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same side of it, then if the two sides A C and A D, terminating at A, were equal to each other, the sides

C D

BC and B D, terminating at the other extremity of the base, could not also be equal to each other; or if BC and BD were equal, then A C and AD could not also be equal to each other. It is not meant that neither pair of conterminous sides can be equal, but that both pairs cannot be equal at the same time.

*

A

B

This proposition is proved by showing that if we were to suppose it possible that there could be two such As as are described in the enunciation, we should be compelled to admit something to be true which is obviously impossible.

For the proof of the proposition we require,

1. To be able to draw a straight line from one given point to another. (Post. I.)

2. To produce a given finite right line to any length. (Post. II.)

We must also know,

1. That if a magnitude be less than one of two equal magnitudes, it is also less than the other.

2. That if a magnitude be less than the smaller of two unequal magnitudes, it is also less than the larger.

3. That if two sides of a ▲ are equal, the s

*The beginner must not suppose that an elaborate proof of this proposition is unnecessary, because the inequality of one or other pair of sides is obvious to the eye. Geometrical truths are never to be settled by this test. And, in fact, if a scale sufficiently large were adopted, it would be possible to draw two As placed like those in the diagram, such that the inequality of the sides could scarcely be detected by the most accurate measurement.

D

C D

opposite to them are also equal, and if the equal sides be produced, the s on the other side of the base are likewise equal (Prop. V.).

Let us suppose that there could be two As A C B and A D B, on the same base A B, and on the same side of it, and such that A C and A D are equal to one another, and also B C and B D equal to one another.* First let the As be such that the vertex of each falls outside the other.

A

B

Join the points C and D by the right line C D.

In the AAC D, if A C and A D be equal to one another, the AUD is equal to the ADC

(Prop. V.)

If the sides B C and B D could also be equal to one another, then also in the ▲ B C D the

would be equal to the But the

a part of it).

BD C.

B C D

B C D is less than the A CD (being

Therefore the Z B C D must also be less than the ZAD C, which is equal to the AC D.

But the

A D C is less than the

B D C (being

a part of it). Therefore the

BCD, which is less than the

A D C, must also be less than the ▲ B D C.

But it was shown before, that if the sides B C and BD be equal, the s B C D and B D C are equal to each other.

Thus the supposition that, in the As A B C and A B D, the sides, A C and A D, could be equal to

* The beginner must beware of supposing that the cbject of the proposition is to prove that the sides AC and AD are not equal to the sides B C and B D; a mistake which the writer has found made more frequently than might be supposed.

each other, and that the two sides B C and B D could also be equal to each other, renders it necessary that we should admit that two ≤s (namely B C D and B D C) are both equal and unequal at the same time, -which is an impossibility.

Now the supposition which leads us of necessity to this absurdity, must itself be absurd. That is, it is absurd to suppose that the two pairs of sides, A C and A D, and B C and B D, can be equal at the same time. (If one pair be equal, the other pair must be unequal.)

Next let us suppose it possible that there should be two ▲s on the same base and on the same side of it, having the sides that terminate at one extremity of the base equal, and likewise those that terminate at the other extremity of the base equal, so placed that the vertex of one of the ▲s (A B C) falls within the other ▲ (A B D).

Join the points D and C by the right line D C.

Produce the line A D to any point E, and the line A C to any point F.

F

E

D

B

A

If it were possible that the sides AD and A C should be equal to one another, and also the sides B C and BD equal to one another, then the ADAC would be an isosceles A, and therefore the s E D C and F C D, formed by producing the equal sides A D and A C, would be equal to one another; and, in the ▲ CB D, the s BCD and B D C would be equal to one another.

But the B D C is less than the E D C (being a part of it): therefore the B D C is also less than the FCD, which is equal to ED C. Much more then is the B D C less than the B C D, which is greater than the F C D.

But it was before shown that, if we suppose the sides B C and B D to be equal to each other, as well as A D and A C, then the B D C must be equal to the BC D.

But it is impossible that the same s (BDC and B C D) should be both equal and unequal.

Consequently, the supposition which led to this impossibility must be absurd. That is, it is absurd to suppose that, on the same base, and on the same side of it, there can be two As, having the sides terminating in one extremity of the base equal to one another, and likewise those terminating in the other extremity of the base equal to one another.

PROPOSITION VIII.

If two triangles have the three sides of the one equal to the three sides of the other, each to each, then the triangles will also be equal to each other in every other respect, that is, the angles of the triangles will be equal, each to each, and the triangles will be equal to each other in area.

For proving this proposition we must know,

1. That two straight lines may be so placed one
upon 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 upon the same base, and upon the same
side of it, there cannot be two As having both
pairs of conterminous sides equal. (Prop. VII.)

Let A B C and D E F be two As, having the side A B equal to the side D E, the side A C to the side D F, and the side C B to the side FE.

C

F

A

B

D

E

The proof of the proposition consists in showing that the two As may be placed one on the other, so as to coincide in every part.

Place the point D on the point B, and let the line DE lie along the line A B.

Then, because the line D E is equal to the line A B, the point E will fall upon the point B.*

C F

Let the As lie on the same side of the base A B. Then the line D F must lie along the line A C, and the line E F along B C, the point F coinciding with the point C. For if they did not, but lay in any other position, as, for example, in the annexed figure, we should have two As, A C B and A F B, standing on the same base, and on the same side of it, and having the conterminous sides A C and A F equal, and likewise the conterminous sides B C and B F equal.

A

B

This was proved in the last proposition to be impossible. Consequently, the supposition which leads to this impossibility must itself be impossible. That is, it is impossible, when the side D E has been placed so as to coincide with the side A B, that the side D F should lie in any other position than along A C, and the side E F in any other position than along B C.

Therefore, as they must lie in some position or other, DF must lie along A C, and E F along B C.

Consequently, as the sides F D and D E coincide in direction with C A and A B, the

FDE is equal to

the CA B. In like manner the FED is equal to the CBA, and the

and the area of the one the other.

DFE to the AC B,

is equal to the area of

Beware of saying, "Place the point D upon the point A, and let the line D E lie along the line A B and let the print E fall on the point B." (See p. 9.)

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