An elementary treatise on geometry : simplified for beginners not versed in algebra . rom a given point without a straightline, to let fall a perpendicular upon that line. Solution. Let A be thegiven point, from whicha perpendicular is to bedrawn to the line MN. 1. With any radius suf-ficiently great describe anarc of a circle. 2. From the two points B and C, where this arc cutsthe line MN, draw the straight lines BA, CA. 3. Bisect the angle BAG (see the last Problem), theline AD is perpendicular to the line MN. Demon. The two triangles ABD, ACD, have two sides, AB,AD, in the one, equal to two
An elementary treatise on geometry : simplified for beginners not versed in algebra . rom a given point without a straightline, to let fall a perpendicular upon that line. Solution. Let A be thegiven point, from whicha perpendicular is to bedrawn to the line MN. 1. With any radius suf-ficiently great describe anarc of a circle. 2. From the two points B and C, where this arc cutsthe line MN, draw the straight lines BA, CA. 3. Bisect the angle BAG (see the last Problem), theline AD is perpendicular to the line MN. Demon. The two triangles ABD, ACD, have two sides, AB,AD, in the one, equal to two sides, AC, AD, in the other, each toeach (AC, AB, being radii of the same circle, and the side ADbeing common to both) \ and have the angles included by thesesides also equal (because the angle BAC is bisected); thereforethese two triangles are equal to one another (Query 1, Sect. II.);and the angle y, opposite to the side AB, in the triangle ABD, isequal to the angle x, opposite to the equal side AC, in the triangleACD. Now, as the two adjacent angles x and y, which the straight. GEOMETRY. 145 line AD makes with the straight line MN, arc equal to each other,the line AD must be perpendicular to MN. (Def. of lines.) Problem V. To bisect a given straight Let AB be thegiven straight line. 1. From A, with a radiusgreater than half of AB, de-scribe an arc of a circle; and *from B, with the same radius,another, cutting the first inthe point C. 2. From the point C draw the perpendicular CM, and the line AB is bisected in M. Demon. The two right-angled triangles AMC, BMC, areequal, because the hypothenuse AC and the side CM in the one,are equal to the hypothenuse and the side CM in the other(page 47) ; and therefore the third side AM in the one, is alsoequal to the third side BM in the other; cojasequently the lineAB is bisected in the point M.
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