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Geometrical Analysis, and Geometry of Curve Lines; Being Volume Second of a Course of Mathematics, and Designed as an Introduction to the Study of Nat (Paperback) Loot Price: R405
Discovery Miles 4 050
Geometrical Analysis, and Geometry of Curve Lines; Being Volume Second of a Course of Mathematics, and Designed as an...
Geometrical Analysis, and Geometry of Curve Lines; Being Volume Second of a Course of Mathematics, and Designed as an...

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Geometrical Analysis, and Geometry of Curve Lines; Being Volume Second of a Course of Mathematics, and Designed as an Introduction to the Study of Nat (Paperback)

John Leslie

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Loot Price R405 Discovery Miles 4 050

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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. 1821. Excerpt: ... PROP. VII. PROB. To bisect a given triangle, by a straight line drawn from a given point in one of its sides. Let it be required, from the point D, to draw DF, bisecting the triangle ABC. ANALYSIS. If D be the middle of AC, the line DB drawn to the vertex will obviously (II. 2. El.) divide the triangle into two equal portions. But, if not, bisect (I. 7. EI.) the side AC in E, and join EB, EF and BD. The triangle ABE is equal to EBC, and is consequently the half of ABC, wherefore ABE is equal to AFD, and, taking AFE from both, the remaining triangle EFB is equal to EFD; and since these triangles stand on the same base, they must (U.S. El.) have the same altitude, or EF is parallel to BD. But the points B and v D C J) being given, the straight line BD is given in position, and consequently EF is also given in position. COMPOSITION. Having bisected AC in E and joined BD, draw EF parallel to it, meeting AB in F; the straight line DF divides the triangle ABC into two equal portions. For join BE. Because BD is parallel to EF, the triangle EFB (II. 1. El.) is equal to EFD; and, adding AFE to each, the triangle AFD is equal to ABE, that is, to the half of the triangle ABC. PROP. VIII. PItOB. To find a point within a given triangle, from which straight lines drawn to the several corners will divide the triangle into three equal portions. Let F be the required point, from which the lines FA, FB, and FC trisect the triangle ABC. ANALYSIS. Draw FD, FE parallel to the sides BA, BC, and join BD, BE. Since FD is parallel to AB, the triangle ABF (II. 1. El.) is equal to ABD, which is hence the third part of ABC; and, for the same reason, the triangle BFC is equal to BEC, which is also the third part of ABC. Wherefore the bases AD and EC are each the third part of ...

General

Imprint: General Books LLC
Country of origin: United States
Release date: February 2012
First published: February 2012
Authors: John Leslie
Dimensions: 246 x 189 x 5mm (L x W x T)
Format: Paperback - Trade
Pages: 96
ISBN-13: 978-1-150-44436-4
Barcode: 9781150444364
Categories: Books
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LSN: 1-150-44436-3

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