« ΠροηγούμενηΣυνέχεια »
On A B describe the equilateral A ACB.
On A B also describe the equilateral A A E B, on the side remote from C.
Join the points C and E by the line CE,
A B is bisected at the point D.
To draw a straight line at right angles * to a given straight line from a given point in the same.
To make the construction we must be able,--
cut off a part equal to the less. (Prop. III.) 3. On a given straight line to construct an equi
lateral A. (Prop. I.) To prove that the construction is correct we must,
know that, If two As have the three sides of the one equal to
the three sides of the other, each to each, then
the <s of the As are equal. (Prop. VIII.) Let A B be the given right line, and C the given
* That is, so that the adjacent Z8 which it makes with the givon line may be equal to each other. (Def. XI.)
point in it. (The line A B may be of any length, and
may be produced either way if necessary.)
In the line C A take any point-as
CE, equal to CD. (Prop. III.)
Join the points F and C by the right line F C, which will be the line required.
Proof. In the As DFC and EF C the sides DF and F E are equal, being sides of an equilateral A. The sides D C and E C are equal, C E having been cut off of the same length as C D. The side F C is common to both. Thus the three sides of the one A are equal to the three sides of the other, each to each.
Consequently the <s of the two As are equal. (Prop. VIII.)
Among these, the Zs FC D and F C E are equal.
Therefore F C stands upon the line A B in such a manner as to make the adjacent Zs equal. In other words, it is perpendicular to A B. (See Def. XI.)
Let the proposition next be gone through with the
figure in the following position :
There is usually appended to this proposition a
corollary, to the effect that by the aid of it it may be proved that two straight lines cannot have a common segment, that is, cannot coincide for a certain distance and then diverge from one another. This ought to be set down as an
* Sea the note on the last proposition.
axiom. It is assumed and made use of long before we come to the eleventh proposition. In the fourth proposition, for example, where certain lines are placed on one another, i.e., so as to coincide in direction, it is necessary that we should assume that, if they coincide for any portion of their length, they cannot afterwards diverge. It is assumed again in the eighth. Moreover, the proof attempted in the corollary is insufficient. It assumes that all right Zs are equal; a proposition for the proof of which the truth of this very axiom is assumed (see p. 12), and which could not itself be laid down as an axiom (as is often done) without the very same assumption. If it were regarded as yet unproved that two right lines cannot have a common segment, there would be no ground for assuming that the two equal _s formed when one line is perpendicular to another must always be of the same size.
To draw a right line perpendicular, or at right angles to a given right line of indefinite length, from a given point without it.*
For the construction in this proposition we must
be able,1. To join two given points by a straight line.
(Post. I.) 2. To describe a circle with a given centre, and at
a given distance from that centre. (Post. III.)
* That is, so that the adjacent <s which it forms with the given line may be equal to each other. (Def. XI.)
3. To bisect a given rectilineal Z. (Prop. IX.) To prove that the construction accomplishes what
is required, we must know that,1. Any two radii of the same circle are equal.
(Def. XIV.) 2. If two As have two sides of the one equal to
two sides of the other, each to each, and have the Zs between the said pair of sides in the one A equal to the between the said pair of sides in the second A, then those As will also be equal in every other respect. (Prop. IV.)
Let A B be the given line.
Let F be the given point without it.
Take a point O on the other side of the line A B.
Join F and C by the right
line F C. With F as centre, and at the distance F C, describe a circle, cutting the line A B in the points D and E.
Join F and D, and F and E, by the right lines FD and FE.
From the point F draw the right line FL, bisecting the ZDFE, and meeting the line D E in the point L. (Prop. IX.)
L is the line required.
Proof. In the As DFL and EFL the sides DF and F E are equal, because they are radii of the same circle;
the line F L is common to the two As; and the ZDFL is equal to the < EFL, because they are the two equal Zs into which the ZDFE is divided by the line F L. That is to say, the two As have two sides of the one equal to two sides of the other, each to each, and the between the said pair of sides in the first A equal to the < between the said pair of sides in the second A.
Consequently, as was proved in the fourth proposition, these two As are also equal in every other respect.
Among these respects is that the < FLD is equal to the <FLE.
The Z8 FLD and F L E are adjacent Zs.
It has thus been shown that the line F L, meeting the line A B, makes the adjacent <s equal to one another. In other words, it is perpendicular to the line A B. (Def. XI.)
The angles which one right line, meeting another, forms with it on the same side, are either two right angles, or together equal to two right angles.
To prove this proposition we must be able to draw
a perpendicular to a given straight line from a
given point in the same. (Prop. XI.) We must also know,1. That magnitudes which are equal to the same
are equal to each other. 2. That if the same quantity be added to each of
two equal quantities, the sums will be equal.
(Ax. II.) Suppose the straight line D O to meet the straight line A B.
The Zs which D C makes with A B must be either equal or unequal.
If they are equal, they are called right Zs.
But suppose they are unequal, then it is easily shown that their sum is the same as that of two right 8.