A Key to the Exercises in the First Six Books of Casey's Elements of Euclid (Paperback)


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...

R362

Or split into 4x interest-free payments of 25% on orders over R50
Learn more

Discovery Miles3620
Delivery AdviceOut of stock

Toggle WishListAdd to wish list
Review this Item

Product Description

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...

Customer Reviews

No reviews or ratings yet - be the first to create one!

Product Details

General

Imprint

Rarebooksclub.com

Country of origin

United States

Release date

March 2012

Availability

Supplier out of stock. If you add this item to your wish list we will let you know when it becomes available.

First published

March 2012

Authors

Dimensions

246 x 189 x 1mm (L x W x T)

Format

Paperback - Trade

Pages

26

ISBN-13

978-1-130-78589-0

Barcode

9781130785890

Categories

LSN

1-130-78589-0



Trending On Loot