sp as sh is to sq, and shall make the angle vsp equal to the angle hsq, and the angle vsh equal to the angle psq, the triangles svh, spq, will be similar, and therefore vh will be to pq as sh is to sq; that is (because of the similar triangles VSP, hsq), as VS is to SP, or as ab to pq. Wherefore vh and ab are equal. But, because of the similar triangles VSH, vsh, VH is to SH as vh to sh; that is, the axis of the conic section now described is to the distance of its foci as the axis ab to the distance of the foci sh; and therefore the figure now described is similar to the figure aph. But, because the triangle PSH is similar to the triangle psh, this figure passes through the point P; and because VH is equal to its axis, and VS is perpendicularly bisected by the right line TR, the said figure touches the right line TR. Q.E.F.
LEMMA XVI.
- From three given points to draw to a fourth point that is not given three right lines whose differences shall be either given, or none at all.
Case 1. Let the given points be A, B, C, and Z the fourth point which we are to find; because of the given difference of the lines AZ, BZ, the locus of the point Z will be an hyperbola whose foci are A and B, and whose principal axis is the given difference. Let that axis be MN. Taking PM to MA as MN is to AB, erect PR perpendicular to AB, and let fall ZR perpendicular to PR; then from the nature of the hyperbola, ZR will be to AZ as MN is to AB. And by the like argument, the locus of the point Z will be another hyperbola, whose foci are A, C, and whose principal axis is the difference between AZ and CZ; and QS a perpendicular on AC may be drawn, to which (QS) if from any point Z of this hyperbola a perpendicular ZS is let fall (this ZS), shall be to AZ as the difference between AZ and CZ is to AC. Wherefore the ratios 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 ratios 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.
