For let the two first circles FSG, FTH cut one the other in K; join PK, QK, RK, aK, bK, cK, and produce QP to L. The angles FaK, FbK, FcK at the circumferences are the halves of the angles FPK, FQK, FRK, at the centres, and therefore equal to 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, fCi, are equal to the angles FaG, FbH, FcI. And therefore the figure ABCfghi may be completed similar to 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 parts intercepted in a given order, between four right lines given by position, shall have a given proportion among themselves. Let the angles FGH, GHI, be so far increased that the right lines FG, GH, HI, may lie in directum; and by constructing the Problem in this case, a right line fghi will be drawn, whose parts fg, gh, hi, intercepted between the four right lines given by position, AB and AD, AD and BD, BD and CE, will be one to another as the lines FG, GH, HI, and will observe the same order among themselves. But the same thing may be more readily done in this manner. Produce AB to K and BD to L, so as BK may be to AB as HI to GH; and DL to BD as GI to FG; and join KL meeting the right line CE in i. Produce iL to M, so as LM may be to iL as GH to HI; then draw MQ parallel to LB, and meeting the right line AD in g, and join gi cutting AB, BD in f, h; I say, the thing is done. For let Mg cut the right line AB in Q, and AD the right line KL in[Pg 152] S, and draw AP parallel to BD, and meeting iL in P, and gM to Lh (gi to hi, Mi to Li, GI to HI, AK to BK) and AP to BL, will be in the same ratio. Cut DL in R, so as DL to RL may be in that same ratio; and because gS to gM, AS to AP, and DS to DL are proportional; therefore (ex æquo) as gS to Lh, so will AS be to BL, and DS to RL; and mixtly, BL - RL to Lh - BL, as AS - DS to gS - AS. That is, BR is to Bh as AD is to Ag, and therefore as BD to gQ. And alternately BR is to BD as Bh to gQ, or as fh to fg. But by construction the line BL was cut in D and R in the same ratio as the line FI in G and H; and therefore BR is to BD as FH to FG. Wherefore fh is to fg as FH to FG. Since, therefore, gi to hi likewise is as Mi to Li, that is, as GI to HI, it is manifest that the lines FI, fi, are similarly cut in G and H, g and h. Q.E.F. In the construction of this Corollary, after the line LK is drawn cutting CE in i, we may produce iE to V, so as EV may be to Ei as FH to HI, and then draw Vf parallel to BD. It will come to the same, if about the centre i with an interval IH, we describe a circle cutting BD in X, and produce iX to Y so as iY may be equal to IF, and then draw Yf parallel to BD.