a perpendicular ZS is let fall, this (ZS) ſhall be to AZ as the difference between AZ and CZ is to AC. Wherefore the ratio's of ZR and ZS to AZ are given, and conſequently the ratio of ZR to ZS one to the other; and therefore if the right lines RP SQ meet in T, and TZ and TA are drawn, the figure TRZS will be given in ſpecie, and the right line TZ, in which the point Z is ſomewhere placed, will be given in poſition. There will be given alſo the right line TA, and the angle ATZ; and becauſe the ratio's of AZ and TZ to ZS are given, their ratio to each other is given alſo; and thence will be given likewiſe the triangle ATZ whoſe vertex is the point Z. Q. E. I.
Case 2. If two of the three lines, for example AZ and BZ, are equal, draw the right line TZ ſo as to biſect the right line AB; then find the triangle ATZ as above. Q. E. I.
Case 3. If all the three are equal, the point Z will be placed in the centre of a circle that paſſes thro' the points A, B, C. Q. E. I.
This problematic lemma is likewiſe ſolved in Apollonui's Book of Tactions reſtored by Victa.
Proposition XXI. Problem XIII.
About a give focus to deſcribe a trajectory, that ſhall paſs through given points and touch right lines given by poſition.
Let the focus S, (Pl. 8. Fig. 2.) the point P, and the tangent TR be given, ſuppoſe that the other focus H is to be found. On the tangent let fall the perpendicular ST, which produce to T, ſo that TY
