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Lastly, let one of the third angles ACB, DFE be a right

angle.

If ACB be a rt. 4,

then

BGC is also a rt. 4;

LS BCG, BGC together two rt. 4s, which is impossible.

Again, if DFE be a rt. 4,

then

AGB is a rt. 4, and. BGC is a rt. 4.

Hence BCG is also a rt. 4,

I. 17.

:. 48 BCG, BGC together=two rt. 4s, which is impossible.

Hence AC is equal to DF,

I. 17.

and the As ABC, DEF are equal in all respects.

Q. E. D.

COR. From the first case of this proposition we deduce the following important theorem:

If two right-angled triangles have the hypotenuse and one side of the one equal respectively to the hypotenuse and one side of the other, the triangles are equal in all respects.

NOTE. In the enunciation of Prop. E, if, instead of the words if one of them be a right angle, we put the words both right angles, this case of the proposition would be identical with our 1. 26.

SECTION II.

The Theory of Parallel Lines.

INTRODUCTION.

WE have detached the Propositions, in which Euclid treats of Parallel Lines, from those which precede and follow them in the First Book; in order that the student may have a clearer notion of the difficulties attending this division of the subject, and of the way in which Euclid proposes to meet them.

We must first explain some technical terms used in this Section :

If a straight line EF cut two other straight lines AB, CD, it makes with those lines eight angles, to which particular names are given.

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The angles numbered 1, 4, 6, 7 are called Interior angles.

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The angles marked 1 and 7 are called alternate angles. The angles marked 4 and 6 are also called alternate angles.

The pairs of angles 1 and 5, 2 and 6, 4 and 8, 3 and 7 are called corresponding angles.

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 48 AGH, GHD equal.

Then must AB be || 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.

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.. 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.

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.

I.

II.

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48 BGH, GHD together two rt. s. Then, in either case, AB must be || to CD.

LEGB is given = L GHD,

and EGB is known to be = LAGH,

..

LAGH= L GHD;

Hyp.

I. 15.

II.

..

and these are alternate 48;

.. AB is to CD.

LS BGH, GHD together=two rt. 4s,

and 48 BGH, AGH together=two rt. 4s,

I. 27.

Hyp.

I. 13.

4 s BGH, AGH together = 48 BGH, GHD together;

...

LAGH= L GHD;

.. AB is to CD.

I. 27.

Q.E. D.

NOTE V. On the Sixth Postulate.

We explained in Note Iv., page 32, that Euclid's Sixth Postulate is the converse of the 17th Proposition.

In the place of this 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 ▲ 8 BGH, GHD together less than two rt. 4 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

LS AGH, BGH together = two rt. 48,

I. 13.

and 4s GHD, BGH are together less than two rt. 4s, .. LAGH is greater than 4 GHD.

Make

MGH= ▲ GHD, and produce MG to N.

Then the alternate 8 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. F. D.

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