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is parallel to the secant PQ, the section again, as in L, and in the case of fig. 2, seeing that DG is parallel to PQ cutting the opposite sections, meeting the opposite section, as in L, and meeting in both cases, ZH in X, and PK in G.

And since PQ, see fig. 1 and 2, is bisected by the diameter HZ, it is ordinately applied to that diameter (32. 1 Sup.), therefore DL, parallel to PQ, is ordinately applied to the same diameter, and is therefore bisected by it (31. 1. Sup.), and so DX is equal to LX; and because of the equiangular triangles QHZ and DXZ, QH is to HZ, as DX is to XZ (4. 6 Eu.), and because of the equiangular triangles ZHP and ZXG, HZ is to HP, as XZ is to XG (by the same), therefore, by equality, QH is to HP, as DX is to XG (22. 5 Eu.) ; whence, QH being equal to HP, DX is equal to XG (Cor. 13. 5 Eu.) ; but DX is above proved to be equal to LX, therefore XG and XL are equal (. x. 1. 1. Eu.), part and whole, which is absurd ; therefore no right line drawn through H, except KH is a diameter of the section, and of course KH is a diameter.

Cor. 1. Hence a diameter of a conick section, passing through the concourse of two tangents, bisects the right line joining their contacts.

For if the diameter divided it unequally, a right line drawn from the concourse to bisect it, being (by this proposition), a diameter, there would be two diameters passing through the concourse of the tangents, and therefore two centres, which is absurd.

Cor. 2. If two right lines, touching a conick section or opposite sections, meet each other; their concourse is in the diameter of the section, which bisects the right line joining their contacts.

For the right line joining the contacts is not a diameter, for if it were, the tangents would be parallel (30. 1 Sup.), if therefore the diameter, bisecting the right line joining the contacts, did not pass through the concourse of the tangents, the diameter passing through the concourse of the tangents, would not bisect the right line joining the contacts, contrary to the preceding corollary.

39

PROP. L. THEOR.

If from a vertex of each of two conjugate diameters (CP and Co,

see fig. 1 and 2), ordinates (PH and OT), be drawn to a third diameter (DG); the square of the segment (CT), of the third diameter (DG), between the centre and either ordinate in ellipses, or between the centre and that drawn from the vertex of the second diameter in hyperbolas, is equal to the rectangle ( DHG), under the segments of that third diameter, between the other ordinate and its vertices. And, in hyperbolas, the square of the segment (CH, see fig. 2), of the same third diameter between the centre and ordinate drawn from the vertex of the transverse diameter ( CP), is equal to the sum of the squares of the semidiameter (CG), to which the ordinates are drawn, and the segment (CT), of the same diameter, between the centre and the other ordinate.

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Though the vertex P of either of the conjugate diameters draw a tangent XPK (48. 1 Sup.), meeting the diamøter DG in K, and its conjugate in X; and (in fig. 1.), because of the parallels PK and CÓ, PH and OT, the triangles PHK and OTC are equiangular (Cor. 3. 9. 1 Sup.), and, because of the parallels CX and PH, the rectangles XPK and CHK are similar (2. 6 and Def. 1. 6 Eu.); therefore the rectangle XPK is to CHK, as the square of PK is to the square of HK (22. 6 Eu.), or, because of the equiangular triangles, as the square of CO is to the square of CT; but the rectangle XPK is equal to the square of co (47.1 Sup.), therefore the square of CT is equal to the rectangle CHK (14.5 Eu.), or, which is equal (45.1 Sup.), the rectangle DHG under the segments of the diameter DG between the ordinate PH and its vertices. Taking away these equals from the square of CG, the excess of the square of CG above

the rectangle DHG, or, which is equal (5. 2 Eu.), the square of CH, is equal to the excess of the square of CG above that of CT, or (5. 2 Eu.), the rectangle DTG.

And, in the case of fig. 2, the demonstration of the equality of the square of CT to the rectangle DHG, in the preceding paragraph, is applicable to this case and its figure, without variation. Adding to these equals the square of CG, the rectangle DHG with the square of CG, or which is equal (6.2 Eu.), the square of CH, is equal to the squares of CG and CT together.

Cor. Hence in the ellipse, the squares of the segments (CH and CT), of the diameter (DG), to which ordinates are drawn from the vertices of two conjugate diameters, between the centre and ordinates, are together equal to the square of the semidiameter (CG) to which they are so drawn. For since the square of CH is, by this proposition, equal to the rectangle DTG, the square of CH within the square of CT, is equal to the rectangle DTG and the square of CT, or which is equal (5. 2 Eu.), the square of CG.

PROP. LI. THEOR.

If from a vertex of each of two conjugate diameters ( CP and Co, see fig. 2 to the preceding proposition ), of a hyperbola, ordinates (PH and OS) be drawn to two other conjugate diameters (CG and CR); the segments (CH and CS) of the diameters, to which the ordinates are drawn, between the centre and the ordinates, are directly, and the ordinates themselves (PH and OS) inversely, as the semidiameters ( CG and CR) to which they are drawn.

For the sum of the squares of CG and CT, or which is equal (50. 1 Sup.), the square of CH, is to the square of OT or CS, as the square of CG is to the square of CR (40. 1 Sup.), therefore CH is to CS, as CG is to CR (22. 6 Eu).

And the rectangle DHG, or which is equal (50. 1 Sup), the square of CT or OS, is to the square of PH, as the square of CG to the square of CR, (40. 1 Sup.), therefore OS is to PH, as CG to CR (22. 6 Eu).

PROP. LII. THEOR.

If one extreme (D), of a right line (DP), equal to the principal

semiaxis (CB) of an ellipse, be in the second axis (MN, und the segment thereof (DG) between the axes, be equal to the difference of the semiaxes (CB and CM); its other extreme (P) is in the perimeter of the ellipse.

Let fall the perpendiculars PH and

M PL on CB and CM. ecause of the equiangular triangles

c PHG and DLP, the square of PH is to A the square of GP or CM, as the square

DG H of DL, or, (47. 1 Eu.), the difference of the squares of DP and LP, or of CB

N and CH, or, (5. 2 Eu.), the rectangle AHB is to the square of DP or CB; therefore the point P is in the perimeter of the ellipse Cor. 3. 40. 1 Sup.).

Cor. Hence, if two unequal right lines AB and MN, of which AB is the greater, bisect each other at right angles in C, and a right line DG be placed between AB and MN equal to the difference of their halves CB and CM, and on DG produced, GP be taken equal to CM, and the right line DGP be so moved through the four right angles, that the point D may be always in the right line MN, and G in AB; the point P would describe an ellipse, whose transverse axis is AB, and second axis MN. And hence an ellipse is described, by means of an instrument, called an Elliptick Compass, as to any one viewing its structure, may hence easily appear.

PROP. LIII. THEOR.

A parallelogram (LSIZ), described about two conjugate diame

ters (QP and RO), of an ellipse or hyperbola, by drawing through their vertices, four right lines, touching the ellipse or conjugate hyperbolas, is equal to the rectangle under the axes (ÅB and MN.

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Let C be the centre, and SPL produced if necessary, meet AB in K, and MN in X, let PH and PT be drawn, at right angles to the axes, and PD and CG at right angles to SPL.

l'hen the right angled triangles (see fig. 1), PHD and CGX, being, because of the parallelism of the sides forming the angles DPH and GCX, equiangular (Cor. 3. 9. 1 Sup.), PH is to PD, as CG is to CX (4. 6 Eu.), therefore the rectangle under PD and GC is equal to the rectangle under PH and CX (16. 6 Eu.), or to the rectangle TCX, or, which is equal (44. 1 Sup. and 17. 6 Eu.), the square of CM; and therefore CG, CM and PD are continually proportional (17. 6 Eu.); therefore the square CG is to the square of CM as CG is to PD (Cor. 2. 20. 6 Eu.), or, because of the equiangular triangles CGK and DPK, as GK is to PK, or, which is equal (1. 6 Eu.), as the rectangle GKP, or, because of the equiangular triangles CGK and PHK, the rectangle CKH (4 and 16. 6 Eu.), or (46. 1 Sup.), AKB to the square of PK, or, which is equal (41. 1 Sup.), as the square of CB to the square of CO ; therefore CG is to CM'as CB is to CO (22. 6 Eu.), and so the rectangle under CO and CG, or, which is equal (35. 1 Eu.), the parallelogram COLP, is equal to the rectangle under CB and CM (16. 6 EU.); whence the parallelogram LSIZ being fourfold the parallelogram COLP (23. 6 Eu.), and the rectangle under AB and MN fourfold that under CB and CM (by the same), the parallelogram LSIZ is equal to the rectangle under AB and MN.

The reasoning of the preceding paragraph, applies to the case of fig. 2 without variation.

Cor. 1. All parallelograms, described about conjugate diameters, of a given ellipse or hyperbola, by drawing tangents

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