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The following simple properties respecting the equal and unequal division of a line are worthy of being remembered.

I. Since AB = 2BC = 2(BD + DC) = 2BD + 2DC. (fig. Prop. v.)

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That is, if a line AB be divided into two equal parts in C, and into two unequal parts in D, the part CD of the line between the points of section is equal to half the difference of the unequal parts AD and DB.

II. Here AD = AC + CD, the sum of the unequal parts, (fig. Prop. v.)

and DB = AC
AC CD their difference.

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Hence by adding these equals together,

.. AD + DB = 2AC,

or the sum and difference of two lines AC, CD, are together equal to twice the greater line.

And the halves of these equals are equal,

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or, half the sum of two unequal lines AC, CD added to half their difference, is equal to the greater line AC.

III. Again, since AD = AC + CD, and DB = = AC - CD,

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or, the difference between the sum and difference of two unequal lines, is equal to twice the less line.

And the halves of these equals are equal,

.. §. AD — 1.DB = CD,

or, half the difference of twò lines subtracted from half their sum is equal to the less of the two lines.

IV. Since

AC - CD = DB the difference,

.. AC = CD + DB,

and adding CD the less to each of these equals,

.. AC + CD = 2CD + DB,

or, the sum of two unequal lines, is equal to twice the less line together with the difference between the lines.

Prop. v. The rectangle contained by AD and DB, and the square on BC are each bounded by the same extent of line, but the spaces enclosed differ by the square on CD.

Bearing in mind that AC, CD are respectively half the sum and half the difference of the two lines AD, DB; the corollary to this proposition may be expressed in the following form: "The rectangle contained by two straight lines is equal to the difference of the squares on half their sum and half their difference."

Prop. v. Algebraically.

Let AB contain 2a linear units,

its half BC will contain a linear units.

And let CD the line between the points of section contain m linear units. Then AD the greater of the two unequal parts, contains a + m linear units; and DB the less contains a m units.

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That is, If a number be divided into two equal parts, and also into two unequal parts, the product of the unequal parts together with the square of half their difference, is equal to the square of half the number.

Prop. VI. A given straight line is said to be produced when it has its length increased in either direction, and the increase it receives, is called the part produced,

If a point be taken in a line or in a line produced, the line is said to be divided internally or externally, and the distances of the point from the ends of the line are called the internal or external segments of the line, according as the point of section is in the line or the line produced. Props. v. and vI., also. IX. and x., are, in fact, the same, if the division of the lines be regarded with respect to their external and internal division.

Prop. vi. Algebraically.

Let AB contain 2a linear units, then its half BC contains a units; and let BD contain m units.

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That is, If a number be divided into two equal numbers, and another number be added to the whole and to one of the parts; the product of the whole number. thus increased and the other number, together with the square of half the given number,. is equal to the square of the number which is made up of half the given number increased.

The algebraical results of Prop. x and Prop. vI. are identical, as it is obvious that the difference of a+m and am in Prop. v. is equal to the difference of 2a + m and m in Prop. vI., and one algebraical result expresses the truth of both propositions.

This arises from the two ways in which the difference between two unequal lines may be represented geometrically, when they are in the same direction,

In the diagram (fig..to Prop. v.), the difference DB of the two unequal lines AC and CD is exhibited by producing the less line CD, and making CB equal to AC the greater,

Then the part produced DB is the difference between AC and CD,. for AC is equal to CB, and taking CD from each,

the difference of AC and CD is equal to the difference of CB and CD. In the diagram (fig. to Prop. vI.), the difference DB of the two unequal lines CD and CA is exhibited by cutting off from CD the greater, a part CB equal to CA the less.

Prop. VII. Either of the two parts AC, CB of the line AB may be taken: and it is equally true, that the squares on AB and AC are equal to twice the rectangle AB, AC, together with the square on BC,

This Proposition may be enunciated in the following form. The square on the difference of any two lines, is equal to the difference between the sum of the squares on the two lines, and twice their rectangle.

It also appears that the difference between the squares on the sum, and on the

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difference of two lines, is equal to four times the rectangle contained by the lines themselves.

Prop. vII. Algebraically.

Let AB contain a linear ́units, and let the parts AC and CB contain m and n linear units respectively.

Then a = m + n;
squaring these equals,

.. a2 = m2 + 2mn + n2,
add n2 to each of these equals,

•. a3 + n2 = m3 + 2mn + 2n3.

But 2mn + 2n3 = 2 (m + n) n = 2an,

•'•..a2 + n2
+ n2 = m2 + 2an.

That is, If a number be divided into any two parts, the squares of the whole number and of one of the parts, are equal to twice the product of the whole number and that part, together with the square of the other part.

Prop. VIII. As in Prop. VII., either part of the line may be taken, and it is also true in this Proposition, that four times the rectangle contained by AB, AC together with the square on BC, is equal to the square on the straight line made up of AB and AC together.

The truth of this Proposition may be deduced from Euc. 11. 4 and 7.

For the square on AD (fig. Prop. 8.) is equal to the squares on AB, BD, and twice the rectangle AB, BD; (Euc. 11. 4.) or the squares on AB, BC, and twice the rectangle AB, BC, because BC is equal to BD: and the squares on AB, BC are equal to twice the rectangle AB, BC with the square on AC: (Euc. II. 7.) therefore the square on AD is equal to four times the rectangle AB, BC together with the square on AC.

Prop. VIII. Algebraically.

Let the whole line AB contain a linear units of which the parts AC, CB contain m, n units respectively.

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That is, If a number be divided into any two parts, four times the product of the whole number and one of the parts, together with the square of the other part, is equal to the square of the number made of the whole and the part first taken.

Prop. VIII. may be put under the following form: The square on the sum of two lines exceeds the square on their difference, by four times the rectangle contained by the lines.

Prop. IX. The demonstration of this proposition may be deduced from Euc. II. 4 and 7.

For (Euc. II. 4.) the square on AD is equal to the squares on AC, CD and twice the rectangle AC, CD; (fig. Prop. 9) and adding the square on DB to each, therefore the squares on AD, DB are equal to the squares on AC, CD and

twice the rectangle AC, CD together with the square on DB; or to the squares on BC, CD and twice the rectangle BC, CD with the square on DB, because BC is equal to AC.

But the squares on BC, CD arë equal to twice the rectangle BC, CD, with the square on DB. (Euc. 11. 7.)

Wherefore the squares on AD, DB are equal to twice the squares on BC and CD.

From this Proposition, it appears that the square on the sum of two lines AC, CD and the square on their difference DB, are together equal to double of the sum of the squares on the two lines AC, CD.

Prop. IX. Algebraically.

Let AB contain 2a linear units, its half AC or BC will contain a units; and let CD the line between the points of section contain, m units..

Then AD the greater of the two unequal parts contains a + m units,.

and DB the less contains a - m units;

•. (a + m)2 = a3 + 2am + m3,

and (« — m)2 = a2 ¬ 2am + m2.

Hence by adding these equals,
•. (a + m)2 + (a — m)2
m)2 = 2a2 + 2m3.

That is, If a number be divided into two equal parts, and also into two unequal parts, the sum of the squares of the two unequal parts, is equal to twice the square of half the number itself, and twice the square of half the difference of the unequal parts.

The proof of Prop. x. may be deduced from Euc. 11, 4, 7, as Prop. IX.

Prop. x. Algebraically.

Let the line AB contain 2a linear units, of which its half AC or CB; will contain a units;

and let BD contain m units.

Then the whole line and the part produced will contain 2a+m units,
and half the line and the part produced will contain a + m units,

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That is, If a number be divided into two equal parts, and the whole number and one of the parts be increased by the addition of another number, the squares of the whole number thus increased, and of the number by which it is increased, are equal to double the squares of half the number, and of half the number increased.

The algebraical results of Prop. ïx., and Prop. x., are identical, (the enunciations of the two Props, arising, as in Prop. v. and Prop. vI., from the two ways of exhibiting the difference between two lines); and both may be included under the following proposition: The square on the sum of two lines and the square on their difference, are together equal to double the sum of the squares on the two lines.

Prop. XI. The following problem is contained in the construction: Produce a straight line, so that the rectangle contained by the whole line thus produced and the part produced, shall be equal to the square on the original line. For the rectangle contained by AF, FA is manifestly equal to the square on CA or AB. Two series of lines, one series decreasing, and the other series increasing in, magnitude, and each line divided in the same manner may be found by means of this proposition.

(1) To find the decreasing series.

In the fig. Euc. II. 11, ÀB′ = AH -|- BH,.

and since AB.BH AH2, .. (AH + BH). BH = AH3;

.. BH2 = AH2
АН - АН. ВН =

AH.(AH – BH).
If now in HA, HL be taken equal to BH,
then HL2 AH (AH – HL), or AH. AL = HL3 :

that is, AH is divided in L, so that the rectangle contained by the whole line AĦ and one part, is equal to the square on the other part HL. By a similar process, HL may be so divided; and so on, by always taking from the greater part of the divided line, a part equal to the less.

(2) To find the increasing series.

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From the fig. it is obvious that CF . FA = CA2. Hence CF is divided in A, in the same manner as AB is divided in H, by adding AF a line equal to the greater segment, to the given line CA or AB. And by successively adding to the last line thus divided, its greater segment, a series of lines increasing in magnitude may be found similarly divided to AB.

It may also be shewn that the squares on the whole line and on the less segment are equal to three times the square on the greater segment. (Euc. XIII, 4.) To solve Prop. XL. algebraically, or to find the point H in AB such that, the -rectangle contained by the whole line AB and the part HB, shall be equal to the square on the other part AH.

Let AB contain a linear units, and AH one of the unknown parts, contain x units,

then the other part HB contains a units.

-And, a (a

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a (a − x) = x2, by the problem,
or x2 + ax = a
a2, a quadratic equation.

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The former of these values of x determines the point H.

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It may be observed, that the parts AH and HB cannot be numerically expressed by any rational number. Approximation to their true values in terms of AB, may be made to any required degree of accuracy, by extending the extraction of the square root of 5 to any number of decimals.

To ascertain the meaning of the other result x =

In the equation a (a — x) = x2,

√5+1
2

a.

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which, when translated into words, gives the following problem:

To find the length to which a given line must be produced, so that the

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