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.
PROPOSITION XXVII. PROBLEM XIX.
- To describe a trajectory that may touch five right lines given by position.
Supposing ABG, BCF, GCD, FDE, EA to be the tangents given by position. Bisect in M and N, AF, BE, the diagonals of the quadrilateral figure ABFE contained under any four of them; and (by Cor. 3, Lem. XXV) the right line MN drawn through the points of
