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It has now to be proved that the side A B is equal to the side DE; that the < CAB is equal to the <FDE; that the < CBA is equal to the < FED; and that the A8 are equal in area.
The mode of proving this is to show that the one A may be so placed upon the other as to coincide with it in every part.
Apply the A ACB to the ADFE, so that the point A may coincide with the point D, and the straight line AC lie along the straight line DF. (Ax. X.) Then, because the lines A C and DF are equal to one another, the point C will coincide with the point F.*
Next, let the As lie on the same side of the line DF. Then, because the < ACB is equal to the ZDFE, the line C B will lie along the line FE; + and because the point C coincides with the point F, and the line CB lies along F E, and is equal to it, the point B will coincide with the point E. * You must not say,
“ Let the point C coincide with the point F;" because that would imply that you could let it coincide with it, or not, as you pleased. When one end, A, has been fixed, and the position of the line AC determined, the length of the line determines where the other end will fall. See p. 11. It will do just as well if you place the point C upon the point F, and let the line C A lie along the line FD. Then because the lines CA and FD are equal, the point A will coincide with the point D. Either way leads us to the result that the line AC will coincide exactly with the line D F.
+ You must not say, “ Let the line C B lie along the line FE;" for that would imply that you could let or make it fall in some other direction if you chose. When the position of one of the lines that contain an < is fixed, it is the size of the < that determines in what direction the other side will fall. See p. 12.
Beginners often make the mistake of saying that the It was shown before that the point A coincides with the point D. As the point B also coincides with the point E, it follows that the straight line A B lies along the straight line DE; otherwise these two straight lines would enclose a space, which is impossible. (Ax. XI.)
Thus it has been shown that the A ABC may be placed upon the A DEF in such a manner that it will coincide with it in every part. In other words, it has been shown that the side A B is equal to the side DE; that the 2 CAB is equal to the < FDE; that the < CBA is equal to the < FED; and that the As are equal in area.
Instead of first placing the line AC on the line DF, and then showing that because the <8 A CB and DFE are equal, the line C B will lie along the line F E, it would have done just as well if we had placed the line C B on the line FE, and then shown that the line CA will lie along the line F D. But it is necessary to begin by making two of those sides coincide, which we know at starting to be equal. It would be of no use to begin by placing the point A upon the point D, and letting the line A B lie along the line D E; because, if we did so, we should be unable to show that the line AC would lie along the line DF; for, we have no right to assert that the <BAC is equal to the EDF, until we have proved that it is so.
When the proposition has been mastered with the figure given above, let it be supposed that, instead of knowing, to start with, that the sides A C and C B are equal respectively to D F and F E, and the ACB to the DFE, we know
point B will coincide with the point E, because the AC B is equal to the ZDFE. The size of the < which one line makes with another can only determine the direction in which the lines lie with respect to each other. We must know the length of the lines as well, in order to be able to say where the ends of them will fall.
that the sides C B and B A are equal respectively to the sides FE and ED, and the < CB A to the < FED. By changing the position of the letters to the following,
the proof may be gone through word for word as before.
Then (in the original figure) let it be supposed that we know that the sides C A and A B are equal respectively to FD and DE, and the < CAB to the < FDE. By altering the position of the letters to the following, the proof remains word for word the same:
A F Then let As of a different shape be employed, and different letters placed at the corners.
PROPOSITION V. The angles at the base of an isosceles triangle are equal; and if the equal sides be produced, the angles at the other side of the base are likewise equal.
An isosceles A is a A which has two sides equal.
In this proposition, therefore, we are supposed to know, to start with, that two sides are equal, and this is all that we know about the A to begin with. What we have to prove is, that, if that be the case, then the s opposite the equal sides are equal to one another, and that if the equal sides be produced, the Zs at the other side of the base are equal to each other.
For the construction used in proving this pro
position we must be able,– 1. To join two given points by a straight line.
(Post. I.) 2. To cut off from the longer of two given straight
lines a piece equal to the shorter. (Prop. III.) For the proof of the proposition we must know,1. That if equals be taken from equals, the re
mainders are equal. (Ax. III.) 2. That if two As have two sides of the one equal
to two sides of the other, each to each, and have likewise the x8 contained by these two pairs of sides equal, then the As will also be
equal in every other respect. (Prop. IV.) Suppose A B C to be a A, having the sides A B and B C equal to each other. We have to prove that the
<s B A C and B C A are equal to B each other, and likewise that the s
FA C and LC A are equal to each other.
To prove this proposition, the fol
lowing construction is made:A
In B A produced take any point, 2 (as F), and from B E cut off a part,
BL, equal to B F (Prop. III.) Join
F and O by the right line F C, and A and L by the right line A L. (Post. I.)
The proof of the first part of the proposition (namely that the <BAC is equal to the < B C A) consists of three main steps. First, it is shown that the <BAL is equal to the angle BCF. Secondly, it is shown that the < CAL is equal to the < ACF. Thirdly, it is inferred that if the equal Z8 CAL and ACF be taken from the equal Z8 BAL and BCF, the remainders, namely the Z8 BAC and BCA, will be equal.
1. To prove that the <BAL is equal to the BCF. There are two As, BAL and BCF (shaded respectively by parallel lines), which overlap each other (where the lines of shading cross).
To prove that the angles B A L and BC F are equal to each other, it is necessary to show that the triangles BAL and BCF are equal to each other in every respect.
Now we know that the side A B of the first is equal to the side B C of the second; and the side B L of the first was made equal to the side B F of the second. Moreover, the < included between the sides A B and BL of the first, is the same as the _ included between the sides C B and B F of the second. So that, in these two As, two sides and the between them in the one, are equal respectively to two sides and the between them in the other. Therefore (according to the fourth proposition) the two As are also equal in all other respects. Therefore, the < BAL is equal to the BCF;* the ZBL A is is equal to the < BFC; and the side A L is equal to the side F C.
2. To prove that the _CAL is equal to the < ACF:
These Zs are parts of the triangles CAL and ACF, which overlap one another. To prove that the angles CAL and ACF are equal, it is necessary to prove that the triangles CAL and AC F are equal in every respect.
Because BF is equal to B L, and BA is equal to
* We have thus reached the first main step of the proof.