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the triangle A en the two fides A E, D e are supposed to be equal. The angles e AD, EDA [by 5. 1.] will be equal. By the same reason, beca re the two sides A F, Fd of the triangle AFD are equal, the angles F AD, F D A upon the base ad will be equal. These therefore being taken away from the former, and there will remain [by axiom 3.] the angles e A F, E D F equal. Which was to be demonstrated.
PROP. IX. PROBL.
To biseEt or cut a given right-lined angle into two
equal parts a. Let the given right-lined angle be BAC; it is required to bisect or cut the same into two equal parts. Let there be taken any point, as D in the right line A,
and from the right line А.
A C [by: prop: 3.] cut off the right line A è equal, to the right line AD;, and draw, the right. line DE; and upon the right line D E [by: prop. 1.] describe the equilateral triangle
DEF, and draw the right line A F: I tay, D
the angle B A c is cut into two equal parts
by the right line A F. B
For becaule the right line A D is equal
to the right line A F, and A F is commoni, therefore the two sides DA, AF will be equal to the two sides E A, A F, each to each; and the base D F is equal to the base E F. Therefore the angle DAF is equal to the angle E AF [by prop. 8.].
Therefore the given right-lined angle Bac is cut into two equal parts ' by the right line A F. Which was to be done.
a The operation would be somewhat more elegant, if an isosceles triangle Def were described or constructed upon the right line DE, having one of its equal fides DF or E F equal to AD or A E. Hence it appears, how any given angle may be divided into 4, 8, 16, 32, &c. equal parts, viz. by bisecting each part again. But from Euclid's postulatums cannot be obtained the division generally of an angle into any other number of equal parts, as 3, 5, 6, 7, c.
PRO P. X. PROBL. To cut a given finite right line into two equal parts b.
Let A B be the given finite right line, which is to be cut into two equal parts.
Upon A B let there be made the equilateral triangle ABC [by prop. 1.) and cut the angle A CB into two equal parts [by prop. 9.] by the right line c D. I say, the right line AB is cut into two equal parts at D.
For because the fide ac is equal to the side o B, and the side cd is common; the two sides A C, C D are equal A D B to the two sides B C, C D, each to each ; and the angle A c d is equal to the angle B C D: therefore the base A D [by prop. 4. ) is equal to the base BD.
Therefore the given finite right line A B is cut into two equal parts in the
point D. Which was to be done.
b The folution of this problem is really no more than de: scribing two equilateral triangles, the one on one fide the given right line a B, and the other on the other side that right line ; and joining the angles of each of these triangles, which are opposite to the given right line A B, by a right line, which will cut the given right line A B into two equal parts.
PROP. XI. PROBL. From a point given in a given right line, to draw
a right line at right angles to it. Let there be a given right line a B, and let the given point in the fame be c; it is required to draw a right line from the given point c at right angles to the right line A B.
In the right line A B let there be taken any point D; and [by prop. 3. ]put the right line c E equal to the right line CD; and (by prop. 1.] upon the right line D E describe the equilateral triangle F DE, and draw the right line Fc: I say, from the given point c in the given right line AB is drawn the right line c r at right angles to A B.
For because the right line CD is equal to the right line C E, and the right line c F is common; there are two right
I lines D C, GF, equal to the two right lines E C, C F, each
to each, and the base D F is equal to the base EF: therefore
the angle DCF is equal (by prop. 8.] to the angle ECF; and these are adjacent angles. But when a right line standing upon a right line makes
the adjacent angles equal to one anA D с E B other, each of these equal angles is a right angle [by def. 10.]. Therefore each of the angles DCF, FCE is a right angle.
Therefore from a given point c, in a given line A B, the right line F c is drawn at right angles to the right line
PRO P. XII. PROBL.
. From a given point, without a given infinite right line,
to draw a right line perpendicular to it e Let the given infinite right line be A B, and let the given point without the same be c; it is required from the given point c without A B to draw a right line perpendicular to it. Let any point o be taken on the other side of the line
AB; and about the centre c, F
with the distance c D, let the circle E F G be described [by poft. 3.) and let the right line E G be bisected in the point H [by prop. 10.] and draw the
right lines CG, CH, CE: I A
say, that the right line c H is H E B drawn from the
given point c, D
without the given infinite right line A B perpendicular to it.
For because G H is equal to the right line H E, and the right line h cis common; there are two right lines G H, HC, equal to two right lines E H, HC, each to each; and the base cg is equal to the base ce [by def. 15.] Therefore the angle c G H is equal to the angle E H c [by prop. 8.); and these are adjacent angles. But when a right line standing upon a right line makes the adjacent angles equal to one another, each of the equal angles is a right angle, and the standing right line is called a perpendicular to that upon which it Itands.
Therefore upon the given infinite right line A B, from a given point c without it, is drawn the perpendicular cH to it. Which was to be done.
• It was very right in Euclid to suppose an infinite right line in this problem ; for if it were a finite right line, a right line could not always be drawn from a given point out of it perpendicularly upon it.
There are several practical ways of folving this problem, as well as the last, but the most expeditious of all is by a fquare.
PROP, XIII. THEO R. If a right line standing upon a right line makes an
gles, these angles fall either be two right angles, or [both together ] equal to two right angles d.
For let any right line A B, standing upon the right line DC, make the angles C B A, A BD; I say, these angles will either be right angles, or [both together) equal to two right angles. For if the angle CB A be equal
E to the angle ABD; then will they
A be both right angles [by def. 10.] but if not, let the right line E B be drawn from the point B at right angles to the right line DC [by
c prop. 11.] then the angles CBE,
B E BD are two right angles. And because the angle C B E is equal to both the angles C B A, A B E, if the common angle EB D be added, the angles C B E, IB D are equal to the three angles CBA, A BE, EBD. Again, fince the angle DBA is equal to both the angles D BE, E B A, if the common angle A B C be added, the angles D B A, A B C will be equal to the three angles D B E, EBA, A B C. But the angles CBE, EBD have been proved to be equal to the fame three angles; and things which are equal to the fame thing, are equal to one another. Therefore the angles CBP, EBD are equal to the angles DBA, ABC; but the angles C B E, EBD are two right angles. Therefore the angles D BA, A B C are equal to two right angles.
If therefore a right line standing upon a right line makes angles, these angles shall either be two right angles, or [both together) equal to two right angles. Which was to be demonstrated.
. This proposition seems to depend upon a certain axiom or common notion, viz. that so much as the angle A B D exceeds the right angle E B D, by so much is the remaining angle A B C exceeded by the right angle E BC; for as there the angle A B E is the excess, so is the same angle A B E here the deficiency. Wherefore it necessarily follows, that both the angles A B D, A B C are equal to two right angles, the one being just as much above a right angle, as the other wants of one.
PROP. XIV. THEOR. If at a point in any right line two right lines drawn
contrary ways do make, with the first line, the adjacent angles equal to two right angles, those right lines will be direitly placed to one another, or both fall into one right line e.
For at the point B in the right line A B let two right lines B C, B D drawn contrary, ways, make, with A B, the adjoining angles A B C, A B D equal to two right angles : I say, the right lines B D, C B will lie directly to one another, or both fall into one right line.
For if B D lies not in the fame direction with B c, let [by post. 2.] C B, B e both lie in the same direction. Then because the right line A B stands upon the right A1
line C B E, the angles A B C, A BE are [by prop. 13.] equal to two, right angles; but [by the suppo
fition] the angles A B C, A B D are I also equal to two right angles :
therefore the angles CBA, A BE c B
are equal to the angles C B A, Take
away the common angle A BC, then the remaining angle A B E will be equal to the remaining angle ABD, the less equal to the greater; which is impossible: therefore, the right lines B E, B C are not both in one direction.
It is demonstrated after the same manner, that no right line but B D 'can have the same direction with the right line c B. Wherefore B C and B D lie both in the same right line.