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For, from the point C draw a straight line CE perpendicular to A B—that is, making the adjacent 8 ECA and E C B equal to one another.

E D

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Therefore, the sum of the two s

ACE and ECB is the same as that of the three Zs ACE, E CD, and D C B.

Again the

ECB.

ACD is the sum of the sACE and

Therefore, the sum of the two s ACD and DCB is the same as the sum of the three

and D C B.

The two right

s A CE, E CD,

s ACE and ECB were before

shown to be equal to the sum of the same three

s.

Therefore, the sum of the 8 A CD and D C B is

the same as that of the right

s ACE and E C B;

and as all rights are equal (see p. 13), the sum of the ZSA C D and D C B is the same as that of any two rights.

PROPOSITION XIV.

If two straight lines meet a third at the same point, but on opposite sides of the line, and make the adjacent angles together equal to two right angles, then those two straight lines are in one and the same straight line.

In other words, we have to prove that if either of the straight lines be produced, the produced part will coincide in direction with the other of the two.

For the proof of this proposition, we must know that,

1. If one straight line meets another, the adjacent Zs which it forms with it are together equal to two rights. (Prop. XIII.)

2. Quantities that are equal to the same are equal to each other. (Ax. I.)

3. If the same quantity be taken away from each of two equal quantities the remainders are equal. (Ax. III.)

The proof of this proposition is of the indirect kind, and consists in showing that it leads to an absurdity to suppose that, if one of the two straight lines be produced, the produced part does not coincide in direction with the other line.

A

Let DC and E C be two straight lines, meeting the line A B at the point C, but on opposite sides of the line A B, and making the s DCA and ACE together equal to two rights.

D

B

We have to show that if one of the lines (as D C) be produced, the produced part coincides in direction with C E. This is shown as follows:

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F

E

Suppose it possible that the line D C produced did not coincide in direction with C E, but lay in some other direction, as CF.

We should then have the line AC meeting the straight line D F at the point C.

The adjacents DC A and A CF, so formed, would together be equal to two rights, as was proved in the last proposition.

Now we know, to start with, that the s A CD and A C E are together equal to two rights.

We should be compelled to admit, therefore, that the Zs D C A and A C F taken together are equal to the Zs D C A and A CE taken together.

If this were so, it would follow that, on taking the

D

ZDCA away from both these equal sums, the remainders, A C E and A C F are equal.

But this is impossible, for the AC F is only a part of the ACE.

This absurdity is a necessary inference from the supposition that the line D C, when produced, does not coincide in direction with CE.

But a supposition which leads of necessity to an impossibility is itself absurd.

Consequently, it is impossible that the line D C when produced should not coincide in direction with CE. Therefore, as D C produced must lie in some direction or other, it must coincide with C E.

PROPOSITION XV.

the

If two straight lines cut one another, vertically opposite angles will be equal to one another.

To prove this proposition we must know,—

1. That if one straight line meets another, the adjacents which it forms with it are together equal to two rights.

2. If the same quantity be taken from two equal quantities, the remainders are equal. (Ax. III.)

Let A B and CD be two straight lines intersecting in the point E. We have to prove

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B

A E D is equal to the

ZCE B, and the

DEB.

A E C to the

Proof. A E is a straight line

meeting the line C D in the point E, and forming

adjacents AEC and A E D.

Consequently, these two ▲s are together equal to two rights.

Again, C E is a straight line, meeting the straight line A B in the point E, and forming adjacent SAEC and C E B.

Consequently, the sA EC and C E B are together equal to two rights. (Prop. XIII.)

But quantities that are equal to the same are equal to each other.

Therefore, the sum of the

SA E C and AED is equal to the sum of the s A E C and C E B, each sum being equal to the sum of two rights.

Take away the AEC from each of these equal sums, and the remainders will be equal, i.e., the A E D is equal to the CE B.

Again, since the straight line DE meets the line A B, the adjacent s AED and D E B are together equal to two rights, and it was before shown that thes A E D and A E C are together equal to two rights.

Consequently, the sum of the s A E D and DE B is equal to the sum of the s A E D and A E C.

Take away the ▲ A E D from each of these equal sums, and the remainders will be equal, i.e., the ▲ AEC is equal to the D E B.

PROPOSITION XVI.

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

For the construction employed in this proposition we must be able,—

That is, not adjacent to the exterior angle.

A

1. To join two given points by a right line. (Post. I.)

2. To produce a given finite right line to any
length. (Post. II.)

3. From the greater of two given lines to cut off
a part equal to the smaller. (Prop. III.)
4. To bisect a given finite straight line. (Prop. X.)
For the demonstration of the truth of the propo-
sition by the aid of the construction we must
know,-

1. That if two straight lines cut one another, the
vertically opposites are equal. (Prop. XV.)
2. That if two As have two sides of the one equal
to two sides of the other, each to each, and also
included by the said pair of sides in the

the
first

equal to the included by the corresponding pair of sides in the second ▲, then the As will also be equal in every other respect. (Prop. IV.)

Let A B C be a ▲, of which one side, A B, has been produce 1. We have to

C

B

F

show that each of the s BCA and B A C (which are not adjacent to the CBD), is less than the

exterior CBD.

We must first take the ACB, which is opposite to the side A B

which has been produ

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In order to compare it with the CBD, make the following construction :

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Bisect the line CB in the point E.

Join the points A and E by the right line AE. (Post. I.)

Produce the line A E to any length. (Post. II.) From the produced part of A E cut off a part, EF, equal to A E. (Prop. III.)

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