Take away the common angle ABC, and the *3 Ax. 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.

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, AED, these

3.1. angles are together equal * to two right angles. Again, because the straight line DE makes with AB the angles

*

* 13. 1. AED, DEB, these also are together equal to two right angles; and CEA, AED, have been demonstrated to be equal to two right angles; +1 Ax. wherefore the angles CEA, AED, are equal † to the angles AED, DEB. Take away the common angle AED, and the remaining angle CEA is *3 Ax. equal 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. I. 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 lines meeting in one point, are together equal to four right angles.

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.

Bisect BE and produce it to F, and make EF equal

AC in E, join

BE, and join FC.

to

Because AE is equal to

EC, and BE† to EF; AE, B

EB are equal to CE, EF, each to each; and the angle AEB is equal to the angle

*

A

* 15. 1.

CEF, because they are opposite vertica, angles; therefore the base AB is equal to the base CF, *4. 1. and the triangleAEB to the triangle CEF, and the remaining, angles to the reinairing 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 then fo ̊Ax. the angle ECF, therefore the angle ACD is greater than BAE. In the sanje manner if the side BC be bisected, and AC be produced to G, it may be demonstrated that the angle BCG, that is, the angle* ACD, is greater than the *15. 1. angle ABC. Therefore, if one side, &c. Q. E. D.

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.

B

Produce BC to D; and because ACD is the exterior angle of the triangle ABC, *16 1. ACD is greater than the interior and opposite angle ABC; to each of these add the angle ACB; therefore the angles 14 Ax. ACD, ACB are greater+ than the angles ABC, *13.1. ACB: but ACD, ACB are together equal to two right angles; therefore the angles ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as also CÁB, ABC, are less than two right angles. Therefore any two angles, &c. Q. E. D.

*3. 1.

5. 1.

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 BCA.

2

Because AC is greater than

A

D

B

AB, make* AD equal to AB, and join BD : and because ADB is the exterior angle of the triangle 16. 1. BDC, it is greater than the interior and opposite angle DCB but ADB is equal* to ABD, + Const. 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.

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 shall be 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* * 5.1 to the angle CAB; but it is † not; therefore † Hyp AC is not equal to AB: neither is it less; because 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

B

been shown that it is not equal to AB; therefore AC is greater than AB.

greater angle, &c. Q. E. D.

Wherefore the

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.

Produce BA to the point D, and make* AD equal to AC; and join DC.

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

B

D

18. 1.

* 3.1.

*5.1.

BCD is greater than the angle ACD: there- †9 Ax. fore 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 19.1. greater side; 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

Because AD is equal† to AC, add BA to each, therefore † Const the whole BD is equal † to the two BA and AC. +2 Ax.

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 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 ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle: BD, and DC shall be less than the other two sides BA, AC of the triangle, but shall contain an angle BDC greater than the angle BAC.

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

+4. Ax. the sides BA, AC are greater+

than BE, EC. Again, because the two sides CE, ED of the † 20. 1. triangle CED are greater+ than CD, add DB to each of these; therefore the sides CE, B 14 Ax. EB, are greater than CD,

16. 1.

D

E

DB: but it has been shown that BA, AC are greater than BE, EC; much more then are BA, AC, 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.