But ED

EC, therefore OK OH. Consequently all straight lines drawn from the point O, where the axis meets the parallel plane GOL, to terminate in the line of common section GKHL, are equal to one another, and GKHL is the circumference of a circle, of which O is в the centre.

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2. Let MBDCN be a cylindric surface, of which BCD is the base, and AE the axis, and let it be cut by the plane GOL, parallel to BCD, the line of common section GKHL is the circumference of a circle, having its centre in AE.

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For, let any two planes, ABC, AFD, touching the axis AE, cut the surface in the straight lines MB, NC; LF, KD (Cor. 6 to Def.), the base BCD, in the diameters BEC, DEF, and the parallel section GOL, in the straight lines GOH, KOL. Then BC is parallel to GH, and FD to KL. But NC, KD, are each parallel to AE. Therefore OD, OC, are parallelograms, and OKED ECOH. Whence GKHL is the circumference of a circle, of which O is the centre.

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COR. 1.-If a conic surface be cut by a plane, parallel to the base, the solid betwixt that plane and the vertex is

a cone.

COR. 2.-If a cylindric surface be cut by two planes parallel to the base, the solid betwixt them, and the solids between each of them and the base, are cylinders. COR. 3.-If a cone or cylinder be cut by a plane parallel to the base, the section is a circle.

COR. 4.-Any plane touching the axis of a cone or cylinder, cuts every parallel section in its diameter.

PROPOSITION II.

Every subcontrary section of an oblique cone or cylinder is a circle.

Let the cone or cylinder MBCN be cut by the plane MC, perpendicular to the base, touching the axis, and meeting the surface in the straight lines MB, NC, and the base in the diameter BC. Let it be cut by another plane DFG, perpendicular to the former, so that their line of common sec

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C B

tion DF may form, with one of the lines MB, NC, the angle NFD, equal to the angle MBC, which BC forms with the other on the same side. The latter section DGF, which is called a subcontrary section, is a circle, and DF its diameter.

For, let the parallel section HGI pass through any point K in DF. Because HGI is parallel to the base, it is perpendicular to MC, and therefore GK (So. Ge. I. 18) its line of common section, with DGF, is perpendicular to the same. Thus, GK is at right angles to DF and HI, and since HI is a diameter of the parallel section (IV. 1, Cor. 4), the rectangle HK·KI=GK2. But the triangles HDK, IKF, being similar (for the angles at K are equal, and the angle NFD = MBC= DHK), DK: HK = KI: KF, and the rectangle DK KFHK KI. Consequently the rectangle DK KF GK2, and the section DGF a circle, having DF for a diameter.

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COR.-Every subcontrary section of an oblique cylinder is equal to its base.

For the triangles HDK, IKF, are isosceles.

PROPOSITION III.

Every section of a cone or cylinder, by a plane meeting the conic or cylindric surface on every side, that is neither a parallel nor a subcontrary section, is an ellipse.

Let EHFG be a section of a cone or cylinder MNP, by

a plane meeting the surface on every side, but neither a

parallel nor a subcontrary sec

tion, EHFG is

an ellipse.

For, let C be the middle of

EF, and K any

other point in it; N

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through C and K let planes pass parallel to the base cutting MNP in AB, OR, and EHFG in HG and LD. Also, let the plane MNP cut the base in a diameter NP perpendicular to the line of common section of the plane of the base and of the section EHG. Then HG is parallel to that line of common section (So. Ge. I. 14), and AB to NP; therefore AB is also perpendicular to HG (So. Ge. I. 9). For a similar reason, OR is perpendicular to LD. Now (IV. 1), the sections AHBG, OLRD, are circles, of which AB, OR, are diameters, and therefore AC CB = CG2, and OK KR = DK2.

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Therefore

CF: KF

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AC: OK.

: CB: KR.

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EC2: EK KF = CG2: DK2. Consequently EHFG is an ellipse, of which EF and GH are two conjugate diameters.

PROPOSITION IV.

If a cone be cut by a plane parallel to another, touching the conic surface, the section is a parabola.

Through any point B, in the circumference of the base BGC, let the tangent DE be drawn, the plane ADE, touching that line, and the vertex A, touches the conic surface in the straight line AB (Cor. 4 to Def.). Let the cone be cut by the plane VGN, parallel to ADE, and meeting the base in GN; the section GVN is a parabola.

For, let ABC, the plane of AB and the axis, intersect GVN, in the line VK, and the base in the diameter

BKC. Also, let MHL be a parallel
section, through any point I in VK,
intersecting ABC, in the diameter
ML, and VKG in the straight line
HI. Then, by reason of parallel
planes, the straight line AB is paral-
lel to VK, BD to GK, GK to HI,
and BC to ML (So. Ge. I. 14). E
Hence MK is a parallelogram; and B
since DBC is a right angle, GK is
perpendicular to BČ, and HI to ML

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(So. Ge. I. 9). Now, by similar triangles, VK: VI = KC IL; but BK = MI, therefore VK : VI = BKKC: MI · IL = GK2: HI2. Whence GVN is a parabola, of which VK is a diameter, and GK its semiordinate.

PROPOSITION V.

Every other section of a cone is a hyperbola.

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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, therefore, in the point E, which 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. Then LI is parallel to BC, and MN to DK (So. Ge.

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I. 14), and therefore MN perpendicular to LI (So. Ge. I. 9). Now, by similar triangles,

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Therefore EK · KF : EN · NF BK · KC: LN NI = DK2: MN2. Whence DFO is a hyperbola, of which EF is a diameter, and DK, MN, semi-ordinates

THE END.

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