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Steps of the Demonstration. 1. Prove that the opposite sides of the fig. are equal,
that the 4 sides = each other, 3. that ZS BAD + ADE = 2 right ZS,
that the fig. is rectangular (and :. a square.)
In the construction of the figure it is necessary to show that ga is on the same right line with Ac; and that AB is in the same right line with ah.
Steps of the Demonstration.
that A ABD = A CBF,
be proved that square AK = Om cl, that bc = BAS + Aco.
* It is sufficient to suppose these lines drawn without actually drawing them, (as in Simpson,) which only makes the figure unnecessarily complicated.
PROPOSITION XLVIII. Theorem. If the square described on one side of a triangle be equal to the squares described upon the other two; the angle contained by these two sides is a right angle.
BEFORE entering on the Second Book, the learner should have a clear understanding of the reason why AB X BC is used to denote a rectangle of which AB and BC are two adjacent sides. Now since every rectangle is a parallelogram, and that the opposite sides of parallelograms are equal, it is plain that if we have two adjacent sides given, we have enough to determine the dimensions of a rectangle; the only question, therefore, is, why these two sides should be connected by the algebraical sign of multiplication, which sign can be properly applied to numbers alone.
The answer is that the sign (x) has, in reality, a reference to number in this case, viz., to the number of inches, (or other units of length) in the lines AB, BC. For instance, if Al be 5 inches long, and BC 3 inches, and if on each of the small inch-lines into which AB is divided a square (as Aa) be described, it is evident that since bc is 3 inches long, the whole figure may be divided into 3 rows, each containing 5 square inches; in other words, the area of the figure is equal to 3 times 5 square inches, or is equal to (the number of square inches on AB) x (the number of inches in bc), which, for the sake of brevity, is expressed by saying that the area of the rectangle = AB X BC.
Most of the Demonstrations in this book are very simple, being, in fact, little more than showing that the figure drawn is made up of the rectangles mentioned in the Enunciation. We have, therefore, in most cases, given only two or three steps, namely, those which appeared most likely to direct the understanding and assist the memory. This last point, viz., the aiding the memory, is the chief object in this Second Book, for the Propositions in it, though easy to be understood, are not so easily remembered by the beginner, as those of the other three Books, and he will find it an important assistance carefully to commit to memory the enunciations of all the theorems he can prove; as he will thus have them ready to apply to the demonstration of the subsequent Propositions.
PROPOSITION I. Theorem. If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the G! rectangles contained by the undivided line, and the several parts F of the divided line.
Proved by showing that a X BC* = figures bk + DL + Eh; and that these are = a X BD + A X DE + A X EC.
PROPOSITION II. Theorem. If a straight line be divided A Ç B into any two parts, the rectangles contained by the whole and each of the parts are together equal to the square of the whole line.
Proved by showing that ab% = figures AF + CE; and that these are = AB X AC + AB X BC.
B parts, the rectangle contained by
the whole and one of the parts is equal to the rectangle contained by the two parts, together with
the square of the aforesaid part. FD
Proved by showing that AB X BC = figures AD + CE, and that these are = AC X CB + CB*.
PROPOSITION IV. Theorem. If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts together with twice the rectangle contained by the parts.
* By the rectangle a X BC (the lines A and bc being separate) is meant the rectangle which may be formed with A and BC for its adjacent sides.