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A B and CV are two straight lines met by the

straight line EF, and the
28 BEF and EFD are
together equal to two right
28. It has to be proved
that the right lines AB
and C D are parallel.
Proof. Because

the

straight line FE meets the straight line AB, the adjacent Zs A EF and BEF are together equal to two right Zs. (Prop. XIII.)

But the Z8 BEF and EFD are also together equal to two right Zs.

Therefore the sum of the <s AEF and BEF is equal to the sum of the <s BEF and EFD.

If from each of these equal sums we take away the _ BEF, the remainders are equal. That is to say, the < AEF is equal to the < EFD.

But AEF and EFD are alternate <8.

Therefore, as these alternate s are equal, the lines A B and CD will never meet when produced; or in other words are parallel. (Prop. XXVII.)

Thus far there is no difficulty whatever with respect to parallel lines. The proof of the next proposition, however, requires that we should assume the truth of the 11th axiom. Now this axiom, though no one ever doubted its truth, is by no means so obviously true as the other axioms. In fact there are several propositions of which a detailed proof is given in geometry, the truth of which is more readily perceived than that of the 11th axiom. Numerous attempts have, in consequence, been made to supersede the necessity for the use of this axiom; but entirely without success, as some axiom or other is always assumed which is quite as difficult as the 11th. The difficulty, of course, arises from the fact that the definition of parallel lines is of a negative character, and therefore, without some positive axiom, would not enable us to found any inferences upon it.

PROPOSITION XXIX.

If two straight lines are parallel and a third straight line intersects them, then the alternate angles so formed will be equal, the exterior angle will be equal to the interior and opposite angle on the same side of the intersecting line, and the two interior angles on the same side of the intersecting lino will together be equal to two right angles.

For the proof of this proposition we must know,1. That if two straight lines meet a third in such

a way that the two interior <s on the same side of it are together less than two right Z8, those straight lines will meet if produced.

(Ax. XI.) 2. That magnitudes that are equal to the same are

equal to one another. (Ax. I.) 3. That if equals or the same be taken from equals,

the remainders are equal. (Ax. III.) 4. That if one straight line meets another, the

adjacent s so formed are together equal to

two right 8. (Prop. XIII.) 6. That if two straight lines intersect one another,

the vertically opposite s are equal. (Prop. XV.) The 29th proposition in reality consists of three distinct propositions, classed together under one heading or enunciation, and corresponding respectively to the 27th and the two parts of the 28th proposition, of which they are the converse.

As all three cases of the 29th proposition depend upon the application of the 11th axiom, it will be most convenient to take the third case first, as in it the axiom in question is applied most directly.

1. If two straight lines are parallel, and a third meets them, the two interior angles on the same side of the intersecting line are together equal to two right angles.

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The two lines A B and C D are parallel (i.e., how

ever far they may be pro-
duced, they will
meet). The line EH inter-
sects them. We have to
prove that the Zs BFL
and FLD are together equal
to two right Zs.

The proof of this is of the indirect kind. If the Zs BFL and FLD are not together equal to two right Zs, they must be either less than two right Zs, or greater than two right Zs.

If they were less than two right <s, then the lines FB and L D would meet if produced (according to the 11th axiom).

But this is impossible, for we know, to start with, that the lines A B and CD are parallel (i. e., can never meet).

Consequently the supposition that would lead to this impossibility must be itself impossible.

That is to say, it is impossible that the <8 BFL and FLD should be together less than two right s.

Again-suppose the <s BFL and FLD were together greater than two right Zs. Since the <8 BFL and LF A are together equal to two right 28 (according to Prop. XIII.), and the Z8 FLD and FLC are also equal to two right Zs, the four Z8 BFL, LFA, FLD, and FLC are together equal to four right s. Therefore, it would follow that on taking away from the sum of these four Zs the two Zs BFL and FLD, if the latter were together greater than two right Z8,

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the remaining Zs AFL and FLC would be together less than two right Zs.

But if the two straight lines A F and C L meeting a third, FL, made the interior <8 AFL and FLC together less than two right <s, the two straight lines FA and LC would meet if produced. (Ax. XI.)

But this is impossible, because we know, to start with, that FA and L C are parallel.

Consequently, it is impossible that the two Zs AFL and FL C can be together less than two right <8, and therefore it is impossible that the two s BFL and FLD should be together greater than two right s.

Hence, as the <s BFL and FLD can be neither less than two right Zs nor greater than two right <s, they must together be to two right Zs.

Moreover, since BFL and FL D are together equal to two right 8, if they be taken away from the four <s BFL, L FA, FLD, and FLC, the remaining <s LFA and FLC will be together equal to two right Zs.

2. If two straight lines are parallel and a third straight line intersects them, the alternate angles so formed will be equal.

Let A B and C D be parallel, and let E H intersect them in the points and L. Then the < AF L will be equal to the < FLD, and the < BFL to the < FLC.

It has already been proved that the two interior Z8 BFL and FLD are together equal to two right 8.

Also, since the straight line L F meets the straight line A B, the adjacent Z8 AFL and BFL are together equal to two right Zs. (Prop. XIII.)

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Therefore the sum of the <8 BFL and FLD is equal to the sum of the Zs AFL and LFB. (Ax. I.) Take

away the common Z, BFL, and the remainders, the s AFL and FLD will be equal to one another.

Also, it was proved in the last case that the two interior Zs AFL and FLC are together equal to two right Zs.

But the <s AFL and BFL are also together equal to two right Zs. Therefore the <s AFL and FLC are together equal to the sum of the two <8 AFL and LFB. Take

away the common Z, AFL, and the remainders, namely, the <s BFL and FLC, will be equal.

3. If two straight lines are parallel and a third straight line intersects them, each of the exterior angles is equal to the interior and opposite angle on the same side of the intersecting line. Let A B and C D be parallel, and let E H intersect

them in the points F and L. We have to prove that the ZEF B is equal to the 2 FLD, and the < EF A to the < FLC.

Proof. Since the straight lines A B and E H intersect

one another at the point F, the vertically opposite s E F B and A FL are equal. (Prop. XV.)

But it was proved in the last case that the < AFL is equal to the 2 FLD.

Therefore the < EFB is also equal to the < FLD.

Again, the vertically opposite <8 EF A and BFL are equal. (Prop. XV.)

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