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POSTULATES. Let it be granted :

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

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

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

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. 10. Two straight lines cannot inclose a space. 11. All right angles are equal to one another.

12. If a straight line meet two straight lines so as to make the two interior angles on the same side of it, taken together, less than two right angles, these two straight lines being continually produced shall at length meet on that side on which are the angles which are less than two right angles.

PROPOSITION I. 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.

Construction, With A as centre, at the distance AB, describe the circle BCD.

(post. 3) With B as centre, at the distance BA, describe the circle ACE.

(post. 3) From the point C, in which the circles cut each other, draw the straight line CA to the point A, and the straight line CB to the point B. (post. 1) Then shall ABC be an equilateral triangle.

Demonstration.
Because A is the centre of the circle BCD,

therefore AB is equal to AC. (def. 11)
Because B is the centre of the circle ACE,

therefore BA is equal to BC. (def. 11)

Hence AC and BC are each equal to AB, and “things which are equal to the same thing are equal to one

another” (ax. 1),
therefore AC and BC are also equal to each other;

and AB, AC, BC, are all equal :
therefore the triangle ABC is equilateral,
and it has been described on the given straight line. Q.E.F.

Exercises. 1. Shew that an equilateral triangle may also be described on the other side of AB.

2. From the greater of two given straight lines terminated at the same point cut off a part equal to the less.

3. Given two straight lines terminated in the same point, one of them finite, the second of unlimited length ; shew how to cut off from the second a part twice as long as the first.

PROPOSITION II. 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 A a straight line equal to BC.

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Construction. From A to B draw the straight line AB. (post. 1) Upon AB describe the equilateral triangle ABD. (prop. 1) Produce the straight lines DA and DB to E and F. (post. 2) With B as centre, at the distance BC, describe the circle CGH,

(post. 3) and let G be the point where it meets the straight line DF. With D as centre, at the distance DG, describe the circle GLK,

(post. 3) and let L be the point where it meets the straight line DE.

Then shall AL be equal to BC.

Demonstration.
Because B is the centre of the smaller circle CGH,

therefore BC is equal to BG. (def. 11)
Because D is the centre of the larger circle GLK,

therefore DG is equal to DL; (def. 11) but parts of these, DB, DA (being sides of an equilateral triangle)

are equal to each other, therefore the remainder BG is equal to the remainder AL. (ax. 3)

Now AL and BC are each of them equal to BG,

therefore AL and BC are equal to each other. (ax. 1) Wherefore from the given point A a straight line AL has been drawn equal to the given straight line.

Q.E.F.

Exercises.

1. If the radius of the smaller circle be half that of the larger, how far is the given point from the nearer extremity of the given line?

2. Construct a similar figure to the above for the case when the point A is joined to the further extremity of the straight line BC..

3. Construct a similar figure to the above for the case when the equilateral triangle is described on the other side of the straight line AB.

4. From the middle point of a given straight line draw another straight line of the same length.

5. From the extremity of a given straight line draw a straight line double of its length.

PROPOSITION III. PROBLEM.

From the greater of two given straight lines to cut off a part equal

to the less.

Let AB and CD be the two given straight lines, of which AB is the greater;

it is required to cut off from AB the greater, a part equal to OD the less.

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Construction. From A draw the straight line AE equal to CD. (prop. 2) With A as centre, and at the distance AE, describe the circle EFG,

meeting the straight line AB in F. (post. 3)

Then shall AF be equal to CD.

Demonstration.
Because A is the centre of the circle EFG,
therefore AF is equal to AE. (def. 11)

But CD is equal to AE; (const.) therefore AF and CD are each of them equal to AE, and therefore AF and CD are equal to each other. (ax. 1) Wherefore from AB the greater of two straight lines, a part AF has been cut off equal to CD the less.

Q.E.F. Exercises. 1. On the greater of two straight lines describe an isosceles triangle, each of whose sides shall be equal to the less. When is this impossible ?

2. Shew how to produce the less of two given straight lines until it is equal to the greater.

3. Produce the greater of two given straight lines until the part produced is equal to the less.

4. From a given point draw a straight line four times the length of a given straight line.

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If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to each other ; they shall also have the bases or third sides equal, and the two triangles shall be equal, and their other angles shall be equal, each to each, namely, those to which the equal sides are opposite.

* In this proposition the student is for the first time introduced to the consideration of the equality of angles, and it is very important that he should understand the test by which Euclid compares them. An angle being the inclination of two straight lines to each other, it is clear that its magni. tude is in no way dependent upon the length of the lines which form it; it will be increased or diminished only by the incli. nation of one line to the other being made greater or less. The equality of two angles ABC, DEF is tested by supposing one of them to be taken up, and superposed upon the other; if when B is placed upon E, A and BA in the direction of D, it is found that the line BC also falls upon EF, then the angles are equal.

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