. Mathematical recreations and essays. Mathematical recreations; Geometry; Bees; Cryptography; Ciphers; String figures; Magic squares. 50 GEOMETRICAL RECREATIONS [CH. Ill OG common, hence (by Euc. I. 8) the angle OGB is equal to the angle OGH. Hence the angle BCD is equal to the angle BGH, that is, 7J-/4 is equal to tt/3, which is absurd. Sixth Fallacy*- To prove that, if two opposite sides of a quadrilateral are equal, the other two sides must be parallel. Let ABGB be a quadrilateral such that AB is equal to BG. Bisect AB in M, and through M draw MO at right angles to AB. Bisect BG in N, and
. Mathematical recreations and essays. Mathematical recreations; Geometry; Bees; Cryptography; Ciphers; String figures; Magic squares. 50 GEOMETRICAL RECREATIONS [CH. Ill OG common, hence (by Euc. I. 8) the angle OGB is equal to the angle OGH. Hence the angle BCD is equal to the angle BGH, that is, 7J-/4 is equal to tt/3, which is absurd. Sixth Fallacy*- To prove that, if two opposite sides of a quadrilateral are equal, the other two sides must be parallel. Let ABGB be a quadrilateral such that AB is equal to BG. Bisect AB in M, and through M draw MO at right angles to AB. Bisect BG in N, and draw NO at right angles to BG. If MO and NO are parallel, then AB and BG (which are at right angles to them) are also parallel. If MO and NO are not parallel, let them meet in 0; then 0 must be either inside the quadrilateral as in the left-hand O. \ B N C diagram or outside the quadrilateral as in the right-hand diagram. Join OA, OB, OG, OB. Since OM bisects AB and is perpendicular to it, we have OA = OB, and the angle 0AM equal to the angle OBM. Similarly OB = OG, and the angle OBN is equal to the angle OGN. Also by hypothesis AB = BG, hence, by Euc. I. 8, the triangles OAB and OBG are equal in all respects, and therefore the angle AOB is equal to the angle BOG. Hence in the left-hand diagram the sum of the angles AOM, AOB is equal to the sum of the angles BOM, BOG; and in the right-hand diagram the difference of the angles AOM, AOB is equal to the difference of the angles BOM, BOG; and therefore in both cases the angle MOB is equal to the angle MOG, OM (or OM produced) bisects the angle BOG. But the angle NOB is equal to the angle NOG, ON bisects the angle BOG; hence OM and ON coincide in direction. * Mathesit, October, 1893, series 2, vol. m, p. Please note that these images are extracted from scanned page images that may have been digitally enhanced for readability - coloration and appearance of these illustrations may not perfectly resemble the original wo
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Keywords: ., bookcentury1900, bookdecade1920, booksubjectgeometry, bookyear192
