For any other would divide the angle BAC into two angles, one greater than the half and the other less; which cannot be equal.

PROPOSITION XV.

THEOREM.-If two straight lines cut one another, the vertical angles shall be equal.

Let the two straight lines AB, CD, cut *INTERC.12. one another in the point* E. The angle AEC shall be equal to the angle DEB, and CEB to AED.

Cor. 2.

+ I. 12.

C

D

Because the straight line AE makes with CD the angles AEC and AED, these angles are togethert equal to two right angles. Again, because the straight line DE makes with AB the angles AED and DEB, these also are together equal to two INTERC. 1. right angles. Wherefore the angles AEC and AED are‡ together equal to the angles AED and DEB. Take away the common angle *INTERC. 1. AED, and the remaining angle AEC is* equal to the remaining angle DEB. In the same way may be shown, that the angles CEB, AED are equal.

Cor. 7.

And by parity of reasoning, the like may be proved of all other straight lines under the same conditions. Wherefore, universally, if two straight lines cut one another, &c. Which was to be demonstrated.

+ I. 9.

1.3.

Constr.

† I. 15.

11. 4.

PROPOSITION XVI.

THEOREM.-If one side of a triangle be prolonged, the exterior angle is greater than either of the interior opposite angles.

Let ABC be a triangle, and let its side BC be prolonged to D. The exterior angle ACD is greater than either of the interior opposite angles CAB, CBA.

Bisect+ CA, in E; join BE, and prolong it to F; and make‡ EF equal to EB. Join also FC; and prolong AC to G.

Because EA is equal to EC, and EB to* EF, EA and EB are equal to EC and EF respectively. Also the angle AEB ist equal to CEF, because they are vertical angles;

G

therefore the third side AB is‡ equal to the third side CF, and

Cor. 2.

the remaining angles of the triangle AEB are equal to the remaining angles of the triangle CEF respectively, viz. those to which the equal sides are opposite. Wherefore the angle ECF is equal to the angle EAB. But the angle ACD is greater than *INTERC. 1. ECF; therefore it is also* greater than EAB which is equal to ECF. In the same manner if the side BC be bisected, may be shown that the angle BCG is greater than CBA. But the angle BCG ist equal to ACD, for they are vertical angles; therefore INTERC. 1. the angle ACD is also greater than CBA. And in the same way may be shown that any other exterior angle made by prolonging a side of the triangle ABC, is greater than either of the interior opposite angles.

+ I. 15.

Cor.2.

*INTERC.12. Cor. 6.

† I. 16.

Cor. 6. * I. 12. Cor. 2.

And by parity of reasoning, the like may be proved in every other triangle. Wherefore, universally, if one side of a triangle be prolonged, &c. Which was to be demonstrated.

PROPOSITION XVII.

THEOREM.-Any two angles of a triangle are together less than two right angles.

Let ABC be a triangle. Any two of its angles are together less than two right angles.

Prolong* BC to D.

A

C D

Because ACD is the exterior angle of the triangle ABC, it ist greater than the interior oppo- B site angle ABC. To each of these add the angle ACB; and INTERC. 1. the angles ACD, ACB are together‡ greater than ABC, ACB. But the angles ACD, ACB are together* equal to two right +INTERC. 1. angles; therefore the angles ABC, ACB are together less than two right angles. In the same manner may be shown that the angles BAC, ACB are together less than two right angles; and by prolonging a side adjacent to another angle, as for instance AB, may be shown that the angles ABC, BAC are together less than two right angles.

And by parity of reasoning, the like may be proved in every other triangle. Wherefore, universally, any two angles of a triangle are together less than two right angles. Which was to be demonstrated.

COR. 1. A triangle cannot have more than one right angle;

• I. 3.

+ I. 16.

Constr. • 1.5.

+ INTERC. 1. Cor. 2.

nor more than one obtuse angle; neither can it have one right angle and also one obtuse angle.

For if it had any of these, two of its angles would be together not less than two right angles. Which (by the Prop. above) cannot be.

COR. 2. A right or an obtuse angle in any triangle, is greater than either of the remaining angles of the triangle.

For if it was not, there would be two angles which are together not less than two right angles.

COR. 3. Every triangle has at the least two acute angles.

For if it had not, it must either have two right angles, or two obtuse, or one right angle and one obtuse. Which (by Cor. 1 above) cannot be.

PROPOSITION XVIII.

THEOREM. The greater side in any triangle has the greater angle opposite to it.

Let ABC be a triangle, of which the side

AC is greater than the side AB. The angle
ABC is greater than the angle ACB.

Since AC is greater than AB, make* AD B
equal to AB; and join BD.

Because ADB is the exterior angle of the triangle BDC, it ist greater than the interior opposite angle DCB. But because AD is equal to AB, the angle ABD is* equal to ADB; therefore the angle ABD ist greater than DCB; still more is the angle ABC (which is greater than ABD) greater than DCB.

And by parity of reasoning, the like may be proved in every other instance. Wherefore the greater side in any triangle &c. Which was to be demonstrated.

PROPOSITION XIX.

THEOREM.-The greater angle in any triangle has the greater side opposite to it.

Let ABC be a triangle, of which the angle B is greater than the angle C. The side AC is greater than the side AB.

For, if it be not greater, AC must either be equal to AB, or less. It is not equal; be

B

A

I. 5.

+ Hyp. ↑ I. 18.

• Hyp.

+1.17.Cor.2.

Cor. 6.

* I. 3.

† I. 5.

* I. 19.

cause then the angle B would be equal to the angle C; but the angle B is not equal to the angle C, for it is† greater. Neither is it less; because then the angle B would be‡ less than the angle C; but it is not less, for it is* greater. And because AC is neither equal to AB nor less, it is greater.

And by parity of reasoning, the like may be proved in every other instance. Wherefore the greater angle in any triangle &c. Which was to be demonstrated.

COR. In any right-angled or obtuse-angled triangle, the side opposite to the right or obtuse angle is the greatest. For the angle ist the greatest.

PROPOSITION XX.

THEOREM-Any two sides of a triangle are together greater than the third side.

Let ABC be a triangle. Any two sides of it are together greater than the third side; viz. the sides BA and AC, greater than BC; and AC and CB, greater than BA; and CB and BA, greater than AC.

B

A

D

INTERC.12. Prolong‡ BA; and make* AD equal to AC. And join DC. Because AD is equal to AC, the angle ACD ist equal to ADC. But the angle BCD is greater than ACD; therefore it is also‡ INTERC. 1. greater than ADC. And because in the triangle BCD the angle Cor. 2. BCD is greater than the angle BDC, the side BD is* greater than the side CB. But BD is equal to BA and AD, or to BA and AC; therefore the sides BA and AC are† together greater than CB. In the same way by prolonging AC, may be shown that the sides AC and CB are together greater than BA; and by prolonging CB, may be shown that CB and BA are together greater than AC.

+ INTERC. Cor. 2.

1.

And by parity of reasoning, the like may be proved in every other triangle. Wherefore, universally, any two sides of a triangle &c. Which was to be demonstrated.

COR. Three straight lines, any two of which are not together greater than the third, cannot form a triangle.

For if they did, there would be two of its sides together not greater than the third.

See Note.

*INTERC.12. Cor. 6.

+ I. 20.

Cor. 6.

PROPOSITION XXI.

THEOREM.-If from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall together be less than the two sides of the first triangle, but shall make a greater angle.

From B and C the ends of any side of a triangle ABC, let the two straight lines BD, CD be drawn to the point D within it. BD and CD are together less than BA and CA; but make an angle BDC greater than the angle BAC.

Prolong* BD till it meets the side CA in E.

B

Because two sides of a triangle are togethert greater than the third side, BA and AE are together greater than BE. To each of INTERC. 1. these add EC; therefore BA, AE, EC are together‡ greater than BE, EC. But AE, EC are equal to AC; therefore BA, AC are together greater than BE, EC. Again, because two sides of a triangle are together greater than the third side, EC and ED are together greater than CD. To each of these add DB; therefore EC, ED, DB are together greater than CD, DB. But ED, DB are equal to BE; therefore BE, EC are together greater than CD, DB. Still more, then, are BA, AC (which were shown to be together greater than BE, EC) together greater than CD, DB. Again, because the exterior angle of a triangle is* greater than the interior opposite angle, the exterior angle DEC of the triangle EAB is greater than the angle EAB. And for the same reason the exterior angle BDC of the triangle DEC is greater than the angle DEC. Still more, then, is it greater than the angle EAB, which was shown to be less than DEC.

* I. 16.

And by parity of reasoning, the like may be proved in every other triangle. Wherefore, universally, if from the ends of the side of a triangle, &c. Which was to be demonstrated.