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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1864 Excerpt: ... equal; if the parallelogram be a square or a rhombus, the diagonals bisect each other at right angles. The converse of this Prop., namely, "If the opposite sides or opposite angles of a quadrilateral figure be equal, the opposite sides shall also be parallel; that is, the figure shall be a parallelogram," is not proved by Euclid. Prop. xxxv. The latter part of the demonstration is not expressed very intelligibly. Simson, who altered the demonstration, seems in fact to consider two trapeziums of the same form and magnitude, and from one of them, to take the triangle ABE; and from the other, the triangle DCF; and then the remainders are equal by the third axiom: that is, the parallelogram A BCD is equal to the parallelogram EBCF. Otherwise, the triangle, whose base is l)E, (fig. 2.) is taken twice from the trapezium, which would appear to bo impossible, if the sense in which Euclid applies the third axiom, is to be retained here. It may be observed, that the two parallelograms exhibited in fig. 2 partially lie on one another, and that the triangle whose base is BC is a common part of them, but that the triangle whose base is DE is entirely without both the parallelograms. After having proved the triangle ABE equal to the triangle DCF, if we take from these equals (fig. 2.) the triangle whose base is DE, and to each of the remainders add the triangle whose base is BC, then the parallelogram ABCD is equal to the parallelogram EBCF. In fig. 3, the equality of "the parallelograms ABCD, EBCF, is shewn by adding the figure EBCD to each of the triangles ABE, DCF. In this proposition, the word equal assumes a new meaning, and is no longer restricted to mean coincidence in all the parts of two figures. Prop, Xxxvin. In this proposition, it is to be und...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1864 Excerpt: ... equal; if the parallelogram be a square or a rhombus, the diagonals bisect each other at right angles. The converse of this Prop., namely, "If the opposite sides or opposite angles of a quadrilateral figure be equal, the opposite sides shall also be parallel; that is, the figure shall be a parallelogram," is not proved by Euclid. Prop. xxxv. The latter part of the demonstration is not expressed very intelligibly. Simson, who altered the demonstration, seems in fact to consider two trapeziums of the same form and magnitude, and from one of them, to take the triangle ABE; and from the other, the triangle DCF; and then the remainders are equal by the third axiom: that is, the parallelogram A BCD is equal to the parallelogram EBCF. Otherwise, the triangle, whose base is l)E, (fig. 2.) is taken twice from the trapezium, which would appear to bo impossible, if the sense in which Euclid applies the third axiom, is to be retained here. It may be observed, that the two parallelograms exhibited in fig. 2 partially lie on one another, and that the triangle whose base is BC is a common part of them, but that the triangle whose base is DE is entirely without both the parallelograms. After having proved the triangle ABE equal to the triangle DCF, if we take from these equals (fig. 2.) the triangle whose base is DE, and to each of the remainders add the triangle whose base is BC, then the parallelogram ABCD is equal to the parallelogram EBCF. In fig. 3, the equality of "the parallelograms ABCD, EBCF, is shewn by adding the figure EBCD to each of the triangles ABE, DCF. In this proposition, the word equal assumes a new meaning, and is no longer restricted to mean coincidence in all the parts of two figures. Prop, Xxxvin. In this proposition, it is to be und...
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