1

In like manner it may be proved, that the square of RL is to the rectangle DPL, in a ratio, compounded of the ratios of the square of LH to the difference of the squares of LH and LD, and of the square of DF to the square of DP.

But in these two compounded ratios, that of the square of DF to the square of DP is common to both, and, because of the equiangular triangles KGD and LHD, the ratios of KG to KD and of LH to LD are equal (4. 6. Eu.), and therefore those, of the square of KG to the square of KD, and of the square of LH to to the square of LD (20. 6. and cor. 3. 22. 5. Eu.), and therefore those, of the square of KG to the difference of the squares of KG and KD, and of the square LH to the difference of the squares of LH and LD (Schol. 18. 5 Eu.) therefore these compound ratios being compounded of equal ratios, are equal (22. 5 Eu.), therefore the ratios equal to them of the rectangle SKT to the rectangle DPK, and of the square of RL to the rectangle DPL are equal, and, by alternating, the rectangle SKT is to the square of RL, as the rectangle DPK, is to the rectangle DPL (16. 5 Eu.), or, the side DP being common to both rectangles, as PK is to PL (1. 6. Eu).

In like manner, if, instead of the tangent RL and secant SKT, the tangent rl and secant skt be used, the truth of the proposition may be shewn, a right line being supposed to be drawn from k to F, by substituting the small letters s, t, k, g, r, 1 and h for the corresponding capitals, as is manifest.

And, by a similar reasoning, it may be proved, in both figures, that the rectangles SKT and skt, are to each other, as the seg ments KP and kP, of the right line parallel to the asymptote, between the parallels ST and st, and the point P.

PROP. XVIII. THEOR.

If a right line, touching a hyperbola, or cutting a hyperbola or opposite hyperbolas in two points, meet two right lines parallel to an asymptote; the squares of the segments of the tangent, or rectangles under the segments of the secant, between the parallels, and the point or points, wherein the tangent or secant meets the section or sections, are to each other, in a ratio, compounded of the ratios, of the segments of the parallels, between the tangent or secant, and the points wherein the same parallels meet the hy perbola, and of the segments of the same parallels, between a right line passing through the foeus adjacent to them, and perpendicularly cutting them, and the adjacent directrix.

Let a right line XZ, see fig. 1, touching a hyperbola in R, or ST, see fig. 1 and 2, cutting a hyperbola or opposite hyperbolas in S and T, meet two right lines DK and GL, parallel to an asymptote CH, and meeting the adjacent directrix DG in D and G; from the focus F adjacent to the parallels, let the right line FBA be drawn, perpendicularly meeting these parallels in A and B, the squares of XR and ZR, or rectangles SKT and SLT, as the case may be, are to each other in a ratio compounded of the ratios of XP to ZQ or PK to QL, and of DA to GB.

First, the rectangles SKT and SLT, see fig. 1 and 2, are to each other, in a ratio compounded of the ratios of PK to QL and of DA to GB.

Join FD, FK, FL and FP; let U and u be the points in which ST and XZ meet the directrix DG; take AV equal to AD, and through K draw KM parallel to PF, meeting DF produced in M.

The right line FV is equal to FD (4. 1 Eu.), and PF to PD (15. 1 Sup.); therefore the triangles DFV and DPF are isosceles, and, having the angle PDF common, are equiangular, therefore DV is to DF, as DF is to DP (4. 6 Eu.), or, which is equal, because of the parallels PF and KM, as FM is to PK; (2. 6 and 16. 5 Eu.), therefore the rectangle under DV or twice DA and PK, is equal to the rectangle DFM (16. 6 Eu.): but, because of PF equal to PD, and parallel to KM, the right lines KD and KM are equal, and a circle described from the centre K, at the distance KD, would pass through M, and so the rectangle DFM is equal to the difference of the squares of KD and KF (Cor. 3. 36. 3 Eu.); therefore the rectangle under twice DA and PK is equal to the difference of the squares of KD and KF.

In like manner, if BN be taken equal to BG, and FQ, FG and FN be joined, and GF produced meet a right line drawn through L parallel to QF, it may be proved, that the rectangle under twice GB and QL is equal to the difference of the squares of LG and LF.

But the ratios of KD and LG, to perpendiculars let fall from K and L on DG, are equal to the determining ratio (Cor. 16. 1 Sup.), therefore the rectangle SKT is to the difference of the squares of KD and KF, as the square of KU is to the difference of the squares of KU and KD (13. 1 Sup.); for a like reason, the rectangle SLT is to the difference of the squares of LG and LF, as the square of LU is to the difference of the squares of LU and LG ; but, because of the equiangular triangles KUD and LUG, the ratios of KU to KD and of LU to LG are equal (4. 6 Eu.), and therefore those, of the square of KU to the square of KD, and of the square of LU to the square of LG (20. 6 and Cor. 3. 22. 5 Eu.), and therefore those, of the square of KU to the difference of the squares of KU and KD, and of the square of LU to the difference of the squares of LU and LG (Schol. 18. 5 Eu.) ; therefore the ratios of the rectangle SKT to the difference of the squares of KD and KF, and of the rectangle SLT to the difference of the squares of LG and LF, which are equal to these

equal ratios, are equal (11. 5 Eu.); and the difference of the squares of KD and KF is above proved equal to the rectangle under twice DA and PK, and the difference of the squares of LG and LF to the rectangle under twice GB and QL, therefore the rectangle SKT is to the rectangle under twice DA and PK, as the rectangle SLT is to the rectangle under twice GB and QL, and, by alternating, the rectangle SKT is to the rectangle SLT, as the rectangle under twice DA and PK is to the rectangle under twice GB and QL (16. 5 Eu.); or, which is equal (23. 6 Eu.), in a ratio compounded of the ratios of PK to QL, and of twice DA to twice GB; or, twice DA being to twice GB, as DA to GB (15. 5 Eu.), in a ratio compounded of the ratios of PK to QL and of DA to. GB.

In like manner it may be proved, in the case of the tangent XZ, see fig. 1, by drawing through X and Z, instead of through K and L, right lines parallel to PF and QF, and otherwise constructing and reasoning as above, that the rectangles under twice DA and XP, and under twice GB and ZQ, are equal to the differences of the squares of XD and XF, and of ZG and ZF; and so the square of RX being, for the like reason as above, to the difference of the squares of XD and XF, as the square of RZ is to the difference of the squares of ZG and ZF; by substituting for the differences of the squares of XD and XF, and of ZG and GF, their equals as above, the rectangles under twice DA and XP, and under twice GB and ZQ, the square of RX is to the rectangle under twice DA and XP, as the square of RZ is to the rectangle under twice GB and ZQ, and, by alternating, the square of RX is to the square of RZ, as the rectangle under twice DA and XP is to the rectangle under twice GB and ZQ (16. 5 Eu.), or, in a ratio, compounded of the ratios of XP to ZQ and twice DA to twice GB (23. 6 Eu.) or, twice DA being to twice GB, as DA to GB (15. 5 Eu.), in a ratio compounded of the ratios of XP to ZQ and of DA to GB,

PROP. XIX. THEOR.

If, in a diameter of a parabola, any point be taken; the rectangle under the segments of the diameter, between its vertex and that point, and its vertex and the directrix, is to the difference of the squares, of the distances of the assumed point, from the focus, and from the directrix, as the square of the segment of the diameter, "between its vertex and the directrix, is to the square of a right line, joining the focus, to the point, in which the diameter meets the directrix.

From K draw KM parallel to FP, meeting DF, produced, if necessary, in M. Because of the equiangular

F

K

M

triangles DKM and DPF, KM is to KD, as PF to PD (4. 6 Eu.), or in a ratio of equality (Def. 8. 1 Sup.): from the centre K, at the distance KD, let a circle be described, which, because of the equality of KD and KM, passes through M ; and, because of the parallels PF and KM, the rectangle DPK is to the rectangle DFM, or, which is equal (Cor. 3. 36. 3 Eu.), the difference of the squares of KF and KD, as the square of PD is to the square of FD (2 and 20. 6 and Cor. 3. 22. 5 Eu).

PROP. XX. THEOR.

If tivo right lines, parallel to each other, both cut in two points, or one of them touch, and the other so cut, a parabola, and meet a diameter; the rectangles under the segments of the secants, or square of the segment of the tangent, between the diameter, and the point or points, wherein they meet the parabola, are to each other, as the segments of the diameter, between its vertex and the parallels.

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