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POSTULATES.

Let it be granted

I. That a straight line may be drawn from any one point to any other point.

2. That a terminated straight line may be produced to any length in a straight line.

3. That a circle may be described from any centre at any distance from that centre.

In practice this amounts to demanding the use of a ruler or straight-edge and a pair of compasses.

Neither of these instruments, however, is to be used for carrying distances (see note 2 on Prop. 2].

When Geometers speak of a problem such as to divide a given angle into three equal angles as 'impossible, they mean "impossible under the restrictions they have placed upon themselves as to the elementary constructions they take for granted.'

They only take for granted the three demanded by the Postulates.

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Axioms.

AXIOMS.

1. Things which are equal to the same thing are equal to one another.

2. If equals be added to equals, the wholes are equal.

3. If equals be taken from equals, the remainders are equal.

4. If equals be added to unequals, the wholes are unequal.

5. If equals be taken from unequals, the remainders are unequal.

6. Things which are double of the same thing are equal to one another.

7. Things which are halves of the same thing are equal to one another.

8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.

9. The whole is greater than its part.
IO. Two straight lines cannot enclose a space.
II. All right angles are equal to one another.

DEF.-A triangle with its three sides equal to eac,

other is called an Equilateral Triangle.

PROPOSITION 1. PROBLEM.

To describe an equilateral triangle on a given finite

straight line. Let AB be the given straight line; it is required to describe an

equilateral triangle on AB.

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From the centre A, at the distance AB, describe the circle BCD.

[POST. 3. From the centre B, at the distance BA, describe

the circle ACE. From the point C, at which the circles cut one another, draw

the straight lines CA, CB to the points A and B. Because the point A is the centre of the circle BCD, therefore AC is equal to AB;

(DEF. 15. And because the point B is the centre of the circle ACE,

therefore BC is equal to BA;
therefore AC, BC, AB are all equal to one another ;

[ax. 1.
therefore ABC is equilateral.

(DEF. 15.

NOTES.

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1. The words 'join CA' are afterwards used as an abbreviation for 'draw a straight line from C to A.'

2. If we join AD and BD (D being the other point in which the two circles cut each other) another equilateral triangle ADB will be described on the given straight line AB.

3. 'A triangle is sometimes regarded as standing on a selected side, which is then called its base, and the intersection of the other two sides is called the vertex.' (Syllabus.)

4. The quadrilateral (four-sided) figure ACBD has all its sides equal. Such a quadrilateral is called a rhombus.

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PROPOSITION 2. PROBLEM.

From a given point to draw a straight line equal to

a given straight line.

Let A be the given point, and BC the given straight line;

it is required to draw from the point A a straight line equal to BC.

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Н.

G

Join AB.

(POST. 1. On AB describe the equilateral triangle DAB. [I. 1. From the centre B, at the distance BC, describe the circle CGH cutting DB produced in G.

(POST. 3. From the centre D, at the distance DG, describe the circle GKL cutting DA produced in L.

[POST. 3. Then AL shall be equal to BC. Because D is the centre of the circle GKL, therefore DL is equal to DG;

(DEF. 15.

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