. The New York coach-maker's magazine. rts, although designed to carry but one man andthe driver, are as large as those of our strongest drays inthe United States, and the wheels as strong and full of riv-ets as the wheels inEzekiels vision were of eyes. Throughthese ponderous hubs the axles project for a distance of GEOMETRICAL EXERCISE. BY P. B. J. {Continued from Page 181, Volume Eleven.)As before promised, I here give the method to bisecta given line, or divide it into two equal parts. Let thisgiven line be A B, which it is required to bisect or divideinto equal parts. From A,with any spre
. The New York coach-maker's magazine. rts, although designed to carry but one man andthe driver, are as large as those of our strongest drays inthe United States, and the wheels as strong and full of riv-ets as the wheels inEzekiels vision were of eyes. Throughthese ponderous hubs the axles project for a distance of GEOMETRICAL EXERCISE. BY P. B. J. {Continued from Page 181, Volume Eleven.)As before promised, I here give the method to bisecta given line, or divide it into two equal parts. Let thisgiven line be A B, which it is required to bisect or divideinto equal parts. From A,with any spread of the compassesgreater than the half of A B, describe the portion of acircle as C F D, then by the same operation, making B thecenter, describe the arc C G D,—cutting the former arcin C and D. Join the points C and D, by the line C ED ; then is A E equal to E B, and the line A B bisectedas required; for joining AC, CD, B D, and B A weshall have a parallelogram whose sides are all equal t0 1S70. THE NEW YORK COACH-MAKERS MAGAZINE. each other, thereby forming a complete square withoutC the aid of that tool, as shown. The ra-dius with which thearcs are described,and which are A Band C D, bisect eachother, and thereforeA E is equal to EB. Moreover C Dis perpendicular orsquare to A B, forthe triangles ACEare DCE are iden-tical, or have theirangles as well astheir sides, equal tothe other; that is, A E is equal to B E, and A C equal toB C, and C E, having a common side to both triangles;therefore the angle A E C is equal to B E C. The angleA E C, added to the angle B E C, is equal to 180 degrees,or two right angles. As these angles are proved equalto each other, they must be equal to ninety or the halfof one hundred and eighty degrees, as C E is perpendic-ular to A B, and the line C E continued to 6, D E isalso perpendicular or square to A B. Hence, also, we haveanother method of drawing a square line from any givenpoint, for if E is made the given point, we need onlyset off E
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