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right angles; and the angle CBE is equal to the two angles CBA, ABE together. Add the angle EBD to each of these equals; then the angles CBE, EBD are equal* #2 Ax. to the three angles CBA, ABE, EBD. Again, the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC, then the angles DBA, ABC are equalf to the three angles DBE, 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 thing are equal to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC: but CBE, EBD are two right angles; therefore DBA, ABC are together equal+to two right angles. Wherefore, when a straight line, &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.

*1 Ax.

+1 Ax.

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.

PROP. XIV. THEOR.

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 line 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 equal together to two right angles: BD shall be in the same straight line with CB.

. For, if BD be not in the same straight

B

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#13. 1.

line with CB, let BE be in the same straight line with it: therefore, because the straight line AB makes with the straight line CBE, upon one side of it, the angles ABC, ABE, these angles are together equal* to two right angles; but the angles ABC, ABD are likewise together +Hyp. equal to two right angles; therefore the angles 12 Ax. ABC, ABE are equal to the angles ABC, ABD: Take away the common angle ABC, and the remain#3 Ax. ing 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.

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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 AEC, AED, these angles are together C

*13. 1.

equal to two right angles. Again, because the straight line DE A makes with AB the angles DEA, DEB, these also are together equal to

#13. 1.

two right angles; and AEC, AED, have been demonstrated to be equal to two right angles; wherefore the angles AEC, AED, are together equal to the angles +1 Ax. AED, DEB. Take away the common angle AED,

and the remaining angle A EC is equal to the remain#3 Ax. ing angle DEB. In a similar manner, it can be demonstrated, that the angles CEB, AED are equal. Therefore, if two straight lines, &c. Q. E. D.

This proposition and the preceding convey the sense in which "adjacent" and "opposite" are used when applied to angles formed by two intersecting straight lines. Indeed always, in Geometry, when an angle is referred to, and, in connexion with it, an opposite angle is mentioned, an angle non-adjacent to the former is always meant.

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 shall be greater than either of the interior opposite angles CBA, BAC.

*10. 1.

+3. 1.

† Const.

#15. 1.

Bisect AC in E, draw BE, which produce, and make EF equalt to BE: join F,C.

Because AE is equal to EC,

and BE† to EF; AE, EB are

equal to CE, EF, each to each; and B the angle AEB is equal to the angle CEF, because they are opposite angles; therefore the base

A

G

AB is equal to the base CF, and the triangle AEB *4. 1. 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 EAB is equal to the angle ECF: but the angle ACD is greatert than the angle *19 Ax. ECF, therefore the angle ACD is greater than BAE. In like manner, if AC be produced to G, and BC be bisected instead of AC, and a line be drawn from A to the point of bisection, and then produced (like BE was) till the produced part be equal to it, and lastly the extremity of this produced part joined to C, it may be demonstrated that the angle BCG, that is, the angle* ACD, is greater than the angle ABC. Therefore, if one side, &c. Q. E. D.

#15. 1.

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 shall be less than two right angles.

#16. 1.

Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater than the interior and opposite angle ABC; to each of these add the angle ACB; therefore the angles ACD, ACB are greatert than the angles ABC, ACB: but

†4 Ax.

#13. 1.

A

C

ACD, ACB are together equal to two right angles; therefore the angles ABC, ACB 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.

NOTE. This proposition is unnecessary, as its proof is involved in that of the 32d of this book; and it is not required in any of the intervening propositions.

PROP. XVIII. 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 ABC shall be greater than the angle ACB. From AC, which is greater than AB, cut

#3. 1.

#5. 1.

off* AD equal to AB, and draw B

BD: then because ADB is the exterior angle of the triangle BDC, it is greater than the interior and *16. 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: therefore much more is the angle ABC greater than ACB. Therefore the greater side, &c. Q. E. D.

+ Const.

PROP. XIX. THEOR.

1

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

D

greater than the angle BCA: the side AC shall be greater than the side AB.

For, if it be not greater, AC must either be equal to AB, or less than it if it were equal, the angle ABC would be #5. 1. equal to the angle ACB; but it ist not; therefore tHyp. AC is not equal to AB: if it were

less, the angle ABC would be less* *18. 1. 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. B 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 shall be 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.

3. 1.

5. 1. 19 Ax.

B

D

Produce B Ato the point D, make* AD equal to AC; and join D C, Because DA is equal to AC, the angle ADC is equal to ACD; but the angle BCD is greatert 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* angle is subtended by the greater side; therefore the side DB is greater than the side BC: but DB is equal to BA and AD, that is, to BA and AC; therefore the sides BA, AC, are greater than BC. In a similar 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.

#19. 1.

Some Editors subjoin to this proposition the corollary that "the difference of any two sides of a triangle is less than the third side," and reason as follows: "For, since BA and AC are together

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