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B C, the remainders AF and CL are equal (Ax. 3). Moreover, the line AL was proved equal to the line CF, and the CLA (which is the same as BLA) to the AFC (which is the same as B F C), when the As BAL and BCF were shown to be equal in every respect. So that in the two As CLA and AFC, the two sides CL and LA of the one are equal respectively to the two sides A F and F C of the other, and the included CLA of the one, is equal to the included AFC of the other. Therefore (according to the fourth proposition) these two As are also equal in all other respects. Among these is that the CA L is equal to the ▲ A C F.

3. If from the equals B AL and B C F, the equals CAL and A C F be taken away, the remainders, namely the s B A C and B CA, will be equal to each other (Ax. 3). But these are the s opposite the equal sides of the isosceles ▲ A B. C. Therefore the first part of the proposition has been proved.

The second part has also been proved. For the Zs at the other side of the base, namely F A C and LCA, are parts of the As AFC and CL A, which have been proved to be equal in every respect.*

Cor. All the s of an equilateral ▲ are equal.

The proof of the next proposition is of the kind called indirect. An indirect proof is of the following nature:-If we know that out of any number of assertions, some one must be true; and if we know, or can prove, that all

* When this proposition has been mastered as given above, it should next be proved without the aid of the accessory figures; then with different letters in the diagram; and then with figures of different shape, and drawn in different positions.

but one are false, we are then able to affirm that the remaining one must be the true one. If, for example, it could be demonstrated that one or other of three men-A, B, and C-committed a certain crime, and we knew or could prove, that neither A nor B committed it, it would follow that C must have committed it. Of the propositions, A is equal to B, A is less than B, A is greater than B (A and B being two magnitudes that admit of comparison), it is clear that one or other must be true. If we can show that A is not equal to B, and also that A is not less than B, it follows that A is greater than B.

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In order to prove that a proposition or assertion is false, we may either prove that something which contradicts it is true, or we may show that some falsity or absurdity would be a nec consequence of it. Now it is assumed as an axiom, nothing which is absurd or false can be a necessary consequence of anything that is true." If a false assertion be a necessary consequence of some other assertion, that other assertion must itself be false. When an assertion or proposition is demonstrated to be false, by showing that if we admitted it we should be compelled also to admit something false or absurd, the proof is called reductio ad absurdum (Reduction to an absurdity). Proofs of this kind are very common in mathematics. Some people have a fancy that they are not so cogent or conclusive as direct proofs; but this is altogether a mistake.

PROPOSITION VI.

If two angles of a triangle are equal, the sides which are opposite the equal angles are also equal.

This proposition is the converse of the fifth. In the fifth we started by supposing that we knew two sides of a A to be equal, and proceeded to prove that the s opposite those equal sides are also equal. In the sixth we are not supposed to know anything about the sides of the A, to

start with, but to know something about two angles of the ▲—namely, their equality. The equality of the sides has to be demonstrated. The proof of this proposition is of the indirect kind. Instead of its being shown at once that the two sides are equal, it is shown that they cannot be unequal.

To prove this proposition we require to be able,---1. To draw a straight line from one given point to another. (Post. I.)

2. To cut off from the greater of two given lines a part equal to the less. (Prop. III.)

And we require to know,—

1. That a part of any whole is less than that whole.

2. That if, in two As, two sides of the one are equal to two sides of the other, each to each, and the included of the first ▲ also equal to the included of the second A, then those two As are equal in every respect.

Let A B C be a ▲, with respect to which it is known that the CAB is equal

A

D

B

to the CB A.

We have to prove that the side A C is equal to the side C B.

This is proved by showing that AC and CB are not unequal.

That A C and CB are not unequal is proved by showing that it leads to an absurdity to suppose that they are unequal.

Suppose the side A () were not equal to the side CB; and suppose, first, that A C were greater than CB,

We should then be able to cut off from A C a part, A D, of the same length as C B.

Suppose that this were done, and that the points D and B were joined by the straight line D B,

We should then have two As, D A B and C B A, in which the side D A would be equal to the side CB, and the side A B common to the two As, and the

DAB equal to the CBA (we were supposed to know this last point to begin with), so that the two ▲s, D A B and C B A, would have two sides of the one (namely, D A and A B), equal to two sides of the other (namely, C B and B A), each to each, and the included (D A B) of the first ▲ equal to the included (CBA) of the second. Therefore, according to the fourth proposition, these two As would be equal to each other in every respect, and among others, in

area.

But the ADA B is only a part of the ▲ C B A. Therefore, if we were to suppose that, while the ZC A B is equal to the C B A, the side A C could be longer than C B, we should be compelled to admit that a part could be equal to the whole of which it is a part.

Now, a supposition which leads to an absurdity must be itself absurd.

Therefore it is absurd to suppose that A C could be longer than C B, while the s CAB and C B A are equal.

Next, suppose that the side B C were longer than AC,

We should then be able to cut off

from B C a part, B E, equal to A C.

Suppose that this were done, and that the points A and E were joined by the right line A E,

We should then have two As CAB and EBA, in which the sides CA and A B of the one would be

A

C

E

B

equal to the sides E B and B A of the other, each to each, and the included CA B of the first, equal to the included E B A of the other.

Therefore, according to the fourth proposition, these As would be equal in every respect, and, among others, in area.

But the AC A B is only a part of the ▲ E BA.

Thus it appears that if we were to suppose that the side B C is longer than the side A C, while the CAB is equal to the CBA, we should be compelled to admit that a part is equal to the whole of which it is a part.

But a supposition which leads to an absurdity must be itself absurd.

Hence it is absurd to suppose that BC is longer than A C, while the CB A is equal to the CAB.

It has thus been proved that A C cannot be greater than B C, and also that B C cannot be greater than A C-i. e., that A C and B C are not unequal.

But A C and BC must be either equal or unequal; therefore, as they are not unequal, they must be equal.

(Let this proposition be gone through next with the figure in a different position, and then with the aid of different letters to mark the sides, &c.)

Cor. It follows from this that all the sides of an equi-angular ▲ are equal.

PROPOSITION VIL.

Upon the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated in the one extremity of the base equal to one another, and likewise those that are terminated in the other extremity of the base equal to one another.

That is, supposing that there were two As such as A CB and A D B, on the same base A B, and on the

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