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BC, the remainders AF and C L are equal (Ax. 3). Moreover, the line AL was proved equal to the line CF, and the <OL A (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 the two As O L A and AFC, the two sides CL and LA of the one are equal respectively to the two sides AF and FC of the other, and the included < CL A 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 < CAL is equal to the < ACF.

3. If from the equal <s B AL and BC F, the equal Z8 CAL and A C F be taken away, the remainders, namely the <s B A C and B C A, will be equal to each other (Ax. 3). But these are the 8 opposite the equal sides of the isosceles A A B C. Therefore the first part of the proposition has been proved.

The second part has also been proved. For the <s at the other side of the base, namely F A C and LC A, are parts of the As A F C and C L A, which have been proved to be equal in every respect.

Cor. All the <s of an equilateral A are equal.

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The proof of the next proposition is of the kind called inairect. 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 menB, 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.

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 necessary consequence of it. Now it is assumed as an axiom, that “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 thie 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 Zs 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 A-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 Aalso equal to the included < of the second A, then those two As are equal in every respect.

Let A B C be à A, with respect to which it is

known that the < CAB is equal to the LCBA. We have to prove that the side A C is equal to the side C B.

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A

B

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

That A C and C B are not unequal

is proved by showing that it leads to an absurdity to suppose that they are unequal.

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

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 CBA, 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 <D A B equal to the < OBA (we were supposed to know this last point to begin with), so that the two As, D A B and CBA, would have two sides of the one (namely, D A and A B), equal to two sides of the other (namely, 0 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 A D A B is only a part of the AC B A.

Therefore, if we were to suppose that, while the LC A B is equal to the ZOBA, the side AC 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 8 C A B and C B A are equal.

Next, suppose that the side BC 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

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

C

А

B

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 A EBA.

Thus it appears that if we were to suppose that the side B C is longer than the side A C, while the < C A B 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 B C is longer than A C, while the < C B A is equal to the < C A B.

It has thus been proved that A C cannot be greater than B C, and also that B C cannot be greater than AC-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 A are equal.

PROPOSITION VII. Upon the same base, and on the same side of it, there cannot be two triangles having their sides wbich 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 ono another.

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

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