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PROPOSITION XIII. THEOREM.

The angles which one straight line makes with another upon one side of it are either two right angles, or together equal to two right angles.

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Let AB make with CD upon one side of it the s ̊ ABC, ABD.

Then must these be either two rt. Ls,

or together equal to two rt. 4 s.

First, if ABC= ▲ ABD as in Fig. 1,

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Then sum of 4s ABC, ABD=sum of 4 s EBC, EBA, ABD, and sum of ▲s EBC, EBD=sum of 4 s EBC, EBA, ABD ; .. sum of 4s ABC, ABD=sum of 4 s EBC, EBD ;

Ax. 1.

.'. sum of 4 s ABC, ABD=sum of a rt. ▲ and a rt. ▲
:. ¿s ABC, ABD are together=two rt. 4 s.

Q. E. D.

Ex. Straight lines drawn connecting the opposite angular points of a quadrilateral figure intersect each other in 0. Shew that the angles at O are together equal to four right angles.

NOTE (1.) If two angles together make up a right angle, each is called the COMPLEMENT of the other. Thus, in fig. 2, 4 ABD is the complement of ▲ ABE.

(2.) If two angles together make up two right angles, each is called the SUPPLEMENT of the other. Thus, in both figures, ABD is the supplement of ABC.

PROPOSITION XIV. THEOREM.

If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines must be in one and the same straight line.

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At the pt. B in the st. line AB let the st. lines BC, BD, on opposite sides of AB, make ≤s ABC, ABD together=two rt. angles.

Then BD must be in the same st. line with BC. For if not, let BE be in the same st. line with BC. 4s ABC, ABE together=two rt. 4 s.

Then

And

48 ABC, ABD together=two rt. 4 s.
.. sum of 8 ABC, ABE=sum of 4 s ABC, ABD.

Take away from each of these equals the

then ABE 2 ABD,

=

I. 13.

Hyp.

ABC;

Ax. 3.

that is, the less-the greater; which is impossible,

.. BE is not in the same st. line with BC.

Similarly it may be shewn that no other line but BD is in

the same st. line with BC.

.. BD is in the same st. line with BC.

Q. E. D.

Ex. Shew the necessity of the words the opposite sides in the enunciation.

PROPOSITION XV. THEOREM.

If two straight lines cut one another, the vertically opposite angles must be equal.

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Let the st. lines AB, CD cut one another in the pt. E.

Then must 4 AEC ▲ BED and

AED=L BEC.

For AE meets CD,

.. sum of 4s AEC, AED=two rt. 4 s.

I. 13.

And DE meets AB,

.. sum of 4s BED, AED=two rt. 4 s;

.. sum of ¿s AEC, AED=sum of ▲ 8 BED, AED;

. LAEC= L BED.

I. 13.

Ax. 3.

Similarly it may be shewn that

= L

AED BEC.

Q. E. D.

COROLLARY I. From this it is manifest, that if two straight lines cut one another, the four angles, which they make at the point of intersection, are together equal to four right angles.

COROLLARY II. All the angles, made by any number of straight lines meeting in one point, are together equal to four right angles.

Ex. 1. Shew that the bisectors of AED and BEC are in the same straight line.

Ex. 2. Prove that AED is equal to the angle between two straight lines drawn at right angles from E to AE and EC, if both lie above CD.

Ex. 3. If AB, CD bisect each other in E; shew that the triangles AED, BEC are equal in all respects.

NOTE 3. On Euclid's definition of an Angle.

Euclid directs us to regard an angle as the inclination of two straight lines to each other, which meet, but are not in the same straight line.

Thus he does not recognise the existence of a single angle equal in magnitude to two right angles.

The words printed in italics are omitted as needless, in Def. VIII., p. 3, and that definition may be extended with advantage in the following terms :—

DEF. Let WQE be a fixed straight line, and QP a line which revolves about the fixed point Q, and which at first coincides with QE.

W

Then, when QP has reached the position represented in the diagram, we say that it has described the angle EQP.

When QP has revolved so far as to coincide with QW, we say that it has described an angle equal to two right angles.

Hence we may obtain an easy proof of Prop. XIII. ; for whatever the position of PQ may be, the angles which it makes with WE are together equal to two right angles.

Again, in Prop. xv. it is evident that ▲ AED= ▲ BEC, since each has the same supplementary AEC.

We shall shew hereafter, p. 149, how this definition may be extended, so as to embrace angles greater than two right angles.

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If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

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Let the side BC of ▲ ABC be produced to D.
Then must ACD be greater than either

Bisect AC in E, and join BE.

CAB or ABC.

I. 10.

Produce BE to F, making EF=BE, and join FC.

Then in AS BEA, FEC,

·· BE=FE, and EA=EC, and ▲ BEA= ▲ FEC,

I. 15.

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EAB,

that is,

.. ACD is greater than

4 ACD is greater than 4 CAB.

Similarly, if AC be produced to G it may be shewn that

and

4 BCG is greater than ▲ ABC.

LBCGL ACD;

.. ▲ ACD is greater than ▲ ABC.

I. 15.

Q. E. D.

Ex. 1. From the same point there cannot be drawn more than two equal straight lines to meet a given straight line.

Ex. 2. If, from any point, a straight line be drawn to a given straight line making with it an acute and an obtuse angle, and if, from the same point, a perpendicular be drawn to the given line; the perpendicular will fall on the side of the acute angle.

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