146 THE MATHEMATICAL PRINCIPLES [BooK 1. 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 Uie other segment, Let the four sides ML, IK, KL, MI, of the parallelogram ML JK touch the conic section in A, B, C, I) ; and let the fifth tangent FQ cut those sides in F, Q, H, and E : and taking the segments ME, KQ of the sides Ml, KJ, or the segments KH, MF of the sides KL, ML, 1 s/.y, 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 El 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 X ME, as also the rectangle KH X ME 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 Kl. MI in q and e, the rectangle KQ X ME will be equal to the rectangle K X Me, and KQ will be to Me as Kq to ME, and by division ns Q? to Ee. COR. 3. Hence, also, if E, 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 Q 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 jive 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 quadri lateral tained figure under ABFE conany four of them ; and (by Cor. 3, Lem. XXV) the right line MN draAvn through the points (,f