For the rectangle GDH is to the rectangle QDP, as the square of CO is to the square of CP (41. 1 Sup.), and the rectangle SDT is to the sum of the squares of CP and CD, as the square of CR is to the square of CP (by the same); therefore the rectangle GDH is to the rectangle SDT, in a ratio, compounded of the ratios, of the square of CO to the square of CR, and of the rectangle QDP to the sum of the squares of CP and CD (Cor. 6. 23. 6 Eu).

The demonstration is similar, if one or both of the right lines meeting each other, were tangents, or cut opposite hyperbolas.

PROP. XLIV. THEOR.

If from any point of a conick section, an ordinate be drawn to a diameter, and a tangent meeting the same diameter; the semidiameter is, in the case of an ellipse or hyperbola, a mean proportional between the segments of the diameter, between the centre and ordinate, and between the centre and tangent; and, in the case of a parabola, the segment of the diameter, between the ordinate and tangent, is bisected in its vertex.

Let PH, see fig. 1, 2 and 3, of this prop. be an ordinate, drawn from any point P of a conick section, to a diameter HK, which is not a second diameter of a hyperbola, or in the case of a hyperbola, fig. 2, let PT be an ordinate drawn to a second diameter CT, and let PK be a tangent drawn from P, meeting the diameter HK in K, and in fig. 2, the diameter CT in X, CG being in fig. 1 and 2, the semidiameter passing through H and K, and CO in fig. 1, that which passes through T and X; CG is, in fig. 1 and 2, a mean proportional between CH and CK; CO, in fig. 2, between CT and CX; and in fig. 3, KH is bisected in G.

Case 1. When the ordinate PH (see fig. 1 and 2), and tangent PK, meet any diameter of an ellipse, or a transverse one of a hyperbola.

Let D and G be the vertices of that diameter, through which, draw DR and GL parallel to PH, meeting PK in R and L, these touch the section in D and G (Def. 12. 1 Sup.); on CD, produced in fig. 2, take CS equal to CH, DS is equal to GH (Ax. 2 and 3. 1 Eu).

And because the tangent RPK meets the tangents DR and GL, the square of RP is to the square of PL, as the square of DR is

to the square of GL (Cor. 2. 14.1 Sup.), therefore RP is to PL, as DR is to GL (22. 6 Eu.), or, because of the equiangular triangles RDK and LGK, as DK is to GK (4. 6 and 16. 5 Eu.); but, because of the parallels RD, PH and LG, DH is to HG, as RP is to PL (Cor. 2. 10. 6 Eu.), therefore DH is to HG, as DK is to GK (11. 5 Eu.); therefore, by dividing in fig. 1, and compounding in fig. 2, SH is to HG, as DG is to GK (17 and 18 5 Eu.); and taking the halves of the antecedents, CH is to HG, as CG is to GK (Theor. 1. 15. 5 and 22. 5 Eu.), and, by converting, CH is to CG, as CG is to CK.

Case 2. When the ordinate PT (see fig. 2), and tangent PX, meet a second diameter CT of a hyperbola.

Let O be a vertex of the diameter CT, and DG the transverse diameter conjugate to the diameter CT, meeting the tangent PX in K; draw PH an ordinate to the diameter DGH (36. 1 Sup).

By the preceding case, CH, CG and CK are continually proportional, therefore the square of CH is to the square of CG, as CH is to CK (Cor. 2. 20. 6 Eu.), and, by dividing, the excess of the square of CH above that of CG, or, which is equal (6. 2 Eu.), the rectangle DHG, is to the square of CG, as KH is to CK (17.5 Eu.; whence, the square of PH being to the square of CO, as the rectangle DHG is to the square of CG (40. 1 Sup. and 16. 5 Eu.), the ratios of the square of PH to the square of CO, and of KH to CK, being each equal to the ratio of the rectangle DHG to the square of CG, are equal to each other (11. 5 Eu.); but, because of the equiangular triangles HKP and CKX, PH is to CX, as KH is to CK (4. 6 and 16. 5 Eu.), therefore the square of PH or CT is to the square of CO, as PH or CT is to CX (11. 5 Eu.); therefore CT, CO and CX are co and continually proportional (20. 6 Eu).

Case 3. When the ordinate PH and tangent PK meet a diameter GH of a parabola.

Let the ordinate PH be produced to meet the parabola again in Q, and let GR, HS and QT, drawn parallel to PK, meet the diameter PT drawn through P; K GR and QT are ordinates to the diameter PT (Def. 12. 1 Sup).

And since PQ is double to PH,

(31. 1 Sup.), PT is double to PS (2. 6 Eu.), and QT to HS (4. 6 and 16.5 Eu.), or GR, therefore the square of QT is fourfold the square of GR (Cor. 4. 2 Eu.), and therefore the abscissa PT, fourfold the abscissa PR (Cor. 2. 40. 1 Sup), and PS or KH double to PR or KG, and so KH is bisected in G.

Cor. If from the vertices (P and G, see fig 3), of two diameters (PT and GH) of a parabola, ordinates (PH and GR) be drawn to the same diameters; the abscissas (GH and PR) are equal.

For, the tangent PK being drawn from P, meeting GI in K, because of the parallelogram PGR, PR is equal to KG (34. 1 Eu.), or, which is equal (by this prop.), to GH.

PROP. XLV. THEOR.

If from any point of an ellipse or hyperbola, an ordinate be drawn to any diameter, and a tangent from the same point, meet the same diameter; the rectangle under the segments of the diameter, between the ordinate and centre, and between the ordinate and tangent, is, in the case of an ellipse, or transverse diameter of a hyperbola, equal to the rectangle under the segments of the same diameter, between the ordinate and its vertices; and, in the case of a second diameter of a hyperbola, to the sum of the squares of the second semidiameter, and segment of the same diameter, between the centre and ordinate.

Let PH, see fig. 1 and 2 of the prec. prop. be an ordinate drawn from any point P of a conick section, to any diameter of an ellipse, or a transverse one of a hyperbola, let this diameter be DG, let PT, see fig. 2, be an ordinate drawn from P to a second diameter CT of a hyperbola, and let a tangent Ps drawn from P, in fig. 1, meet DG produced in K, and, in fig. 2, meet

DG in K, and CT in X ; the rectangle CHK is, in both figures, equal to the rectangle DHG, and, in fig. 2, the rectangle CTX is equal to the sum of the squares of CO and CT.

Part 1. The rectangle CHK in fig. 1 and 2, is equal to the rectangle HG.

For the rectangle HCK is equal to the square of CG (44. 1 Sup. and 17. 6 Eu.), therefore, in the ellipse, fig. 1, taking from each the square of CH, the excess of the rectangle HCK above the square of CH, or, (3. 2 Eu.), the rectangle CHK, is equal to the excess of the square of CG above that of CH, or (5. 2 Eu.), the rectangle DHG; and, in the hyperbola, fig. 2, taking these equals from the square of CH, the excess of the square of CH above the rectangle HCK, or (2. 2 Eu.), the rectangle CHK, is equal to the excess of the square of CH above that of CG, or (6. 2 Eu.), the rectangle DHG.

Part 2. In fig. 2, the rectangle CTX is equal to the sum of the squares of CO and CT.

For the rectangle TCX is equal to the square of CO (44. 1 Sup. and 17. 6 Eu.), adding to each the square of CT, the rectangle TCX with the square of CT, or (3. 2 Eu.), the rectangle CTX is equal to the sum of the squares of CO and CT.

PROP. XLVI. THEOR.

The same things being supposed; the rectangle under the segments of the diameter, between the tangent and centre, and between the tangent and ordinate, is, in the case of an ellipse, or transverse diameter of a hyperbola, equal to the rectangle under the segments of the same, between the tangent and its vertices; and, in the case of a second diameter of a hyperbola, to the sum of the squares of the second semidiameter, and the segment of the same diameter, between the centre and tangent,

Part. 1. In fig. 1 and 2 of the 44th prop. the rectangle CKH is equal to the rectangle DKG,

For the rectangle HCK is equal to the square of CG (44. 1 Sup. and 17.6 Eu.), therefore, in the ellipse, fig. 1, taking each from the square of CK, the excess of the square of CK above the rectangle HCK, or (2. 2. Eu.), the rectangle CKH, is equal to the excess of the square of CK above that of CG, or (6. 2 Eu.), the rectangle DKG; and, in the hyperbola, fig. 2, taking

from these equals, the square of CK, the excess of the rectangle HCK above the square of CK, or (3. 2 Eu.), the rectangle CKH, is equal to the excess of the square of CG above that of CK, or (5. 2 Eu.), the rectangle DKG.

Part 2. In fig. 2, the rectangle CXT is equal to the sum of the squares of CO and CX.

For the rectangle TCX is equal to the square of CO (44. 1 Sup. and 17. 6 Eu.), adding to each the square of CX, the rectangle TCX with the square of CX, or (3. 2 Eu.), the rectangle CXT, is equal to the sum of the squares of CO and CX.

PROP. XLVII. THEOR.

If two parallel right lines (DR and GL, see fig. 1 and 2 to prop. 44), touching an ellipse or opposite hyperbolas, meet another tangent (RLK); the rectangle under the segments (DR and GL) of the parallels, between their contacts and the tangent which they meet, is equal to the square of the semidiameter (CO), to which they are parallel. And the rectangle (RPL), under the segments of the tangent (RL), which the parallels meet, between its contact (P), and the parallel tangents, is equal to the square of the semidiameter (CZ), which is parallel to it; as is the rectangle (XPK), under the segments of any tangent (RP), meeting two conjugate diameters (CO and DG), between the contact (P) and the diameters.

Part 1. The rectangle under DR and GL is equal to the square of CO.

The right line DG joining the contacts D and G is a diameter, for if G were not the other vertex of the diameter passing through D, a right line drawn from G parallel to DR, to the diameter passing through D, would meet that diameter within the section, for it is parallel to the tangent drawn through the vertex of the diameter remote from D (30. 1 Sup. and 30. 1 Eu.), which tangent falling wholly without the section (Def. 10. 1 Sup.), if the right line so drawn from G, did not meet that diameter within the section, it would meet the tangent drawn through the vertex remote from D, contrary to the definition of parallel right lines; therefore if DG were not a diameter, a right line drawn through G parallel to DR would enter the section and not be a tangent, contrary to the supposition.