An elementary treatise on geometry : simplified for beginners not versed in algebra . -M- 1/ ^E 2d Case. Let ABCDEF be the given figure; let thevertex of the required triangle be situated in the point Mwithin the figure, and let the base fall upon AF. 1. From M to any angle of the figure, say D, draw theline MD, and draw the lines MA, MF, by which meansthe figure ABCDEF is divided into the triangle MAF,and the figures MDCBA, MDEF. 2. Then transform MDCBA and MDEF into the tri-angles McA, McF, whose bases are in the continuationof AF; the triangle cMe is equal to the figure ABCDEF. The demonstr
An elementary treatise on geometry : simplified for beginners not versed in algebra . -M- 1/ ^E 2d Case. Let ABCDEF be the given figure; let thevertex of the required triangle be situated in the point Mwithin the figure, and let the base fall upon AF. 1. From M to any angle of the figure, say D, draw theline MD, and draw the lines MA, MF, by which meansthe figure ABCDEF is divided into the triangle MAF,and the figures MDCBA, MDEF. 2. Then transform MDCBA and MDEF into the tri-angles McA, McF, whose bases are in the continuationof AF; the triangle cMe is equal to the figure ABCDEF. The demonstration follows from those of the last three problems. Problem XXVII. To transform a given rectangleinto a square of equal Solution. Let ABCD be the given rectangle. 1. Extend the greater side, AB, of the rectangle, makingBM equal to BD. 2. Bisect AM in O, and, from the point O as a centre,with a radius AO, equal to OM, describe a semicircle. GEOMETRY. 165 3. Extend the side BD of the rectangle, until it meetsthe circle in E. 4. Upon BE construct the square BEFG, which is thesquare sought. Demon. The perpendicular BE is a mean proportional be-tween AB and BM (see Problem XVIII.); therefore we have the proportion AB : BE = BE : BM ; and as, in every geometrical proportion, the product of the means equals that of the extremes (Theory of Prop., Principle 10, page 65), we have the product of the side BE multiphed by itself, equal to the product of the side AB of the parallelogram, multiphed by the adjacent side BD (or BM). But the first of these products is the area of the square BEFG, and the other is the area of the rectangle ABCD ; therefore these two figures are, in area, equal to one another. Problem XXV
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