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direction by the angle ABC. Again, suppose it to be turned round the point C, from the position BC to AC, it will then have changed its direction by the angle BCA. Lastly, let it be turned round the point A, from the position CA to BA, it will then have changed its direction by the angle CAB. It has thus changed its position by the B three angles of the triangle ; but after the first change, the end which was at first above A would be at D; after the second change, the same end would evidently be at E; and after the third change of position, it would be below the point B in the straight line AB; and hence its position has just been reversed by describing the three angles of the triangle. Therefore the three angles of any triangle are together equal to two right angles.

Second : Suppose a person standing at B, and looking in the direction BA, turns round till he be looking in the direction BC, he will have turned through the angle ABC; let him now walk along to C without turning, he will now be looking to D in BC produced, let him now turn round till his face be in the direction E in AC produced, he will now have turned round through the angle DCE, which is equal to the angle BCA (I. 15); let him now walk backwards to A, still looking in the direction AE, and then at A turn round through the angle CAB, he will now be looking in the direction AB instead of BA; having now turned through the three angles of the triangle ABC his position is reversed, and he has therefore turned through two right angles; hence all the angles of a triangle are equal to two right angles.

PROPOSITION XXXV. to a inclusive. In these propositions it should be observed that when the parallelograms and triangles are proved equal, they are only proved equal in area, not in any other respect. Some writers on Geometry, instead of saying that the parallelograms and triangles are equal, say they are equivalent; it might probably be better to say that they are equiareal, meaning thereby that they are equal in area; but this word not being in common use, the ordinary phrase has been retained, but the teacher should be careful to explain that the word equal in these propositions does not mean equal in any respect except that of area.

PROPOSITION XLIII. In this proposition, when it is proved that the complements of the parallelograms about the diagonal of any parallelogram are equal, it is only proved that they are equal in area, but it may in this case always be proved that these complements are also equiangular.

PROPOSITION XLV. 9 "There are two other ways in which this construction may be effected more simply than that given in Euclid's Elements; referring to the figure given in Euclid-first, let AB be produced beyond B, and through C draw CG parallel to DB meeting AB produced in G, and join DG; the triangle DBG is equal to the triangle DCB, for they are on the same base DB and between the sane parallels DB and CG (I. 37), to each of these equals add the triangle DAB, and the triangle DAG is equal to the whole figure ABCD. Bisect AG in F, and at the point F make the angle GFH equal to the given angle E, through D draw DHK parallel to AG, and through G draw GK parallel to FH, then it is plain, by Proposition 42, that the parallelogram HFGK is equal to the triangle DAG, and is therefore also equal to the given rectilineal figure ABCD, and it has by construction the angle HFG equal to the given angle E, which was required to be done.

Second: If the diagonal DB be bisected in F, and at the point F an angle be made equal to E, and the line forming this angle be produced both ways to meet parallels to DB through C and A, and a line be also drawn through B parallel to the line which makes the given angle at F, and produced to meet the parallels through C and A, a parallelogram will be formed, which can easily be proved, by Proposition 42, to be equal to the given quadrilateral figure, and to have an angle equal to the given rectilineal angle.

I

SECOND BOOK.

The first eight propositions of this book are all true both algebraically and arithmetically, if for rectangle, product be substituted; for line, number; and for square the second power of the letter or number, which is also called the square of the letter or number.

The reason of this is, that a line can be represented by the number of times that it contains a given line, and by omitting the measuring unit, the line becomes a number. Again, since the rectangle contained by two lines is the area of a right-angled parallelogram, which has these lines for two adjacent sides; and this area is represented by the number of squares on the unit of measure which it contains, and this area is always the product of the length multiplied into the breadth, into the unit of measure which is one, the area is the product of the length into the breadth. Also the square on a line is the area of the figure, which is expressed numerically by the length of the side multiplied into itself and again into the unit of measure, which is taken as one; the area of a square is therefore represented by the length of its side expressed numerically multiplied into itself or the second power of the length of the side; and this is the reason why the second power of a number is called its square. These relations being established, the propositions above referred to may all be proved algebraiacally and numerically, which should make them still more interesting to the student, and give him a more comprehensive idea of their nature and use. In proving the propositions thus, the following axiom will be assumed:

AXIOM.
If equals be multiplied by equals, the products are equal.

1830 PROPOSITION I. THEOREM. Referring to the diagram in the Second Book, assume the length of the line A = a, BC = 6, BD = c, DE = d, and EC = then b=c+dte; and multiplying each of these equals by a ab = ac + ad + ae, but ab = the area of the rectangle BH, ac = the area of the rectangle BK, ad = the area of the rectangle DL, and ae = the area of the rectangle EH; and therefore the fact that ab = ac + ad + ae proves the truth of the proposition. Again, assume a = 6, 6 = 12, c= 5, d= 3, and consequently e = 4; then 12 x 6 = 5 X 6 + 3 6+4 6, or 72 = 30 + 18 + 24 = 72. COR. If there be two numbers, one of which is divided into

any number of parts, the product of the two numbers is equal

to the sum of the products of the undivided into the several , parts of the divided number.

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PROPOSITION XI. PROBLEM. This proposition cannot be demonstrated numerically, for though a line can be divided geometrically, so that the rectangle contained by the whole and the less part may be, exactly equal to the square of the greater, it is impossible to divide a number into two parts, so that the product of the whole number and the less part shall be exactly equal to the square of the greater part; if we solve the problem algebraically by taking the whole line as one, and its

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