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The Solution of Geometrical Exercises, Explained and Illustrated

The Solution of Geometrical Exercises, Explained and Illustrated


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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1879 edition. Excerpt: ...from AB. Hence we have thisresult. The locus of points equidistant from two given points is the perpendicular which bisects the line joining the given points. Prop. XII.--Let AB, AC be the lines, and P the given point. Bisect the Z" BAC by the line AQ, and from P drop the perpendicular PO upon AQ, meeting AC in C and AB in B. PCB is the line required. For if the triangle AOO be turned over about the line AO as an axis, then CO will take the direction OB, because the Ze CO A = the Ze AOB. Also, AC will fall upon.4.8, because the Ze CAO = the Ze 5.40. Therefore, since CO takes the direction 05, and CA the direction.42, they must intersect at B. Hence, C will fall on B, and the angle ACO will coincide with ABO, and be equal to it. Therefore, PBC is equally inclined to AB and AC. Prop. XIII.--For the Le GBA is half the Ze ABD, and Ze ABE is half the Ze ABC; therefore CBH is half the two angles ABD and ABC, i.e., the half of two right angles. Hence GBH is a right angle. Prop. XIV.--Let ABOH, CBOD be equal squares, so placed as to have one side, OB, common. Then AB, BC will be in the same straight line, and also HO, OD. 1.14. If now a third square HOFC, equal to AO, be placed so as to have one side coinciding with HO, then AH, HG will be in the same straight line, and also BO, OF will be in the same straight line. I. 14. Lastly, by placing a fourth square ODEF so as to have one side coinciding with OF, the contiguous side will then coincide with OD, because FOD is a right Ze (I. 13), and DE, CD will be in one straight line, as also GF, FE. 1.14. Then the fig. ACEG has all its angles right angles, and each of its sides is equal to twice the side of one of the given squares; therefore, AE is a square. Prop. XV.--(1.) For the opposite angle...


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Product Details
  • ISBN-13: 9781234368029
  • Publisher: Rarebooksclub.com
  • Publisher Imprint: Rarebooksclub.com
  • Height: 246 mm
  • No of Pages: 24
  • Spine Width: 1 mm
  • Width: 189 mm
  • ISBN-10: 1234368021
  • Publisher Date: 22 Jan 2013
  • Binding: Paperback
  • Language: English
  • Returnable: N
  • Weight: 64 gr


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