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PROPOSITION XXVII: THEOREM.

If a straight line, falling upon two other straight lines, make the alternate angles equal to one another ; these two straight lines must be parallel.

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

make the alternate 2s AGH, GHD equal.

Then must AB be II to CD.

For if not, AB and CD will meet, if produced, either towards B, D, or towards A, C.

Let them be produced and meet towards B, D in K.
Then GHK is a A ;

and :: LAGH is greater than 2 GHD. But

LAGH= GHD, which is impossible.

I. 16.

Нур.

.. AB, CD do not meet when produced towards B, D. In like manner it may be shewn that they do not meet when produced towards A, C. .: AB and CD are parallel.

Def. 26.

Q. E. D.

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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. EGB=corresponding 2 GHD, or
II. 28 BGH, GHD together=two rt. 2 s.
Then, in either case, AB must be || to CD.

I.
:: 2 EGB is given= 2 GHD,

Нур. and 2 EGB is known to be= . AGH,

I. 15. ..LAGH= LGHD; and these are alternate 28; .. AB is || to CD.

I. 27. II. ::25 BGH, GHD together=two rt. 28, Нур.

and zs BGH, AGH together=two rt. 28, I. 13. .:28 BGH, AGH together= 25 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 us BGH, GHD together less than two rt. 7 8. Then must AB, CD meet when produced towards B, D.

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For if not, suppose AB and CD to be parallel.
Then :: 28 AGH, BGH together=two rt. ZS,

I. 13. and 2 s GHD, BGH are together less than two rt. ¿S,

.: LAGH is greater than 2 GHD. Make _MGH= LGHD, and produce MG to N. Then :: the alternate s MGH, GHD are equal, .. MN is || 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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F
Let the st. line EF fall on the parallel st. lines AB, CD.
Then must

I. Ls BGH, GHD together=two rt. Z s.
II. - AGH=alternate 2 GHD.

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

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

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

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

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

Post. 6 which cannot be, for they are parallel, :: Z8 BGH, GHD together=two rt. Z s. II. : %s BGH, GHD together=two rt. Z S, and Zs BGH, AGH together=two rt. ZS,

I. 13. .: 28 BGH, AGH together= ls BGH, GHD together, and ..LAGH= LGHD.

Ax. 3. III. : LAGH= LGHD, and L AGH= 2 EGB,

I. 15. ... EGB= LGHD.

Ax. 1. Q. E. D.

EXERCISES,

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