Lemma 17

Lemma 17

If from any point of a given conic section, to the four produced sides AB, CD, AC, DB, of any trapezium ABDC inscribed in that section, as many right lines PQ, PR, PS, PT are drawn in given ang ei, each line to each side ; the rectangle PQ, X PR of those on the opposite sides AB, CD, will be to the rectangle PS X PT of those on tie other two opposite sides AC, BD, in a given ratio.

CASE 1

PQ and PR are

PS and PT ; as PQ PR to the side and AB. to the side And I s AC, and ^^ p ; T | farther, that one AC and BD, are parallel pair of the opposite sides, as betwixt themselves; then the right line which bisects^ those parallel sides will be one of the diameters of the I3 IQ 1L RQ. Let O be the point in which will be an ordinate to that diameter. Produce conic section, and will likewise bisect RQ is bisected, and PO OK PO to K, so that may be equal on the other side of that diameter. and K to PO, and OK will be an ordinate B P Since, therefore, the points A, ; are placed in the conic section, and cuts in a given angle, PK AB PQK the rectangle (by Prop. XVII., XIX., XXI. and XXI1L, Book III., of Apollonius s Conies) will be to the in a given ratio. rectangle But and are equal, as being the differences of the lines AQB QK OP, and PR OQ, OR equal ; whence the rectangles and therefore the rectangle the rectangle PS X PT PQ X PR PQK is to in a given ratio. and PQ X PR the rectangle Q.E.D A^ OK, are equal B, that Is, to

CASE parallel. Let us next suppose that the oppo and BD of the trapezium are not Draw Be/ parallel to AC, and meeting AC ST in as well the right line in d. Join Cd cutting /, PQ as the conic section in r, and draw DM AB in N. cutting Cd in M, and (because of the similar triangles BTt, parallel to PQ, N Q PQ is to Tt as DN to NB. And ^^ the to AQ or PS as DM to AN. antece- Wherefore, by multiplying DBN), Et so I 2. site sides Then [BOOK Rr is or dents by the antecedents, and the consequents by the consequents, as the X Rr is to the rectangle PS X Tt, so will the rectangle rectangle and (by Case 1) so is the rectangle be i)M to the N rectangle PQ ANB PQ PQ X PR X Pr CASE ; PS X Pt and by division, Q.E.D. rectangle PS X PT. to the rectangle to the

so is the rectangle Let us suppose, lastly, the four lines ?Q, PR, PS, PT, not to be parallel to the sides AC, AB, but any way inclined to them. In their and Ps, Pt place draw Pq, Pr, parallel to AC 3.

AB parallel to triangles tios of PQ</, IQ and because the angles of the ; PRr, PSs, PTt are given, the to Pq, PR PS to Pr, to P*, PT to ra- Pt will b? also given; and therefore the compound to P? X Pr, and PS X X ed ratios given. PT PR Pk to Ps X Pt are before, the ratio of Pq X Pi to PS X and therefore also the ratio of X But from what we have demonstrated Ps X Pt is given PT. to ; PR PQ