as the letters ACEA. Complete the ſegments into entire circles, and let P be the centre of the firſt circle FSG, Q the centre of the ſecond FTH. Join and produce both ways the line PQ, and in it take QR in the ſame ratio to PQ as BC has to AB. But QR is to be taken towards that ſide of the point Q, that the order of the letters P, Q, R may be the ſame, as of the letters A, B, C; and about the centre R with interval RF deſcribe a fourth circle FNc cutting third circle FVI in c. Join Fc cutting the firſt circle in a and the ſecond in b. Draw aG, bH cI, and let the figure abcFGHI be made ſimilar to the figure abcFGHI; and the trpezium fghi will be taht which was requited to be deſcribed.
For let the two firſt circles FSG, FTH cut one the other in K; join PK, QK, EK, aK, bK, cK, and and produce QP, to L. The angles FaK, FbK, FcK at the circumferences, are the halves of the angles LPK, LQK, LRK, the halves of thoſe angles. Wherefore the figure PQRK is equiangular and ſimilar to the figure abcK, and conſequently ab is to bc as PQ to QR, that is, as AB to BC. But by conſtruction, the angles fAg, fBh, fVi are equal to the angles FaG, FbH, FcI. And therefore the figure abcFGHI. Which done, a trapezium fghi will be conſtructed ſimilar to the trapezium FGHI, and which by its angles, f, g, h, i will touch the right lines ABC, AD, BD, CE. Q. E. F.
Cor. Hence a right line may be drawn whoſe part intercepted in a given order, between four right lines given by poſition, ſhall have a given
