This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1885 Excerpt: ... axis CF of the point P, and the O B. From C, where CE, CF intersect, draw tangents CD, CG to A and B. Join CP. C is the centre of the required circle. Dem Since CE is the radical axis of the O" A, B, CG= CD (Lemma); and because CF is the radical axis of the O B and the point P, CG = CP;.-. CG, CD, CP are equal, and therefore the O, whose centre is C, and radius CP, will cut the O' A and B orthogonally ("Sequel," Book III., Prop. xxi.). (3) Let A, B, C be the O'. Find DE, the radical axis of A and B, and DF the radical axis of A and C. From D, where DE, DF intersect, draw tangents to A, B, C. Now these tangents are equal; and the O, with D as centre, and one of them as distance, will pass through the ends of the other two, and will cut the O' A, B, C orthogonally (" Sequel," Book III., Prop. xxi.). equal to two right L," and BMC, DMC together equal two right /";.-. DMC = BAC, and is an L of an equilateral A; and because MC = MD, MCD is an equilateral A. Again, because BMCA is a cyclic quadrilateral, the L MBC = MAC, and ABC = AMC; but ABC = MDC, since each is an L of an equilateral A;.-. AMC = MDC; hence (I. xxvi.) the A" AMC, BDC are equal;.-. AM = BD; that is, AM = MB + MC. 46. (1) Let ABC be a A, the sum of whose sides AB, AC is given, and the L BAC, both in magnitude and position. About the A ABC describe a O. It is required to prove that the locus of its centre F is a right line. Dem.--Bisect the arc BC in D. Join AD. Let fall a 1 DE on AB. From F let fall a 1 FG on AD, and produce it both ways to meet the circumference. Now AE = (AB + AC) (xxx., Ex. 4); hence AE is a given line;.-. E is a given point. And since DE is 1 to a given line AE, at a given point, DE is given in position; and because the L BAD = BAC, BA...