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Case 2. Let us next suppose that the opposite sides AC and BD of the trapezium are not parallel. Draw Bd parallel to AC, and meeting as well the right line ST in t, as the conic section in d. Join 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 $\times$ Rr is to the rectangle PS $\times$ Tt, so will the rectangle NDM be to the rectangle ANB; and (by Case 1) so is the rectangle PQ $\times$ Pr to the rectangle PS $\times$ Pt; and by division, so is the rectangle PQ $\times$ PR to the rectangle PS $\times$ PT.   Q.E.D.

Case 3. Let us suppose, lastly, the four lines PQ, PR, PS, PT, not to be parallel to the sides AC, AB, but any way inclined to them. In their place draw Pq, Pr, parallel to AC; and Ps, Pt parallel to AB; and because the angles of the triangles PQq, PRr, PSs, PTt are given, the ratios of PQ to Pq, PR to Pr, PS to Ps, PT to Pt will be also given; and therefore the compounded ratios PQ $\times$ PR to Pq $\times$ Pr, and PS $\times$ PT to Ps $\times$ Pt are given. But from what we have demonstrated before, the ratio of Pq $\times$ Pr to Ps $\times$ Pt is given; and therefore also the ratio of PQ $\times$ PR to PS $\times$ PT.   Q.E.D.

LEMMA XVIII.

The same things supposed, if the rectangle PQ $\times$ PR of the lines drawn to the two opposite sides of the trapezium is to the rectangle PS $\times$ PT of those drawn to the other two sides in a given ratio, the point P, from whence those lines are drawn, will be placed in a conic section described about the trapezium.

Conceive a conic section to be described passing through the points A, B, C, D, and any one of the infinite number of points P, as for example p; I say, the point P will be always placed in this section. If you deny the thing, join AP cutting this conic section somewhere else, if possible, than in P, as in b. Therefore if from those points p and b, in the given angles to the sides of the trapezium, we draw the right lines pq, pr, ps, pt, and bk, bn, bf, bd, we shall have, as bk $\times$ bn to bf $\times$ bd,