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b

add the angle ACB; therefore the angles ACD, ACB are greater than the angles ABC, ACB; but ACD, ACB are together equal to two right angles; therefore b 13. 1. the angles ABC, BCA are 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.

PROP. XVIII. THEOR.

The greater side of every triangle has the greater angle opposite to it.

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From AC, which is greater than AB, cut off a AD equal B

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C

a 3. 1.

to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater than the inte- b 16. 1. rior and opposite angle DCB; but ADB is equal to c 5. J. ABD, because the side AB is equal to the side AD; therefore the angle ABD is likewise greater than the angle ACB; wherefore much more is the angle ABC greater than ACB. Therefore the greater side, &c. Q. E. D.

Book I.

a 5. 1.

b 18. 1.

a 3. 1.

b 5. 1.

c 19. 1.

PROP. XIX. THEOR.

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 AC is likewise greater than the side AB.

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 to the angle ACB; but it is not; therefore AC is not equal to AB; neither is it less; be- B

A

cause then the angle ABC would be less than the angle ACB; but it is not; therefore the side AC is not less than AB; and it has been shewn that it is not equal to AB; therefore AC is greater than AB. Wherefore the greater angle, &c. Q. E. D.

PROP. XX. THEOR.

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 AB, BC greater than AC; and BC, CA greater than AB.

Produce BA to the

point D, and make a AD
equal to AC; and join
DC.

Because DA is equal
to AC, the angle ADC
is likewise equal to
ACD; but the angle B

b

C

D

BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater side is opposite to the

c

greater angle: therefore the side DB is greater than the Book 1. side BC; but DB is equal to BA and AC together; therefore BA and AC together are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC CA greater than AB. Therefore any two sides, &c. Q. E. D.

PROP. XXI. THEOR.

If from the ends of one side of a triangle, there be drawn two straight lines to a point within the triangle, these two lines shall be less than the other two sides of the triangle, but shall contain a greater angle.

Let the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA AC of the triangle, but contain an angle BDC greater than the angle BẶC.

N.

Produce BD to E; and because two sides of a triangle are greater than the third side, the two sides BA, a 20, 1. AE of the triangle ABE are greater than BE. To each

of these add EC; therefore the sides BA, AC are greater than BE, EC: Again, because the two sides CE, ED, of the triangle CED are greater than CD, if DB be added to each, the sides CE, EB, will be greater than CD, DB; but it has been shewn that BA, AC are greater than BE, EC; much more then are BA, ACB greater than BD, DC.

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D

Α

Again, because the exterior angle of a triangle is b 16.1. greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than

le

Book I. the angle BAC. Therefore, if from the ends of, &c. Q. E. D.

a 20. 1.

a 3. 1.

PROP. XXII. PROB.

To construct a triangle of which the sides shall be equal to three given straight lines; but any two whatever of these lines must be greater than the third a.

Let A, B, C be the three given straight lines, of which any two whatever are greater than the third, viz. A and B greater than C; A and C greater than B; and B and C than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each.

Take a straight line DE terminated at the point D, but unlimited to

wards E, and make

a DF equal to A,
FG to B, and GH
equal to C; and
from the centre F, D
at the distance FD,

b 3. Post. describe the circle
DKL; and from
the centre G, at
the distance GH,
describe b another

circle HLK; and

join KF, KG; the

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triangle KFG has its sides equal to the three straight lines, A, B, C.

c

Because the point F is the centre of the circle DKL, c 11. Def. FD is equal to FK; but FD is equal to the straight line A; therefore FK is equal to A: Again, because G is the centre of the circle LKH, GH is equal to GK; but GH is equal to C; therefore, also GK is equal to C; and FG is equal to B; therefore the three straight lines KF, FG, GK, are equal to the three A, B, C: And therefore the triangle KFG has its three sides, KF, FG, GK equal to the three given straight lines, A, B, C. . Which was to be done.

PROP XXIII. PROB.

At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle; it is required to make an angle at the given point A in the given straight line AB, that shall be equal to the given rectilineal angle DCE.

Take in CD, CE

Book I.

G

E

any points D, E, and

F

join DE; and make a D

the triangle AFG, the

sides of which shall

be equal to the three

B

a 22. 1.

straight lines, CD, DE, CE, so that CD be equal to AF, CE to AG, and DE to FG; and because DC CE are equal to FA, AG, each to each, and the base DE to the base FG; the angle DCE is equal to the angle FAG. 68. 1. Therefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done.

PROP. XXIV. THEOR.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides of the other; the base of that which has the greater angle shall be greater than the base of the other.

Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF; the base BC is also greater than the base EF.

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