Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/163

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the right line which biſ€cts thoſe parallel ſides will be one of the diameters of the conic ſection. and will likewiſe biſect RQ. Let O be the point in which RQ is biſected, and PO will be an ordinate to that diameter. Produce PO to K ſo that OK may be equal to PO, and OK will be an ordinate on the other ſide of that diameter. Since therefore the points A, B, P, and K are placed in the conic ſection, and PK cuts AB in a given angle, the rectangle PQK (by prop. 17. 19. 21. & 23. book 3. of Apolloniu's conics) will be to the rectangle AQB in a given ratio. But QK and PR are equal, as being the differences of the equal lines OK, OP, and OQ, OR; whence the rectangles PQK and PQ x PR are equal; and therefore the rectangle PQ x PR is to the rectangle AQB, that is, to the rectangle PS x PT in a given ratio. Q. E. D.


Plate 8, Figure 5
Plate 8, Figure 5

Case 2. Let us next ſuppoſe that the oppoſite ſides AC and BD (Pl. 8. Fig.. 5.) of the trapezium, are not parallel. Draw Bd parallel to AC and meeting as well the right line ST in r, as the conic ſection in d. loin Cd cutting PQ in r, and draw DM parallel to PQ, cutting Cd in M and AB in N. Then (becauſe of the ſimilar triangles BTt, DBN) Bt or PQ is to Tt as DN to NB. And ſo Rr is to AQ or PS as DM to AN. Wherefore. by multiplying the antecedents by the antecedents and the conſequents by the conſequents, as the rectangle PQ x Rr is to the rectangle PS x Tt. ſo will the rectangle NDM be to the rectangle ANB, and (by caſe 1.) ſo is the rectangle PQ x Pr to the rectangle Ps x Pt, and by diviſion, ſo is the rectangle PQ x PT to the rectangle PS x PT. Q. E. D.

Plate 8, Figure 6
Plate 8, Figure 6

Case 3. Let us ſuppoſe laſtly the four lines PQ, PR, PS, PT (Pl. 8. Fig. 6.) not to be parallel