Prop. IV. might have been deduced from the two preceding propositions; but Euclid has preferred the method of exhibiting, in the demonstrations of the second book, the equality of the spaces compared.

In the corollary to Prop. XLVI. Book I, it is stated that a parallelogram which has one right angle, has all its angles right angles. By applying this corollary, the demonstration of Prop. iv. may be considerably shortened.

If the two parts of the line be equal, then the square on the whole line is equal to four times the square on half the line.

Also, if a line be divided into any three parts, the square on the whole line is equal to the squares on the three parts, and twice the rectangles contained by every two parts.

Prop. IV. Algebraically. (fig. Prop. iv.)

Let the line AB contain a linear units, and the parts of it AC and BC, m and n linear units respectively.

Then am + n,

squaring these equals, .. a2 = (m + n)2,

or a2 = m2 + 2mn + no..

That is, if a number be divided into any two parts, the square of the number is equal to the squares of the two parts together with twice the product of the two parts.

From Euc. II. 4, may be deduced a proof of Euc. 1. 47. In the fig. take DL on DE, and EM on EB, each equal to BC, and join CH, HL, LM, MC. Then the figure HLMC is a square, and the four triangles CAH, HDL, LEM, MBC are equal to one another, and together are equal to the two rectangles AG, GE.

Now AG, GE, FH, CK are together equal to the whole figure ADEB; and HLMC, with the four triangles CAH, HDL, LEB, MBC also make up the whole figure ADEB;

Hence AG, GE, FH, CK are equal to HLMC together with the four triangles:

but AG, GE are equal to the four triangles.

wherefore FH, CK are equal to HLMC,

that is, the squares on AC, AH are together equal to the square on CH. Prop. v. It must be kept in mind, that the sum of two straight lines in Geometry, means the straight line formed by joining the two lines together, so that both may be in the same straight line.

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.) and AB AD + DB;

.. 2CD + 2DB = AD + DB,

and by subtracting 2DB from these equals,

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 - CD their difference.

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,

...AD + . DB = AC,

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

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 two lines subtracted from half their sum is equal to the less of the two lines.

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. 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.

Also m is half the difference of a + m and a

=

·m;

.. (a + m) (a - m) a2 - m2, to each of these equals add m2; :. (a + m) (a − m) + m2 = a2.

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.

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 on the squares of half their sum and half their difference."

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.

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. Prop. vi. Algebraically.

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

Then AD contains 2a + m units,
and.. (2a + m) m = 2am + m2;

to each of these equals add a2,
:. (2a + m) m + a2 = a2 + 2am + m3.
But a2+2am + n2 = (a + m)2,

.. (2a + m) m + a2 = (a + m)2.

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. v. 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.

Prop. vII. Algebraically.

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

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 con

tained 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. II. 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. II. 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.

Then m + n = a,

and subtracting or taking n from each,

.. m = a - n,

squaring these equals,

.. m2 = a2 - 2an + n2,

and adding 4an to each of these equals,

.. 4an + m2 = a2 + 2an + n2.

But a2+2an + n2 = (a + n)2,

.. 4an + m2 = (a + n)2.

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. 11. 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 are 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.

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.

Also AD the greater of the two unequal parts contains a + m units, and DB the less contains am units.

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. II. 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, .. (2a + m)2 = 4a2 + 4am + m2,

add m2 to each of these equals,

.. (2a + m)2 + m2 = 4a2 + 4am + 2m2.
Again, (a + m)2 = a2 + 2am + m3,
add a2 to each of these equals,
•. (a + m)2 + a3 = 2a2 + 2am + m2,
and doubling these equals,

.. 2 (a + m)2 + 2a2 4a2 + 4am + 2m3.

But (2a + m)2 + m2 = 4a2 + 4am + 2m3.

Hence.. (2a + m)2 + m2 = 2a2 + 2 (a + m)3.

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. Ix, 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. 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, AB = AH + BH,

and since AB. BH = 4H2, .. (AH + BH). BH = AH2,

If now in HA, HL be taken equal to BH,

then HL2 = AH (AH – HL), or AH. AL = HL2:

that is, AH is divided in L, so that the rectangle contained by the whole line AH 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.

From the fig. it is obvious that CF. FA CA3, 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