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to the base FC, and the triangle ABD to the triangle FBC. [I. 4.]

Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and

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between the same parallels BD, AL. [I. 41.] And the square GB is double of the triangle FBC, because they are on the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal to one another. [Ax. 6.] Therefore the parallelogram BL is equal to the square GB.

In the same manner, by joining AE, BK, it can be demonstrated, that the parallelogram CL is equal to the square HC. Therefore the whole square BDEC is equal to the two squares GB, HC. (Ax. 2.] And the square BDEC is described on BC, and the squares GB, HC on BA, AC.

Therefore the square described on the side BC is equal to the squares described on the side BA, AC'. Therefore, in any right-angled triangle, &c.

Prop. 47. The following proof of this prop. is ascribed to the Astronomer Royal.

Let BC, BD be any two squares having their bases in the

Q.E.D.

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А same straight line. Take Co=AB. Join GO, OD. Make FH-ED the side of the greater square. Then the four triangles AGO, ODC, DEH, GFH, may all be proved equal to each other by Euclid Í. 4. Therefore the fig. GODH is equal to the two squares AF and EC, and it is equilateral. Also by I. 32, it may be shewn to be rectangular. Therefore it is a square, and it is the square described on the hypotenuse of a right-angled triangle, whose sides are DC and OC=AB. Therefore the square on the hypotenuse of a right-angled triangle, &c. Ex. 1. Make a square which shall be equal to the sum of two

given squares. 2. If two right-angled triangles have the hypotenuse and one

side equal in each triangle, the triangles are equal in all

respects. 3. The hypotenuses of three right-angled isosceles triangles are

80 placed as to form a right-angled triangle. Construct the figure and shew that the area of the triangle on the hypotenuse is equal to the area of the other two.

PROPOSITION 48. THEOREM. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right anyle.

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Let the square described on BC, one of the

sides of the triangle ABC, be equal to the squares described on the other sides, , AC: the angle BAC shall be a right angle.

From the point A draw 2.

AD at right angles to AC;[I. 11.) and make AD equal to BA; [I. 3.) and join DC.

Then because DA is

equal to BA, the square on DA is equal to the square on BA. To each of these add the square on AC. Therefore the squares on DA, AC are equal to the squares on BA, B AC. [Ax. 2.] But because the angle DAC is a right angle, [Const.) the square on DC is equal to the squares on DA, AC. (1. 47.] And, by hypothesis, the square on BC is equal to the squares on BA, AC. Therefore the square on DC is equal to the square on BC. [Ax. 1.] Therefore also the side DC is equal to the side BC.

And because the side DA is equal to the side AB; [Const.] and the side AC is common to the two triangles DAC, BAC; the two sides DA, AC are equal to the two sides BA, AC, each to each; and the base DC has been proved equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8.) But DAC is a right angle ; [Const.] therefore also BAC is a right angle. [Ax. 1.]

Therefore, if the square, &c. Q.E.D. Ex. 1. Shew that if the squares on BA and AC are less than the

square on BC the angle BAC is acute; but if greater

than BAC is obtuse. 2. Make a square equal to the difference of two given squares. BOOK II.

INTRODUCTION.

above ways.

THE Second Book of Euclid treats of the properties of rectangles contained by straight lines divided into segments in various ways. When a line is divided into two parts or segments it may be cut either equally or unequally. If the point of division be in the line produced, the distances of that point from the two ends of the line may still be called 'segments of the line, which in this case is said to be cut externally. The first ten propositions in this book demonstrate the relations between rectangles formed by the segments of lines divided in one of the

When the line is divided both equally and unequally the point of unequal division may be either internal or external, i.e., either in the line itself or in the line produced ; and the relations between the rectangles formed by the segments thus made will be really the same in the two cases and capable of expression in one proposition. Thus propositions 5 and 6 are so related, as also propositions 9 and 10. The first ten propositions may therefore be reduced to eight. These have necessarily great similarity, especially as to the mode in which they are demonstrated. In fact, the same principle underlies the demonstration of them all, viz., that any area is equal to the sum of the several areas into which it may be divided.'

These eight propositions have also very striking resemblances to certain elementary truths in Arithmetic and Algebra. Indeed each one of these geometrical propositions has a numerical proposition corresponding to it, and so striking is the similarity between the two sets of propositions, that it has frequently been proposed to make the one depend for its proof upon the demonstration of the other. The nature of the connection between them will be understood from what follows.

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BOOK II. INTRODUCTION.

Measurement of Lines. In order to express the length of a line by a number we must first fix upon a unit of length,i.e., the length of the line which is to be called one. Any other line will then be measured by the number of times which this unit is contained in it. Thus if the length of a foot rule be taken as the unit, any line along which this unit can be set off exactly three times will be 3 feet long, while one which contains it four times will be 4 feet long. Such lines may be represented by the numbers 3 and 4, and are said to be commensurable. The unit may be any length we choose to fix upon, but it must be the same throughout the discussion. But when the unit is thus settled, it may so happen that of two lines one may contain the unit an exact number of times and the other may not. In this case the lines are said to be incommensurable with that unit, and, as will be seen hereafter, it may so happen that this will be the case whatever the unit fixed upon may be. Such lines are incommensurable ; i.e., they cannot both be represented by numbers because they do not contain the same unit an exact number of times.

Measurement of Areas. To measure areas we begin by fixing upon a certain area to be called one, and this unit is generally the square described upon a length-unit. This forms the square unit. We may now prove this proposition.

The product of the numbers representing the adjacent sides of a rectangle will give the number of square units contained in the area of the rectangle.

Let the unit of length be contained exactly four times in the length of the rectangle ABCD, and three times in its breadth. Then the area of the rectangle will be 12 square units.

For apply the unit of length to AB and AD, setting it off 4 times in AB and 3 times in AD, and through the points of division draw lines parallel to the sides of the rectangle. Then AO is a “square unit ;' for it is a parallelogram with equal sides and having one right angle. Hence it is a square, and is described on the unit of length; therefore it is a square unit. In like manner each of the other parts into which AC is divided is a square unit. Now the number of such units in AP is

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