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The Elements of Euclid, the parts read in the University of Cambridge [book 1-6 and parts of book 11,12] with geometrical problems, by J.W. Colenso - Softcover
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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. 1846 Excerpt: ... of the sides AB, AD in the points F, E, and through E draw EH parallel to AB or DC, and through F draw FK parallel to AD or BC. Then each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram; and thenopposite sides are equal: And because AD is equal to AB, and that AE is the half of AD, and AF the half of AB, _____ therefore AE is equal to AF, and there-B H fore also the sides opposite to these are equal, viz. FG to GE: In like manner it may be shewn that GH, GK are each of them equal to GF or GE: Therefore the four straight lines GE, GF, GH, GK are equal to one another, and the circle described from the centre G, at the distance of any one of them, shall pass through the extremities of the other three, and touch the straight lines AB, BC, CD, DA, because the angles at the points, E, F, H, K, are right angles (3. 16): Therefore each of the straight lines AB, BC, CD, DA touches the circle, which is therefore inscribed in the square ABCD. O..E.F. PROP. IX. Prob. To describe a circle about a given square. Let ABCD be the given square: it is required to describe a circle about it. Join AC, BD, cutting one another in E: Then, because AB is equal to AD, and AC common to the two triangles BAC, DAC, the two sides BA, AC are equal to the two DA, AC, each to each, and the base BC is equal to the base DC--therefore the angle BAC is equal to the angle DAC, and the angle BAD is bisected by the straight line AC: In like manner, it may be shewn that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC: Therefore, because the angle DAB is equal to the angle ABC, and that the angle EAB is the half of DAB, and EBA the half of ABC, therefore the angle EAB is equal to the angle EBA, and the side EA to the side EB: And, in like manner, it ma...
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- PublisherRarebooksclub.com
- Publication date2012
- ISBN 10 1130818853
- ISBN 13 9781130818857
- BindingPaperback
- Number of pages66
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