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

Prop. VI. Algebraically.

Let AB contain 2 a 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 = 2 am + m2;

to each of these equals add a”,
o's (2a + m) m + ao = a2 + 2am + m2.

But a’ + 2am + n2 = (a + m),

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

=ma + 2mn + 919,
add na to each of these equals,

.. a? + na = m2 + 2mn + 2n2.
But 2mn + 2 n2 = 2 (m + n) n= : 2 an,

... al + n2 = 2 an + m2. That is, If a number be divided into any two parts, the square of the whole number and of one of the parts is equal to twice the product of the whole number and that part, together with the square of the other part.

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,

... m= al — 2an + na, and adding 4an to each of these equals,

4an + m= a2 + 2an + n?.

But a2 + 2an + na = (a + n)?,

.. 4an + me = (a + n). 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 up of the whole and that part.

Prop. ix. Algebraically.

Let AB contain 2 a 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 a - m units.

Then (a + m)2 = a2 + 2am + m2,
and (a m)? = a2 + 2am + m2,

Hence by adding these equals,

.. (a + m)2 + (a m)2 = 2a2 + 2 mo. 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 and twice the square of half the difference of the unequal parts.

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 = 4 a2 + 4am + m2,

add ma to each of these equals,
s'. (2a + m)2 + m2 = 4a2 + 4am + 2 mo.
Again (a + m)2 = a2 + 2am + m2,

add a? to each of these equals,
(a + m)2 + a2 = 2a? + 2 am + m2,

and doubling these equals,
.. 2 (a + m)2 + 2 a? = 4a? + 4 am + m2.
But (2a + m)2 + m2 = 4a2 + 4am + m2.

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

To solve Prop. XI algebraically, or to find the point H in AB such that the rect. angle contained by the whole line AB and the part HB shall be equal to the square of 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 – x units.
And a (a x) = xo, by the problem,
or xu+ a x = a', a quadratic equation.

ta V5
Hence

2

a

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

So that
V5 - 1

AH, one part,
2

AB =

3 - 15

and

a - X = a - AH

AB

HB, the other part.

. a.

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.

V5 + 1 To ascertain the meaning of the other result x =

2 In the equation a (a – x) = x”,

for x write – x, then a (a + x) = x?, which when translated into words gives the following problem.

To find the length a given line must be produced so that the rectangle contained by the given line and the line made up of the given line and the part produced, may be equal to the square of the part produced.

Or, the problem may also be expressed as follows :

To find two lines having a given difference, such that the rectangle contained by the difference and one of them may be equal to the square of the other.

Prop. XII. Algebraically.
Assuming the truth of Prop. 47, Book 1, Algebraically.
Let BC, CA, AB contain a, b, c linear units respectively,

and let CD, DA, contain m, 12 units,

then BD contains a + m units.
And therefore, c? = (a + m) + n?, from the right-angled triangle ABD,

also 62 = m2 + n° from ACD;
s'. c? – 62 = (a + m)2 m2

aľ + 2am + mo ma

+ 2am,

62 + a2 + 2am,

that is, c is greater than 62 + a2 by 2 am. Prop. XIII. Case 11 may be proved more simply as follows, in the same manner as Case 1.

Since BD is divided into two parts in the point D, therefore the squares of CB, BD are equal to twice the rectangle contained by CB, BD and the square of CD; (11. 7.)

add the square of AD to each of these equals ; therefore the squares of CB, BD, DA are equal to twice the rectangle CB, BD, and the squares of CD and DA, but the squares of BD, DA are equal to the square of AB, (1. 47.)

and the squares of CD, DA are equal to the squares of AC, therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD.

That is, &c. Prop. XIII. Algebraically.

Let BC, CA, AB contain respectively a, b, c linear units, and let BD and AD also contain m and n units. Case 1.

Then DC contains a - m units.

Case II.

Therefore c2 = n2 + m2 from the right-angled triangle ABD,

and b2 = n2 + (a m)2 from ADC; .. c? 72 = mo (a m)?

= m? - a2 + 2am - ma

-a2 + 2am,
.. a2 + c = 6 + 2 am,
or 6+ 2am = a2 + c,

that is, 62 is less than aa+c2 by 2 am.
DC = m - a units,
.. c? = m% + no from the right-angled triangle ABD,

and 12 = (m a)? + m2 from ACD,
.. c? = ma – (m a)

= m- ma + 2am - a?

= 2am – a, .. a? + C2

12 + 2 am, or 72 + 2 am

that is, zo is less than a2 + c by 2am. Here m is equal to a. And 72 + a2 = c>, from the right-angled triangle ABC.

Add to each of these equals ař,

... 32 + 2a? = c2 + a,
that is, 62 is less than c? + a2 by 2 a, or 2aa.

a2 + c,

Case III.

BOOK III.

DEFINITIONS.

I. Equal circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumferences are equal.

This is not a definition, but a theorem, the truth of which is evident; for, if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centres are equal.

II. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it.

III. Circles are said to touch one another, which meet, but do not cut one another.

IV. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal.

V. And the straight line on which the greater perpendicular falls, is said to be farther from the centre.

VI. A segment of a circle is the figure contained by a straight line, and the circumference which it cuts off.

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