Lemma XX.
If the two oppoſite angular points A and P (Pl. 9. Fig. 2.) of any parallelogram ASPQ touch any conic ſection in the points A and P; and the ſides AQ, AS of one of thoſe angles, indefinitely produced, meet the ſame conic ſection in B and C; and from the points of concourſe B and C to any fifth point D of the conic ſection, two right lines BD, CD are drawn meeting the two other ſides PS, PQ of the parallelogram, indefinitely produced, in T and R; the parts PR and PT, cut off from the ſides, will always be one to the other in a given ratio. And vice verſa, if thoſe parts cut of are one to the other in a given ratio, the locus of the point D will be a conic ſection, paſſing through the the four points A, B, C, P.
Case 1. Join BP, CP, and from the point D draw the two right lines DG, DE, of which the firſt DG ſhall be parallel to AB, and meet PB, PQ, CA in H, I, G; and the other DE ſhall be parallel to AC, and meet PC, PS, AB, in F, K, E; and (by Lem. 17.) the rectangle DE x DF will be to the rectangle DG x DH, in a given ratio. But PQ in to DE (or IQ) as PB to HB, and conſequently as PT to DH; and by permutation, PQ
