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d

Describe (i. 1) upon ab an equilateral triangle abc, and bisect (i. 9) the angle a c b by the straight line cd ab is cut into two equal parts in the point d. Because a c is equal to cb, and cd common to the two triangles a cd, bcd; the two sides a c, cd are equal to bc, cd, each to each; and the angle a cd is equal to the angle bcd; therefore the base ad is equal to the base (i. 4) db, and the b straight line ab is divided into two equal parts in the point d. Which was to be done.

PROPOSITION XI.-PROBLEM.

To draw a straight line at right angles to a given straight line, from a given point in the same.

LET a b be a given straight line, and c a point given in it; it is required to draw a straight line from the point c at right angles to a b.

Take any point d in a c, and make (i. 3) ce equal to cd, and upon de

a d

e b

describe (i. 1) the equilateral triangle dfe, and join fc, the straight line fc drawn from the given point c is at right angles to the given straight line a b.

Because dc is equal to ce, and fc common to the two triangles d cf, ecf; the two sides dc, cf, are equal to the two ec, cf, each to each; and the base dfis equal to the base ef; therefore the angle dcf is equal (i. 8) to the angle But, when the adjacent angles

ecf; and they are adjacent angles. which one straight line makes with another straight line are equal to one another, each of them is called a right angle (10 def.); therefore each of the angles dcf, e cf, is a right angle. Wherefore, from the given point c, in the given straight line a b, fc has been drawn at right angles to a b. Which was to be done.

COR. By help of this problem, it may be demonstrated, that two straight lines cannot have a common segment.

If it be possible, let the two straight lines abc, abd have the segment ab common to both of them. From the point b draw be at right angles to ab; and because abc is a straight line, the angle cbe is equal (10 def.) to the angle e ba; in the same manner, because abd is a straight line, the angle dbe is equal to the angle eba; wherefore the angle dbe is equal to the angle cbe, the less to the greater, which is impossible; therefore two straight lines cannot have a common segment.

a

b

d

с

PROPOSITION XII.-PROBLEM.

To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.

LET a b be the given straight line, which may be produced to any length both ways, and let c be a point without it. straight line perpendicular

It is required to draw a

to ab from the point c.
Take any point d upon
the other side of a b, and
from the centre c, at the
distance cd, describe (3
post.) the circle egf meet-
ing a b in f, g; and bisect
(i. 10) fg in h, and join
cf, ch, cg; the straight
line ch, drawn from the
given point c, is perpendicular to the given straight line a b.

a

f

h
d

g

b

Because fh is equal to hg, and he common to the two triangles fhc, ghc, the two sides fh, hc are equal to the two gh, hc, each to each; and the base c f is equal (15 def.) to the base cg; therefore the angle chf is equal (i. 8) to the angle chg; and they are adjacent angles; but when (10 def.) a straight line standing on another straight line makes the adjacent angles equal to one another, each of them is a right angle; and the straight line which stands upon the other is called a perpendicular to it; therefore from the given point ca perpendicular ch has been drawn to the given straight line a b. Which was to be done.

PROPOSITION XIII.—THEOREM.

The angles which one straight line makes with another upon one side of it are either two right angles, or are together equal to two right angles.

LET the straight line ab make with cd, upon one side of it, the angles cba, abd these are either two right angles, or are together equal to two right angles.

:

For if the angle cba be equal to a bd, each of them is a right angle

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(10 def.); but if not, from the point b draw be at right angles (i. 11) to cd; therefore the angles c be, ebd are two right angles (10 def.); and because cbe is equal to the two angles cba, a be together, add the angle e bd

to each of these equals ; therefore the angles cbe, eb d are equal (2 ax.) to the three angles cba, abe, ebd. Again, because the angle dba is equal to the two angles dbe, eba, add to these equals the angle abc, therefore the angles dba, abc are equal to the three angles d be, eba, abc but the angles cbe, ebd have been demonstrated to be equal to the same three angles; and things that are equal to the same are equal (1 ax.) to one another; therefore the angles cbe, ebd are equal to the angles dba, abc; but cbe, ebd are two right angles; therefore d ba, a b c are together equal to two right angles. Wherefore, the angles which one straight line, &c. Q. E. D.

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If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

At the point b in the straight line a b, let the two straight lines bc, bd

с

b

a

e

d

upon the opposite sides of a b, make the adjacent angles abc, abd equal together to two right angles, b d is in the same straight line with c b.

For, if b d be not in the same straight line with cb, let be be in the same straight line with it; therefore, because the straight line ab makes angles with the straight line cbe, upon one side of it, the angles a bc, abe are together equal (i. 13) to two right angles; but the angles a bc, abd are likewise together equal to two right angles; therefore the angles c ba, a be are equal to the angles c ba, abd: take away the common angle cba, the remaining angle a be is equal (3 ax.) to the remaining angle a b d, the less to the greater, which is impossible; therefore be is not in the same straight line with bc. And in like manner it may be demonstrated, that no other can be in the same straight line with it but b d, which therefore is in the same straight line with cb. Wherefore, if at a point, &c. Q. E. D.

PROPOSITION XV.—THEOREM.

If two straight lines cut one another, the vertical, or opposite, angles

shall be equal.

LET the two straight lines a b, cd cut one another in the point e; the angle a ec shall be equal to the angle deb, and ceb to a ed. Because the straight line a e makes with c d the angles cea, a ed, these angles are together equal (i. 13) to two right angles. Again, because the straight line de makes with ab the angles aed, deb, these also are together equal (i. 13) to two right angles; and cea, a ed, have been demonstrated to be equal to two right angles; wherefore the angles cea, aed,

a

b

are equal to the angles aed, deb. Take away the common angle a e d, and the remaining angle ce a is equal (3 ax.) to the remaining angle de b. In the same manner it can be demonstrated, that the angles ceb, aed are equal. Therefore, if two straight lines, &c. Q. E. D.

COR. 1. From this it is manifest, that, if two straight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles.

COR. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles.

PROPOSITION XVI.-THEOREM.

If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

LET a b c be a triangle, and let its side b c be produced to d, the exterior angle a cd is greater than either of the interior opposite angles cba, bac.

Bisect (i. 10) a c in e, join be and produce it to f, and make e f equal to be; join also fc, and produce a c to g.

Because a e is equal to ec, and be to ef; a e, eb are equal to ce, ef, each to each; and the angle a e b is equal (i. 15) to the angle cef, because they are opposite. vertical angles; therefore the base a b is equal (i. 4) to the base c f, and the triangle aeb to the triangle cef, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle bae is equal to the angle ecf; but the angle ecd is greater than the angle e cf; therefore the angle a cd is greater than

b

a

d

g

bae. In the same manner, if the side b c be bisected, it may be demonstrated that the angle bcg, that is, (i. 15) the angle a cd, is greater than the angle abc. Therefore, if one side, &c. Q. E. D.

PROPOSITION XVII.-THEOREM.

Any two angles of a triangle are together less than two right angles.

LET abc be any triangle; any two of its angles together are less than two right angles.

Produce bc to d; and because a cd is the exterior angle of the triangle a bc, acd is greater (i. 16) than the interior and opposite angle abc; to each of these add the angle acb; therefore the angles a cd, acb are greater than the angles a b c, a cb; but a cd, acb are together equal (i. 13) to two right angles; therefore the angles a bc, bca are

a

с

14

less than two right angles.

In like manner it may be demonstrated, that bac, acb, as also cab, abc, are less than two right angles. Therefore any two angles, &c. Q. E. D.

b

PROPOSITION XVIII.-THEOREM.

The greater side of every triangle is opposite to the greater angle.

d

с

LET abc be a triangle, of which the side ac is greater than the side ab; the angle abc is also greater than the angle bca.

Because a c is greater than ab, make (i. 3) cis a d equal to a b, and join bd; and because adb is the exterior angle of the triangle bd c, it is greater (i. 16) than the interior and opposite angle dcb; but a db is equal (i. 5) to a bd, because the side a b is equal to the side ad; therefore the angle abd is likewise greater than the angle acb. angle a bc than a cb. Therefore the greater side, &c. Q. E. D. Much greater then is the

PROPOSITION XIX.-THEOREM.

The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

LET abc be a triangle, of which the angle abc is greater than the angle bca; the side a c is likewise greater than the side a b.

a

For, if it be not greater, ac must either be equal to ab, or less than it; it is not equal, because then the angle abc would be equal (i. 5) to the angle a cb; but it is not; therefore a c is not equal to ab; neither is it less; because then the angle a b c would be less than the angle a cb (i. 18); but it is not; therefore the side a c is not less than ab; and it has been shewn that it is not equal to ab; therefore ac is greater than a b. Wherefore the greater angle, &c.

b

Q. E. D.

PROPOSITION XX. THEOREM.

Any two sides of a triangle are together greater than the third side.

LET abc be a triangle; any two sides of it together are greater than the third side, viz. the sides ba, ac greater than the side bc; and a b, bc greater than ac; and bc, ca greater than a b.

Produce ba to the point d, and make (i. 3) ad equal to ac; and join d c.

Because da is equal to a c, the angle a dc is likewise equal to a c d (i. 5); but the angle bcd is greater than the angle acd; therefore the angle

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