Right Lines
Table of Contents
Lemma 23
If two right lines, as AC, BD given by position, and terminating in given points A, B, are in a given ratio one to the other, and the right line CD, by which the indetermined points C, D are joined is cut in K in a given ratio; I say, that the point K will be placed in a right line given by position.
For let the right lines AC, BD meet in E, and in BE take BG to AE as BD is to AC, and let FD be always equal to the given line EG; and, by construction, EC will be to GD, that is, to EF, as AC to BD, and therefore in a given ratio; and therefore the triangle EFC will be given in kind. Let CF be cut in L so as CL may be to CF in the ratio of CK to CD; and because that is a given ratio, the triangle EFL will be given in kind, and therefore the point L will be placed in the right line EL given by position. Join LK, and the triangles CLK, CFD will be similar; and because FD is a given line, and LK is to FD in a given ratio, LK will be also given. To this let EH be taken equal, and ELKH will be always a parallelogram. And therefore the point K is always placed in the side HK (given by position) of that parallelogram. Q.E.D.
Corollary
Because the figure EFLC is given in kind, the three right lines EF, EL, and EC, that is, GD, HK, and EC, will have given ratios to each other.
Lemma 24
If 3 right lines, two whereof are parallel, and given by position, touch any conic section.
The semi-diameter of the section which is parallel to those two is a mean proportional between the segments of those two that are intercepted between the points of contact and the third tangent.
Let AF, GB be the two parallels touching the conic section ADB in A and B; EF the third right line touching the conic section in I, and meeting the two former tangents in F and G, and let CD be the semi-diameter of the figure parallel to those tangents; I say, that AF, CD, BG are continually proportional.
For if the conjugate diameters AB, DM meet the tangent FG in E and H, and cut one the other in C, and the parallelogram IKCL be completed; from the nature of the conic sections, EC will be to CA as CA to CL; and so by division, EC - CA to CA - CL, or EA to AL; and by composition, EA to EA + AL or EL, as EC to EC + CA or EB.
Therefore, (because of the similitude of the triangles EAF, ELI, ECH, EBG) AF is to LI as CH to BG. Likewise, from the nature of the conic sections, LI (or CK) is to CD as CD to CH; and therefore (ex æquo perturbatè) AF is to CD as CD to BG. Q.E.D.
Corollary 1
Hence if two tangents FG, PQ, meet two parallel tangents AF, BG in F and G, P and Q, and cut one the other in O; AF (ex æquo perturbatè) will be to BQ as AP to BG, and by division, as FP to GQ, and therefore as FO to OG.
Corollary 2
Whence also the two right lines PG, FQ drawn through the points P and G, F and Q, will meet in the right line ACB passing through the centre of the figure and the points of contact A, B.
LEMMA 25
If four sides of a parallelogram indefinitely produced touch any conic section, and are cut by a fifth tangent; I say, that, taking those segments of any two conterminous sides that terminate in opposite angles of the parallelogram, either segment is to the side from which it is cut off as that part of the other conterminous side which is intercepted between the point of contact and the third side is to the other segment.
Let the four sides ML, IK, KL, MI, of the parallelogram MLIK touch the F conic section in A, B, C, D; and let the fifth tangent FQ cut those sides in F, Q, H, and E; and taking the segments ME, KQ of the sides MI, KI, or the segments KH, MF of the sides KL, ML; I say, that ME is to MI as BK to KQ; and KH to KL as AM to MF. For, by Cor. 1 of the preceding Lemma, ME is to EI as (AM or) BK to BQ; and, by composition, ME is to MI as BK to KQ. Q.E.D. Also KH is to HL as (BK or) AM to AF; and by division, KH to KL as AM to MF. Q.E.D.
Cor. 1. Hence if a parallelogram IKLM described about a given conic section is given, the rectangle KQ × ME, as also the rectangle KH × MF equal thereto, will be given. For, by reason of the similar triangles KQH, MFE, those rectangles are equal.
Cor. 2. And if a sixth tangent eq is drawn meeting the tangents KI, MI in q and e, the rectangle KQ × ME will be equal to the rectangle Kq × Me, and KQ will be to Me as Kq to ME, and by division as Qq to Ee.
Cor. 3. Hence, also, if Eq, eQ, are joined and bisected, and a right line is drawn through the points of bisection, this right line will pass through the centre of the conic section.
For since Qq is to Ee as KQ to Me, the same right line will pass through the middle of all the lines Eq, eQ, MK (by Lem. XXIII), and the middle point of the right line MK is the centre of the section.