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Then, if we were to join L and B by the straight line B L, we should have a ▲ LAB, in which the side LA would be equal to the side DF in the A DEF, the side A B to the side D E, and the LAB equal to the FDE.* That is to say, these two As would have two sides and the between them in the one, equal respectively to two sides and the between

them in the other.

Consequently (Prop. IV.) these ▲s would also be equal in every other respect.

Among these respects would be that the LBA would be equal to the

Now the

FED.

FE D is equal to the ▲ CBA. (This is one of the points supposed at first respecting the As A B C and D E F.)

Therefore also the LB A would be equal to the OBA:-That is to say, a part would be equal to the whole of which it is a part—which is an impossibility.

It follows, therefore, that the supposition which would lead of necessity to this impossibility (namely, that the side A C is greater than the side DF) must itself be impossible. That is to say, the side A C is not greater than the side D F.

In a similar manner, it may be shown that the side A C is not less than the side D F.†

*This we know to start with.

†The proof of this at length is as follows:

F

Suppose it possible that DF should be longer than AC, we should then be able to cut off a part, DP, equal to AC. If this were done, and the points P and E were joined by the right line PE, we should have a ▲PDE, in which the side PD would be equal to the side C A, the side D E to the side A B, and the PDE to the CAB.

A

BD

E

But the line AC must be either greater than D F, less than DF, or equal to D F.

It follows, therefore, since A C can be neither greater nor less than D F, that it must be equal to DF.

Hence the two As, CA B and FDE, have two sides of the one (namely C A and A B) equal respectively to two sides of the other (namely F D and D E), and the included CA B in the first equal to the included

FDE in the second.

It follows, therefore (Prop. IV.), that these ▲s are also equal in every other respect.

Case 2. Next suppose that the As BCD and FGH have the DBC in the one

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In the proof of this case also the first step consists in

Consequently (according to the fourth proposition) these As would be equal in every respect.

Among these respects would be that the PED would be equal to the CBA.

But it was supposed, to start with, that the CBA is equal to the FED.

Therefore, the PED would also be equal to the FED.

But this is impossible, for the PED is only a part of the FED.

It follows, therefore, that the supposition which led to this absurdity must itself be impossible.

That is, it is impossible that the side D F should be longer than the side A C, or in other words that the side A C should be less than the side D F.

showing that the As have one other pair of sides equal, namely, the pair consisting of that side in each which, with one of the sides supposed to be equal, includes, in each A, one of the equals (or the pair of sides to be proved equal may be described as being opposite that pair of s which are not known or supposed to be equal, to start with). In the As before us these sides must be BC and FG, which with the equal sides BD and FH, include the equal s DBC and HFG.*

If B C and F G were not equal, one or other would be the greater of the two.

Suppose the side B C were longer than the side FG, we should then be able to cut off a part, BK, equal to FG.

If this were done, and the points D and K were joined by the right line DK, we should get a A, DBK, in which the side D B would be equal to the side HF, the side BK to the side F G, and the DBK to the HFG. That is to say, two sides and the included in the one would be equal respectively to two sides and the included in the other A. It would follow (according to the fourth proposition) that these As would be equal also in every other respect.

Among these respects would be, that the DKB would be equal to the HGF.

Now it was supposed, to start with, that the D C B is equal to the HG F.

* It would be of no use to show that the sides D C and HG are equal, even if it were possible to do so, for we should still be unable to apply the fourth proposition to prove that the As are equal in every other respect. For though we should have two sides in the one, viz., B D and D C, equal to two sides in the other, viz. F H and HG, we are not entitled to affirm (before it is proved), that the BDC is equal to the FHG.

It would follow, therefore, that the D KB would be equal to the DC B.

But this is impossible, for the DKB is the exterior of the ▲ DCK, formed by the production of the side CK, and is therefore greater than the DC B.

Consequently the supposition which led to this impossibility must be itself impossible. That is to say, it is imposible that the side B C should be greater than the side F G.

In the same way it may be proved that the side B C is not less than the side F G.*

D

H

* The proof, at length, is as follows:If FG were greater than B C, we should be able to cut off a part, FL, equal to B C.

B

C F

'L G

If this were done, and the points H and L were joined by the straight line, H L, we should have a ▲ HFL, in which the side HF would be equal to the side DB in the ▲ DBC, the side FL to the side B C, and the included HFL to the included DBC. Consequently (according to the fourth proposition), the As HFL and D B C would also be equal in every other respect.

Among these respects would be that the HLF would be equal to the DC B.

Now the DCB we know, to start with, to be equal to the HGF.

Consequently, the HLF would be equal to the

HGF.

of the

But this is impossible, for HLF is the exterior AHGL, formed by the production of the side G L, and is therefore greater than the interior and opposite, HGF.

It follows, therefore, that the supposition which led to this impossibility must itself be impossible, that is to say, it is impossible that the side F G should be greater than the side B C.

G

But the side B C must be either greater than, less than, or equal to, the side F G.

Consequently, as it is neither greater nor less than FG, it is equal to F G.

Hence, in the As DBC and H FG, two sides of the one, namely D B and B C, are equal to two sides of the other, namely HF and FG, each to each, and the included DBC of the one is equal to the included HFG of the other.

Consequently, according to the fourth proposition, theses are also equal in every other respect.

This case of the proposition may next be proved, supposing that the angles D B C and DCB are equal to the

8 HFG and HG F, each to each, and the side DC to the side HG (in the last figure).

By altering the position of the letters to that indicated in the following diagram, the demonstration of the proposition will remain word for word the same.

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As another variation let the proposition be gone through with the changes which will be occasioned by placing the letters as in the following diagram :—

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The 4th, 8th, and 26th propositions contain all the conditions essential for the equality of two As in all respects,

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