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For, from the point o draw a straight line C E perpendicular to A B—that is, making the adjacent
<s ECA and ECB equal to one another.
It is evident that the ECB is the sum of the two Z8 ECD and DCB.
Therefore, the sum of the two Z8 ACE and ECB is the same as that of the three Z8 A CE, ECD, and D C B.
Again the < AC D is the sum of the ZS ACE and ЕС В.
Therefore, the sum of the two Zs ACD and DOB is the same as the sum of the three ZS A O E, ECD, and DO B.
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 28 A CD and D C B is the same as that of the right Zs ACE and ECB; and as all right 2s are equal (see p. 13), the sum of the <s A CD and D C B is the same as that of any two right Zs.
If two straight lines meet a third at the same point, bụt 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 28. (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 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 Zs D C A and ACE together equal to two right Zs.
We have to show that if one of the lines (as D C) be produced, the produced part coincides in direction with O 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 C F.
We should then have the line AC meeting the straight line D F at the point C.
The adjacent <s D C A and A CF, so formed, would together be equal to two right <s, as was proved in the last proposition.
Now we know, to start with, that the Zs A CD and A CE are together equal to two right Zs.
We should be compelled to admit, therefore, that the 28 D C A and AC F taken together are equal to the 28 DO A and A CE taken together.
If this were so, it would follow that, on taking the
<DCA away from both these equal sums, the remainders, A C E and A C F are equal.
But this is impossible, for the Z A C F is only a part of the ZACE.
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 Zs which it forms with it are together
equal to two right <8. 2. If the same quantity be taken from two equa
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
ZOE B, and the LAE O to the
Proof. A E is a straight line meeting the line C D in the point E, and forming adjacent Zs AEC and A E D.
Consequently, these two Zs are together equal to two right Zs.
Again, CE is a straight line, meeting the straight line A B in the point E, and forming adjacent <s A EC and C E B.
Consequently, the ZSA 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 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 right Zs.
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 D E meets the line A B, the adjacent Zs A E D and D E B are together equal to two right <s, and it was before shown that the <s AE D and A E O are together equal to two right Zs.
Consequently, the sum of the Zs A E D and D EB is equal to the sum of the ZS A E D and A E C.
Take away the ZA E D from each of these equal sums, and the remainders will be equal, i.e., the < AEC is equal to the < DE B.
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
vertically opposite <s 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 28 will also be equal in every other respect. (Prop. IV.)
Let A BO bo 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 Z CBD.
We must first take the / ACB,
which is opposite to the side AB which has been produ
In order to compare it with the < CBD, makė the following construction :
Bisect the line CB 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.)