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Then, if we were to join L and B by the straight line B L, we should have a A L AB, in which the side L A would be equal to the side D F 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 As would also be equal in every other respect.

Among these respects would be that the ZLBA would be equal to the FED.

Now the _ F E 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 _ CBA:-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, tbat the side A C is greater than the side DF) must itself be impossible. That is to say, the side A Cis 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.

с

P

* 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 A C, 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 BD

E

PE, we should have a APDE, 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

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

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 _ C A B in the first equal to the included _ FDE in the second.

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

Case 2. Next suppose that the As B C D and FGH have the < DBC in the one equal to the < HFG in the other, the < D C B in the one equal to the <HGF in the other, and the sides BD and FH (opposite the equal Zs BCD and FGH) equal to one B another.

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

.D

н

K CF

G

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

Among these respects would be that the ZPED would be equal to the ZCB A.

But it was supposed, to start with, that the < CB A is equal to the <FED.

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

But this is impossible, for the <PE D is only a part of the ZFED.

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 A which, with one of the sides supposed to be equal, includes, in each A, one of the equal Zs (or the pair of sides to be proved equal may be described as being opposite that pair of Z8 which are not known or supposed to be equal, to start with). In the As before us these sides must be B C and FG, which with the equal sides B D and FH, include the equal 28 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 I 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 B K to the side F G, and the DBK to the LHFG. That is to say, two sides and the included in the one A 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 ZDKB would be equal to the < HGF.

Now it was supposed, to start with, that the L DC B is equal to the < HGF.

* 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. FH and HG, we are not entitled to affirm (before it is proved), that the < BDC is equal to the < FHG.

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It would follow, therefore, that the <D K B would be equal to the < D C B.

But this is impossible, for the <D KB is the exterior < of the ADCK, formed by the production of the side CK, and is therefore greater than the DCB.

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 FG.

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

D

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LG

* The proof, at length, is as follows: If F G were greater than B C,

н we should be able to cut off a part, F L, equal to BC.

If this were done, and the points H and L were joined by the straight line, HL, we should have

B

CI a AHFL, in which the side HF would be equal to the side D B in the AD BC, the side FL to the side B C, and the included < HFL to the included <D BC. 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 <D CB.

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

Consequently, the < HLF would be equal to the 2 HGF.

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

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 FG.

Hence, in the As D BC and HFG, 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 < DB C of the one is equal to the included < HFG of the other.

Consequently, according to the fourth proposition, these As 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 <s HFG and HGF, 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 :

AA
AN

G

K

H

F D

B

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

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