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 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×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. 22.) and the middle point of the right line MK is the centre of the ſection.
