well as aE cutting the third circle EMF in c. Compleat the figure ABCdef ſimilar and equal to the figure absDEF. I ſay the thing is done.
For drawing Fc meeting aD in n, and joining aG, bG, QG, QD, PD; by conſtruction the angle EaD is equal to the angle CAB, and the angle acF equal to the angle ACB; and therefore the triangle anc equiangular to the triangle ABC. Wherefore the angle anc or FnD is equal to the angle ABC, and conſequently to the angle FbD; and therefore the point n falls on the point b. Moreover the angle GPQ which is half the angle GPD at the centre is equal to the angle GaD at the circumference; and the angle GQP, which is half the angle GQD at the centre, is equal to the complement to two right angles of the angle GbD at the circumference, and therefore equal to the angle Gab. Upon which account the triangles GPQ, Gab, are ſimilar, and Ga is to ab as GP to PQ; that is (by conſtruction) as Ga to AB. Wherefore ab and AB are equal; and conſequently the triangles abc, ABC, which we have now proved to be ſimilar, are alſo equal. And therefore ſince the angles D, E, F, of the triangle DEF do reſpectively touch the ſides ab, ac, be of the triangle abc, the figure ABCdef may be compleated ſimilar and equal to the figure abcDEF, and by compleating it the problem will be ſolved. Q. E. F.
Cor. Hence a right line may be drawn whoſe parts given in length may be intercepted between three right lines given by poſition. Suppoſe the triangle DEF, by the acceſs of its point D to the ſide EF, and by having the ſides DE, DF placed in directum to be changed into a right line
