Ball, W. W. Rouse (Walter William Rouse), 1850-1925 · London, Macmillan · leaf 68
itude; hence, byEuc. VI. 1, l\ABG. AABD = BG:BD, .-. AG:AD=BG : BD. . AG^ _AD•• BG ~ BD • * See a note by M, Ooccoz in UIllustration, Paris, Jan. 12, 1895. CH. Ill] GEOMETRICAL RECREATIONS 47 Hence, by Enc. ii. 13, AB + BG^-2BG.be^ AB^BD-2BD . BEBG ~ BD .\^+BG-2BE=^~ + BD- 2BE,. ^^ BD-^^ EC • ^^^-… … Let ABG be a triangle. Complete the parallelogram ofwhich AB and BG are sides. Divide AB into n + 1 equalparts, and through the points so determined draw n linesparallel to BG. Similarly, divide BG into n + 1 equal parts,and through the points so determined draw n lines parallel toAB. The…
Geometry · Cryptography · Ciphers
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