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To describe an equilateral triangle on a given finite straight line.

Let AB be the given st. line. It is required to describe an equilat. A on AB.

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Constr. From the centre A, at the distance AB, describe O BCD;

(post. 3) from the centre B, at the distance BA, describe O ACE;

(post. 3) and from the pt. C, in which the os cut one another, draw the st. lines CA,CB, to the pts. A,B. (post. 1)

Then ABC shall be an equilat. A. Dem. For, : the pt. A is the centre of o BCD, .. AC= AB,

(def. 15) and :: the pt. B is the centre of o ACE,

.:. BC= AB;
but it has been proved that AC = AB;
.. AC, BC, are each of them

AB;

but things which are equal to the same thing are equal to one another;

(ax. I) .. AC= BC; Thus AB, BC, AC are equal to one another : and J ABC is therefore equilateral,

(def. 24) and it is described on the given straight line AB. Q. E. F.

[Here the data are the given st. line; the quæsita, an equilateral triangle on it. The general enunciation, 'to describe an equilateral triangle on a given finite straight line;' the particular enunciation, 'let AB be the given st. line. It is required to describe an equilat. A on AB.' It may be observed that in Books i. and ii. no properties of the circle are discussed in any way; it is merely employed in constructions, as defined in def. 15, and as granted by the third postulate.)

By a method similar to that used in this problem, describe on a given finite straight line an isosceles triangle, whose sides shall be each double of the base.

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From a given point, to draw a straight line equal to a given straight line.

Let A be the given pt., and BC the given straight line. It is required to draw from the pt. A a st. line equal to BC.

K

H

D

B

E

F

Constr. From the pt. A to B draw the st. line AB;

(post. 1) on AB describe the equilat. A ABD, (i. 1) and produce the st. lines DA, DB to E and F; (post. 2) from the centre B, at the distance BC, describe O CHG;

(post. 3) and from the centre D, at the distance DG, describe O GKL

(do) Then the st. line AL shall be = BC. Dem. : the pt. B is the centre of O CHG, .. BC

(def. 15) and :: D is the centre of O GKL, .. DL= DG

(do) and DA, DB parts of them are equal ; (def. 24)

... the rem. AL- = the rem. BG; (ax. 3) but it has been shown that BC= BG,

BG;

.. AL and BC are each of them equal to BG; and things that are equal to the same thing are equal to one another;

(ax. 1) .. AL= BC. Wherefore, from the given pt. A a st. line AL has been drawn equal to the given st. line BC.

Q. E. F. [It is unnecessary to point out the general enunciation, and particular enunciation; the beginner will easily be able to do this for himself, fronr the explanation given. The data are the given point and the given straight line; the quæsita, a straight line equal to the given one. Observe how this proposition depends on the first (the side reference i. I, means Book i., prop. 1); the learner will find that throughout Euclid the successive propositions are linked to and depend upon the preceding ones, as the successive links of a chain hang one from another.

When the given point is neither in the line nor in the line produced, there are several ways in which the line may be drawn. For (1) the given line has two ends, each of which may be joined to the given point ; (2) the equilateral triangle may be described on either side of this line ; and (3) the side BD may be produced either way. It will be an excellent exercise to draw figures and to work out the proposition for each case : this will give eight different examples of this problem.]

PROPOSITION III.

PROBLEM.

From the greater of two given straight lines to cut off a bart equal to the less.

Let AB and CD be the two given st. lines, of which AB is the greater.

It is required to cut off from AB, the greater, a part equal to CD, the less.

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Constr. From the pt. A draw the st. line AE = CD; (i. 2) with centre A at the distance AE describe O EFH,

(post. 3) cutting AB in the pt. F.

Then will AF = CD.
Dem. :: A is the centre of O EFH,

(def. 15) but the st. line CD= AE; (constr.) whence AF and CD are each of them equal to AE; .. AF = CD.

(ax. I) Wherefore, from AB, the greater of two st. lines, a part AF has been cut off equal to CD, the less. Q. E. F.

.. AF:

AE;

[The data are two given st. lines, the quæsita, to cut off from the greater a part equal to the less.]

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