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It is indifferent which pair of sides be made to coincide to begin with.

The eighth proposition may be proved without having recourse to the seventh, as follows:

We must know,-
1. That two straight lines may be so placed one on

the other as to coincide in direction, and cannot
diverge from one another if they coincide for

any portion of their length. (Ax. B.) 2. That the sums or differences of two pairs of

equal magnitudes are equal. (Ax. II. and III.) 3. That if two As have two sides of the one equal

to two sides of the other, each to each, and have likewise the <s included between those sides equal, the As are also equal in all other respects.

(Prop. IV.) 4. If two sides of a A are equal to each other, the

<s opposite to them are also equal. (Prop. V.)

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Let A B C and D E F be the two As having the three sides of the one equal to the three sides of the other, each to each.

Let the A DEF be applied to the A A B C so that the point D may coincide with the point A, and the line D E lie along the line A B. Then because A B is equal to D Е, the point E will coincide with the point B.

Let the A8 lie on opposite sides of the common line A B.

Join the points C and F by the line C F. In the ACAF, the side A C is equal to the side A F; therefore the < ACF is equal to the < A FC.

In the ACB F, the side B C is equal to the side BF; therefore the < B C F is equal to the < B F C.

If the equal Z8 B C F and BFC be added to the equal Z8 A C F and A F C (as in Fig. 1), or taken from them (as in Fig. 2), the sums or differences AC B and A F B will be equal.

(If the line C F passes through the point B, it follows immediately that the <8 A C B and A F B (i. e., DFE) are equal, as in that case these are s at the base of the isosceles A CAF.)

Therefore the A8 A C B and A F B have two sides (A C and C B), and the included < (A C B) of the one equal to two sides (A F and F B), and the included _ (A F B) of the other, each to each.

Therefore these As are also equal to each other in all other respects.

But the A AF B is the same as the ADFE.

It has, then, been proved that the As A O B and DFE are equal to each other in every respect.

If the eighth proposition be proved in this manner,

the study of the difficult seventh proposition may be postponed till a later period,

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PROPOSITION IX. To bisect a given rectilineal angle that is, to divide it into two equal angles.

For the construction employed in this proposition

we require to be able -
1. To join two given points by a straight line.

2. From the greater of two given lines to cut off a

part equal to the less. (Prop. III.)
3. On a given right line to construct an equilateral

A. (Prop. I.)
To demonstrate that the construction effects what

is required, we must know-
That if two As have the three sides of the one

equal to the three sides of the other, each to each, their s also are equal. (Prop. VIII.)

B

Suppose A B C to be the given
Z.
In the line B A take any point-

as D.

E

From the line B C cut off a part,

B E, equal to B D. (Prop. III.) A F C

Join the points D and E by the

right line D E. On the line D E, and on the side remote from B, construct an equilateral A, D EF. (Prop. I.)

Join the points B and F by the right line B F. Then the right line B F divides the given < into two equal parts.

This is shown as follows :

DBF and EBF are two As, in which the sides D B and B E are equal (having been made so in the construction); the sides D F and F E are equal (D E F being an equilateral A); and the side B F is common to the two As: so that the three sides of the one A are equal to the three sides of the other, each to each.

It was proved in the eighth proposition, that in two such As the <s are equal, each to each.

Therefore, in the As D BF and E BF, the Zs DBF and EB F are equal.

But the <8 D B F and E B F are the parts into which the < D BE is divided by the line B F.

F

It has been proved, therefore, that the line BF bisects* the <D BE.

Let the proposition next be gone through with an < of a different size and in a different position, as in the annexed figure.

D

E

B

PROPOSITION X.

To bisect a given finite straight line, that is, to divide it into two equal parts.

For the construction in this proposition we must

be able,-
1. On a given finite right line to construct an equi-

lateral A. (Prop. I.)
2. To bisect a given rectilineal Z.
To

prove that the construction accomplishes what
is required we must know that,
If two As have two sides of the one equal to two

sides of the other, each to each, and have also the s included between these pairs of sides equal, then the As are also equal in every other respect. (Prop. IV.)

* It is obvious that by successive bisections an angle may be divided into 4, 8, 16, or (generally) 2" equal parts. It is impossible by ordinary geometry to divide an into three equal parts, or into any number of equal parts which is a multiple or power of three, with the exception of a' right Z. By the aid of the 32nd proposition of Book I., as applied to an equilateral A, we can obtain the third part of two right Z8, and by bisecting that, the third part of one right L. Certain propositions in the Fourth Book enable us to find the fifth, and the fifteenth part of a right Z.

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A

D

B

Suppose A B to be the given finite right line.

On A B describe an equilateral* A A B C. (Prop. I.)

From the point C draw a line bisecting the _ A O B. (Prop. IX.)

Let this line meet the line A B in

the point D. The line A B is now bisected at the point D.

For in the As A CD and B C D the sides A C and B C are equal, being sides of an equilateral A. The side C D is common to the two As, and the ACD is equal to the < BCD,—the whole Z A C B having been divided into two equal parts by the line C D. Thus two sides and the included < of the one A are equal respectively to two sides and the included _ of the other A.

These As, therefore, are equal in every other respect. (Prop. IV.)

Among these respects is that the side A D is equal to the side B D.

The line A B, therefore, has been divided into two equal parts at the point D.

Let the proposition now be gone through with the

equilateral A constructed on the other side of

A B. Euclid seems to have supposed that it is a simpler process to bisect an Z, than to bisect a straight line, and that the latter process is dependent on the former. In this he is wrong. If we go through the process which is assumed to be gone through in bisecting the < AC B in the above figure, we shall find that we bisect the line A B, without taking any account of the < ACB. Thus :

* It is enough for the purpose to describe an isosceles A on A B, with sides of any length. But the construction of an equilateral A is simpler than that of any other isosceles A.

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