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For, from the point C draw a straight line CE perpendicular to A B-that is, making the adjacent Z8 ECA and E CB equal to one another.

E D

с

B

It is evident that the ECB is the sum of the two sECD and DCB.

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.

A C D is the sum of the sACE and

Therefore, the sum of the two 8 ACD and D C B is the same as the sum of the three

and D C B.

s A CE, ECD,

The two right s ACE and ECB were before shown to be equal to the sum of the same three Zs.

Therefore, the sum of the s 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 LSA 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 D C 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 right ≤8.

We have to show that if one of

the lines (as D C) be produced,

D

B

the produced part coincides in direction with C E. This is shown as follows:--

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 CE are together equal to two rights.

We should be compelled to admit, therefore, that the 28 D C A and A C F taken together are equal to the Zs DCA 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 CE and A C F are equal.

But this is impossible, for the ACF 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 C E.

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 C E. Therefore, as D C produced must lie in some direction or other, it must coincide with C E.

PROPOSITION XV.

If two straight lines cut one another, the 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 right

s.

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

A

E

-D that the

B

AED is equal to the

ZCE B, and the AE C to the
Z DEB.

Proof. A E is a straight line

meeting the line C D in the point E, and forming adjacent s A E C 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

and C E B.

s AEC

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 s A E C and A E D 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 ED is equal to the ▲ C E B.

Again, since the straight line D E 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 AED and A E C are together equal to two rights.

Consequently, the sum of the s A E D and DEB 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 DE 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

C

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

F

Let A B C be a ▲, of which one side, A B, has been produced. We have to show that each of the Zs BCA and BAC (which are not adjacent to the CBD), is less than the

B

D

exterior CBD.

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

which has been produ

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

Bisect the line CB in the point E.

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

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

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