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16

Book I. BE be in the same straight line with it; therefore, because

the straight line AB makes angles with the straight line CBE, a 13. ). upon one side of it, the angles ABC, ABE are together equal a

to two right angles; but the angles ABC, ABD are likewise together equal to two right angles; therefore the angles CBA,

ABE are equal to the angles CBA, ABD: Take away the b 3. Ax. common angle ABC, the remaining angle ABE is equal to

the remaining angle ABD, 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 BD, which therefore is in the same straight line with CB. Wherefore, if at a point,&c. Q. E. D.

PROP. XV. THEOR.

If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

a

Let the two straight lines AB, CD, cut one another in the point E; the angle A EC shall be equal to the angle DEB, and CEB to AED.

Because the straight line
AE makes with CD the angles

CEA, AED, these angles are a 13. 1. together equal to two right

angles. Again, because the
straight line DE makes with
AB the angles AED, DEB, A

E

B these also are together equal to two right angles; and CEA,

D AED have been demonstrated to be equal to two right angles, wherefore the angles CEA,

AED are equal to the angles AED, DEB. Take away the b 3. Ax. common angle AED, and the remaining angle CEA is equalb

to the remaining angle DEB. 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 which 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 straight lines meeting in one point, are together equal to four right angles.

Book I.

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

Let ABC be a triangle, and let its side BC be produced to
D, the exterior angle ACD is greater than either of the inte-
rior opposite angles CBA, BAC,
Bisect" AC in E, join

a 10. 1. BE and produce it to F,

A and make EF equal to BE;

F
join also FC, and produce
AC to G.
Because AE is equal to

E
EC, and BE to EF; AE,
EB are equal to CE, EF,
each to each; and the angle
AEB is equal to the angle B C С

D

b 15. 1. CEF, because they are opposite vertical angles; therefore the base AB is

G equal to the base CF, and

C 4. 1. 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 ECF; therefore the angle ACD is greater than BAE: In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, that is, d the angle ACD is d 15. 1. greater than the angle ABC. Therefore, “ if one side,&c. Q. E. D.

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

Let ABC be any triangle ; any two of its angles together

A are less than two right angles.

Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater than the interior and

a 16. 1. opposite angle ABC; to each of these add the angle ACB; therefore the angles ACD,

B

С D

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B

Book I. ACB are greater than the angles ABC, ACB; but ACD,

ACB are together equal to two right angles; therefore the b 13. 1.

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

Let ABC be a triangle, of
which the side AC is greater
than the side AB; the angle А.
ABC is also greater than the
angle BCA.

Because AC is greater than a 3. 1. AB, make a AD equal to AB,

D
and join BD; and because
ADB is the exterior angle of B

the triangle BDC, ADB is b 16. 1. greater than the interior and c 5. 1. opposite angle DCB; but ADB is equal to 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, o the greater side," &c. Q. E. D.

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 А.
less than it; it is not equal, be-

cause then the angle ABC a 5. 1.

would be equal a to the angle
ACB; but it is not; therefore
AC is not equal to AB; neither
is it less; because then the an-

B b 18. 1. gle ABC would be less b than

the angle ACB; but it is not: therefore the side AC is not less Book I. than AB; and it has been shown 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 See N. 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;

a 3. 1.

D and join DC. Because DA is equal to AC,

А the angle ADC is likewise

b 5. 1. equal to ACD; but the angle BCD is greater than the angle ACD; therefore the angle B

C 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 c 19. 1. opposite to the greater angle: therefore the side DB is greater than the side BC; but DB is equal to BA and AC; therefore the sides BA, AC 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 that AB. Therefore, any two sides," &c. Q. E. D.

PROP. XXI. THEOR. If, from the ends of one side of a triangle there be See N. drawn two straight lines to a point within the tri. angle, these 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 BAC.

Produce BD to E; and because two sides of a triangled are d 20. 1. greater than the third side, the two sides BA, AE of the tri

Book I. angle ABE are greater than BE. To each of these add EC; therefore the sides BA, AC,

А.
are greater than BE, EC.
Again, because the two sides

E
CE, ED of the triangle CED

D
are greater than CD, add DB
to each of these; therefore
the sides CE, EB are greater
than CD, DB; but it has
been shown that BA, AC are

в
greater than BE, EC; much

more then are BA, AC greater than BD, DC. e 16. 1.

Again, because the exterior angle of a triangle e is 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 ČEB 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 the angle BAC. Therefore, if from the ends of,&c. Q. E. D.

See N.

PROP. XXII. PROB.
To make a triangle of which the sides shall be equal

to three given straight lines, but any two whatever of f 20. 1. these must be greater than the third'.

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 un

limited towards E, and a 3. 1. make a DF equal to A,

FG to B, and GH equal
to C; and from the cen-

tre F, at the distance
b 3. Post. FD, describe the circle D

F

G НЕ
DKL; and from the

L
centre G, at the distance
GH, describe another
circle HKL; and join

B В
KF, KG; the triangle

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

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