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Produce AC to E, and AD to F, and join CD.

Then because AC is equal to AD in the triangle ACD, therefore the angles ECD, FDC upon the other side of the base CD, are equal to one another; (1. 5.)

but the angle ECD is greater than the angle BCD; (ax. 9.) therefore also the angle FDC is greater than the angle BCD; much more then is the angle BDC greater than the angle BCD.

Again, because B C is equal to B D in the triangle BCD, therefore the angle BDC is equal to the angle BCD, (1. 5.)

but the angle BDC has been proved greater than BCD, wherefore the angle BDC is both equal to, and greater than the

angle BCD; which is impossible. Thirdly. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base and on the same side of it, &c. Q.E.D.

PROPOSITION VIII. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other.

Let ABC, DEF be two triangles, having the two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF, and also the base B C equal to the base EF.

D G

Then the angle BAC shall be equal to the angle EDF.

For, if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC on EF;

then because BC is equal to EF, (hyp.) therefore the point C shall coincide with the point F;

wherefore BC coinciding with EF,

BA and AC shall coincide with ED, DF; for, if the base B C coincide with the base EF, but the sides BA, AC, do not coincide with the sides ED, DF, but have a different situation as EG, GF:

then, upon the same base, and upon the same side of it, there can be two triangles which have their sides which are terminated in one extremity of the base, equal to one another, and likewise those sides which are terminated in the other extremity; but this is impossible. (1.7.)

Therefore, if the base BC coincide with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. (ax. 8.)

"Therefore if two triangles have two sides, &c. Q.E.D.

PROPOSITION IX. PROBLEM. To bisect a given rectilineal angle, that is, to divide it into two equal angles.

Let BAC be the given rectilineal angle.

It is required to bisect it.

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In AB take any point D;
from AC cut off AE equal to AD, (1. 3.) and join DE;

on the side of DE remote from A,
describe the equilateral triangle DEF (1. 1.), and join AF.
Then the straight line AF shall bisect the angle BAC.

Because AD is equal to AE, (constr.) and AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal to the two sides EA, AF, each to each;

and the base DF is equal to the base EF: (constr.) therefore the angle DAF is equal to the angle EAF. (1. 8.) Wherefore the angle BAC is bisected by the straight line AF. Q.E.F.

PROPOSITION X. PROBLEM. To bisect a given finite straight line, that is, to divide it into two equal parts.

Let AB be the given straight line. . It is required to divide AB into two equal parts. Upon AB describe the equilateral triangle ABC; (1. 1.)

DB

and bisect the angle ACB by the straight line CD meeting AB in the point D. (1. 9.) Then AB shall be cut into two equal parts in the point D.

Because AC is equal to CB, (constr.) and CD is common to the two triangles ACD, BCD; the two sides AC, CD are equal to the two BC, CD, each to each;

and the angle ACD is equal to BCD; (constr.)

therefore the base AD is equal to the base BD. (1. 4.) Wherefore the straight line AB is divided into two equal parts in the

point D. Q.E.F.

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

Let AB be the given straight line, and C a given point in it.

It is required to draw a straight line from the point C at right angles to AB.

B

D C E In AC take any point D, and make CE equal to CD; (1. 3.) upon DE describe the equilateral triangle DEF (1. 1), and join CF.

Then CF drawn from the point C, shall be at right angles to AB. Because DC is equal to EC, and FC is common to the two triangles

DCF, ECF; the two sides DC, CF are equal to the two sides EC, CF, each to each;

and the base .DF is equal to the base EF; (constr.) therefore the angle DCF is equal to the angle ECF: (1. 8.)

and these two angles are adjacent angles. But when the two adjacent angles which one straight line makes with another straight line, are equal to one another, each of them is called a right angle: (def. 10.)

therefore each of the angles DCF, ECF is a right angle. Wherefore from the given point C, in the given straight line AB, FC has been drawn at right angles to AB. Q.E.F.

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 segment AB be common to the two straight lines ABC, ABD.

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From the point B, draw BE at right angles to AB; (1. 11.)

then because ABC is a straight line, therefore the angle ABE is equal to the angle EBC, (def. 10.)

Similarly, because ABD is a straight line, therefore the angle ABE is equal to the angle EBD;

but the angle ABE is equal to the angle EBC, wherefore the angle EBD is equal to the angle EBC, (ax. 1.)

the less equal to the greater angle, which is impossible. Therefore two straight lines cannot have a common segment.

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

Let AB be the given straight line, which may be produced any length both ways, and let C be a point without it.

It is required to draw a straight line perpendicular to AB from the point C.

B

Upon the other side of AB take any point D, and from the center C, at the distance CD, describe the circle EGF meeting AB, produced if necessary, in F and G: (post. 3.)

bisect FG in H (1. 10.), and join CH. Then the straight line CH drawn from the given point C, shall be perpendicular to the given straight line AB.

Join FC, and CG.
Because FH is equal to HG, (constr.)

and HC is common to the triangles FHC, GHC; the two sides FH, HC, are equal to the two GH, HC, each to each;

and the base CF is equal to the base CG; (def. 15.) therefore the angle FHC is equal to the angle GHC; (1. 8.)

and these are adjacent angles. But when 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. (def. 10.)

Therefore from the given point C, a perpendicular CH has been drawn to the given straight line AB. Q.E.F..

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.

Then these shall be either two right angles, or, shall be together, equal to two right angles.

D B C D в с
For if the angle CBA be equal to the angle ABD,

each of them is a right angle. (def. 10.)
But if the angle CBA be not equal to the angle ABD,
from the point B draw BE at right angles to CD. (1. 11.)
Then the angles CBE, EBD are two right angles. (def. 10.)

And because the angle CBE is equal to the angles CBA, ABE,

add the angle EBD to each of these equals; therefore the angles CBE, EBD are equal to the three angles CBA,

ABE, EBD. (ax. 2.) Again, because the angle DBA is equal to the two angles DBE, EBA,

add to each of these equals the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE,

EBA, ABC. But the angles CBE, EBD have been proved equal to the same three angles; and things which are equal to the same thing are equal to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC;

but the angles CBE, EBD are two right angles ; therefore the angles DBA, AB Care together equal to two right angles. (ax. 1.)

Wherefore, when a straight line, &c. Q.E.D.

PROPOSITION XIV. THEOREM. 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; then these two straight lines shall be in one and the same straight line.

At the point B in the straight line AB, let the two straight lines BC, BD upon the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles.

Then BD shall be in the same straight line with BC.

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For, if BD be not in the same straight line with BC,

it possible, let B E be in the same straight line with it. Then because AB meets the straight line CBE; therefore the adjacent angles CBA, ABE are equal to two right angles ;

(1. 13.) but the angles CBA, ABD are equal to two right angles; (hyp.) therefore the angles CBA, ABE are equal to the angles CBA, ABD:

(ax. 1.)

take away from these equals the common angle CBA, therefore the remaining angle ABE is equal to the remaining angle

ABD; (ax. 3.)
the less angle equal to the greater, which is impossible:

therefore BE is not in the same straight line with BC. And in the same manner it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the same straight line with BC.

Wherefore, if at a point, &c. Q.E.D.

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