Page:The Mathematical Principles of Natural Philosophy - 1729 - Volume 1.djvu/201

 well as aE cutting the third circle EMF in c. Compleat the figure ABCdef imilar and equal to the figure absDEF. I ay the thing is done.

For drawing Fc meeting aD in n, and joining aG, bG, QG, QD, PD; by contruction 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 conequently 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 imilar, and Ga is to ab as GP to PQ; that is (by contruction) as Ga to AB. Wherefore ab and AB are equal; and conequently the triangles abc, ABC, which we have now proved to be imilar, are alo equal. And therefore ince the angles D, E, F, of the triangle DEF do repectively touch the ides ab, ac, be of the triangle abc, the figure ABCdef may be compleated imilar and equal to the figure abcDEF, and by compleating it the problem will be olved. Q. E. F.

Hence a right line may be drawn whoe parts given in length may be intercepted between three right lines given by poition. Suppoe the triangle DEF, by the acces of its point D to the ide EF, and by having the ides DE, DF placed in directum to be changed into a right line