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For, from the point C draw a straight line CE perpendicular to A B—that is, making the adjacent
Z8 EC A and ECB equal to one another.
It is evident that the ZEC B is the sum of the two Zs ECD and DCB.
Therefore, the sum of the two <8 ACE and ECB is the same as that of the three <8 A CE, EC D, and D C B.
Again the < ACD is the sum of the Z8 ACE and ЕСВ.
Therefore, the sum of the two Zs ACD and DCB is the same as the sum of the three xs A CE, ECD, and D C B.
The two right Zs ACE and ECB were before shown to be equal to the sum of the same three s.
Therefore, the sum of the Zs A CD and D C B is the same as that of the right Z8 ACE and ECB; and as all right Zs are equal (see p. 13), the sum of the Zs A C D and D C B is the same as that of any two right 28.
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
1. If one straight line meets another, the adjacent
Zs which it forms with it are together equal to
two right Zs. (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.
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 Zs DC A and A CE together equal to two right <8.
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:
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 adjacent Zs D C A and A CF, so formed, would together be equal to two right Z8, 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 right 8.
We should be compelled to admit, therefore, that the Z8 D C A and A C F taken together are equal to the Zs D C A and AC E taken together.
If this were so, it would follow that, on taking the
LDCA away from both these equal sums, the remainders, ACE and ACF 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 DO 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.
If two straight lines cut one another, the vertically opposite angles will be equal to one another.
To prove this proposition we must know,–
adjacent Z8 which it forms with it are together
equal to two right Zs. 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 D that the _ A E D is equal to the
<CE B, and the <AEC to the ZDE B.
Proof. A E is a straight line meeting the line C D in the point E, and forming adjacent Zs A E C and A ED.
Consequently, these two Zs are together equal to two right Zs.
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 s A EC and C E B are together equal to two right Zs. (Prop. XIII.)
But quantities that are equal to the same are equal to each other.
Therefore, the sum of the Zs AEC and A E D is equal to the sum of the Zs A E C and C E B, each sum being equal to the sum of two right <s.
Take away the ZA E C 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 D E meets the line A B, the adjacent <s A E D and D E B are together equal to two right <s, and it was before shown that the Zs A E D and A E C are together equal to two right Zs.
Consequently, the sum of the Zs A E D and D E B is equal to the sum of the 8 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.
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
vertioally opposite Zs 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 A equal to the < included by the corresponding pair of sides in the second A, then the As will also be equal in every other respect.
(Prop. IV.) Let A B C be a A, of which one side, A B, has
been produce l. We have to show that each of the <s BCA and BAC (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 AB wbich has been produ
In order to compare it with the < CBD, make the following construction :
Bisect the line C B in the point E.
Join the points A and E by the right line A E. (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.)