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in the same straight line with it; therefore, because the straight Book I. line AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal a to two right a 13. 1. 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 common angle ABC, the remaining angle ABE is equal to the remaining angle b 3. Ax. 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.

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

Because the straight line AE makes with CD the angles CEA, C AED, these angles are together equal a to two right angles.

a 13. 1. Again, because the straight line A

B
DE makes with AB the angles
AED, DEB, these also are to-
gether equal a to two right angles;

D and CEA, AED have been demonstrated to be equal to two right angles; wherefore the angles CEA, AED are equal to the angles AED, DEB. the common angle AED), and the remaing angle CEA is equal bb 3. Ax. 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 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.

D

Take away

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 interior opposite angles CBA, BAC.

A
2 10.1.
Bisect a AC in E, join BE

F
and produce it to F, and
make EF equal to BE; join
also FC, and produce AC to
G.

E
Because AE is equal to
EC, and BE to EF; AE,
EB are equal to CE, EF,
each to each ; and the angle

B
с

D b 15. 1. AEB is equal b to the angle

CEF, because they are oppo

site vertical angles; therefore C4. 1. the base AB is equal < to the

G base CF, 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 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, d 15. 1. that is d, the angle ACD, is greater than the angle ABC. There

fore, if one side, &c. Q. E. D.

PROP. XVII. THEOR.

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

A

Let ABC be any triangle ; any two of its angles together are less than two right angles.

Produce BC to D; and because ACD is the exterior angle

of the triangle ABC, ACD is a 16. 1. greater a than the interior and

opposite angle ABC; to each of

B В.

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these add the angle ACB ; therefore the angles ACD, ACB are Book I.
greater than the angles ABC, ACB ; but ACD, ACB are to-
gether equal to two right angles; therefore the angles ABC, b 13. 1.
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 is opposite to the greater angle.

Let ABC be a triangle, of which

A the side AC is greater than the side AB ; the angle ABC is also greater than the angle BCA.

D Because AC is greater than AB, make a AD equal to AB, and join

2 3. 1. BD ; and because ADB is the ex. terior angle of the triangle BDC,

B

C
it is greater b than the interior and opposite angle DCB ; but b 16. 1.
ADB is equal to AÐB, because the side AB is equal to the side c 5. 1.
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.

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 A either be equal to AB, or less than it; it is not equal, because then the angle ABC would be equal a to the

a 5. 1. angle ACB; but it is not ; therefore AC is not equal to AB ; neither is it less; because then the angle ABC would be less than the angle ACB; B

b 18, 1.

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Book I. but it is not ; therefore the side AC is not less 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.

See N.

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

A

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

D a 3. 1. make a AD equal to AC ; and join

DC.

Because DA is equal to AC, the b 5. 1. angle ADC is likewise equal b to

ACD ; but the angle BCD is great-
er than the angle ACD; therefore
the angle BCD is greater than the B

С angle ADC ; and because the angle BCD of the triangle DCB c 19. 1. is greater than its angle BDC, and that the greater c side is op

posite 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 than AB. Therefore, any two sides, &c. Q. E. D.

PROP. XXI. THEOR.

See N.

IF, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, 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 triangle are greater than the third side, the two sides BA, AE of the tri

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

A greater than BE, EC: again, because the two sides CE, ED

E of the triangle CED 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 greater than BE, EC; much more then are BA, AC B

C greater than BD, DC.

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

PROP. XXII. PROB.

To make a triangle of which the sides shall be See N. equal to three given straight lines, but any two what. ever of these must be greater than the third a.

a 20. 1.

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 towards E, and make a DF equal to, A,

a 3. 1.

K
FG to B, and GH equal
to C; and from the centre
F, at the distance FD, de-

D scribe b the circle DKL;

E
F G

H

b. 3. Post. and from the centre G, at

L the distance GH, describe b another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines A, B, C.

Because the point F is the centre of the circle DKL, FD is

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