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PROPOSITION XXVIII. THEOREM.

If a straight line, falling upon two other straight lines, make the exterior angle equal to the interior and opposite upon the same side of the line, or make the interior angles upon the same side together equal to two right angles ; the two straight lines are parallel to one another.

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Let the st. line EF, falling on st. lines AB, CD, make

I. LEGB=corresponding - GHD, or
II. 28 BGH, GHD together=two rt. Z s.
Then, in either case, AB must be lI to CD.

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II. :: 28 BGH, GHD together=two rt. 28, Нур.

and 28 BGH, AGH together=two rt. 28, I. 13. .. 28 BGH, AGH together= 28 BGH, GHD together ;

.: LAGH= LGHD;
.. AB is || to CD.

I. 27.

Q. E. D.

NOTE 5. On the Sixth Postulate.

In the place of Euclid's Sixth Postulate many modern writers on Geometry propose, as more evident to the senses, the following Postulate :

Two straight lines which cut one another cannot both be parallel to the same straight line."

If this be assumed, we can prove Post. 6, as a Theorem, thus :

Let the line EF falling on the lines AB, CD make the zs BGH, GAD together less than two rt. 28.

Then must AB, CD meet when produced towards B, D.

d

G

N
B

H

For if not, suppose AB and CD to be parallel. Then : 28 AGH, BGH together=two rt. 6 s, I. 13. and 68 GHD, BGH are together less than two rt. ZS,

... AGH is greater than 2 GHD. Make 2 MGH= LGHD, and produce MG to N. Then :: the alternate zs MGH, GHD are equal, .:: MN is I to CD.

I. 27. Thus two lines MN, AB which cut one another are both parallel to CD, which is impossible.

.. AB and CD are not parallel. It is also clear that they meet towards B, D, because GB lies between GN and HD.

Q. E. D.

PROPOSITION XXIX. THEOREM. If a straight line fall upon two parallel straight lines, it makes the two interior angles upon the same side together equal to two right angles, and also the alternate angles equal to one another, and also the exterior angle equal to the interior and opposite upon the same side.

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Let the st. line EF fall on the parallel st. lines AB, CD.
Then must

I. 68 BGH, GHD together=two rt. 2 s.
II. . AGH=alternate 2 GHD.

III. . EGB=corresponding 2 GHD.
I. 28 BGH, GHD cannot be together less than two rt. 28,

for then AB and CD would meet if produced towards B and D,

Post. 6. which cannot be, for they are parallel. Nor can zs BGH, GHD be together greater than two

rt. LS,

for then 28 AGH, GHC would be together less than two rt. 28,

I. 13. and AB, CD would meet if produced towards A and C

Post. 6 which cannot be, for they are parallel, :: 28 BGH, GHD together=two rt. Z s. II. :: Z8 BGH, GHD together=two rt. Z S,

and 28 BGH, AGH together=two rt. Z , I. 13. .: 48 BGH, AGH together= 28 BGH, GHD together, and .: LAGH= LGHD.

Ax. 3. III. : LAGH= LGHD, and LAGH= EGB,

I. 15. .. LEGB= LGHD.

Ax. 1. Q. E. D.

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EXERCISES. 1. If through & point, equidistant from two parallel straight lines, two straight lines be drawn cutting the parallel straight lines ; they will intercept equal portions of those lines.

2. If a straight line be drawn, bisecting one of the angles of a triangle, to meet the opposite side ; the straight lines drawn from the point of section, parallel to the other sides and terminated by those sides, will be equal.

3. If any straight line joining two parallel straight lines be bisected, any other straight line, drawn through the point of bisection to meet the two lines, will be bisected in that point.

NOTE. One Theorem (A) is said to be the converse of another Theorem (B), when the hypothesis in (A) is the conclusion in (B), and the conclusion in (A) is the hypothesis in (B). For example, the Theorem I. A. may be stated thus :

Hypothesis. If two sides of a triangle be equal.

Conclusion. The angles opposite those sides must also be equal The converse of this is the Theorem I. B. Cor. :

Hypothesis. If two angles of a triangle be equal.

Conclusion. The sides opposite those angles must also be equal The following are other instances :

Postulate vi, is the converse of I, 17,
I, 29 is the converse of I. 27 and 28.

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PROPOSITION XXX. THEOREM.

Straight lines which are parallel to the same straight line are parallel to one another.

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Let the st. lines AB, CD be each || to EF.

Then must AB be || to CD.

Draw the st. line GH, cutting AB, CD, EF in the pts. O, P, Q.

Then :: GH cuts the || lines AB, EF, ...AOP=alternate PQF.

I. 29. And :: GH cuts the | lines CD, EF, .. extr. 2 OPD=intr. PQF;

I. 29. .: LAOP= _ OPD;

and these are alternate angles ;
.. AB is || to CD.

I. 27.

Q. E. D. The following Theorems are important. They admit of easy proof, and are therefore left as Exercises for the student.

1. If two straight lines be parallel to two other straight lines, each to each, the first pair make the same angles with one another as the second.

2. If two straight lines be perpendicular to two other straight lines, each to each, the first pair make the same angles with one another as the second.

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