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

Let ABC be a triangle: then shall any two of its angles, as ABC, ACB, be together less than two right angles. Construction. Produce the side BC to D.

Proof. Then because ACD is an exterior angle of the triangle ABC,

therefore it is greater than the interior opposite angle

ABC.

To each of these add the angle ACB:

Ax. 4.

I. 16. then the angles ACD, ACB are together greater than the angles ABC, ACB. But the adjacent angles ACD, ACB are together equal to two right angles. Therefore the angles ABC, ACB are together less than two right angles.

I. 13.

Similarly it may be shewn that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles.

Q. E. D.

NOTE. It follows from this Proposition that every triangle must have at least two acute angles: for if one angle is obtuse, or a right angle, each of the other angles must be less than a right angle.

EXERCISES.

1. Enunciate this Proposition so as to shew that it is the converse of Axiom 12.

2. If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles.

3. Shew how a proof of Proposition 17 may be obtained by joining each vertex in turn to any point in the opposite side.

H. E.

3

PROPOSITION 18. THEOREM.

If one side of a triangle be greater than another, then the angle opposite to the greater side shall be greater than the angle opposite to the less.

A

D

B

Let ABC be a triangle, in which the side AC is greater than the side AB :

then shall the angle ABC be greater than the angle ACB. Construction. From AC, the greater, cut off a part AD equal

therefore the angle ABD is equal to the angle ADB.

1. 5.

I. 16.

But the exterior angle ADB of the triangle BDC is greater than the interior opposite angle DCB, that is, greater than the angle ACB. Therefore also the angle ABD is greater than the angle ACB; still more then is the angle ABC greater than the angle ACB.

Q. E. D.

Euclid enunciated Proposition 18 as follows:

The greater side of every triangle has the greater angle opposite to it.

[This form of enunciation is found to be a common source of difficulty with beginners, who fail to distinguish what is assumed in it and what is to be proved.]

[For Exercises see page 38.]

PROPOSITION 19. THEOREM.

If one angle of a triangle be greater than another, then the side opposite to the greater angle shall be greater than the side opposite to the less.

A

B

Let ABC be a triangle in which the angle ABC is greater than the angle ACB:

then shall the side AC be greater than the side AB.
Proof. For if AC be not greater than AB,

it must be either equal to, or less than AB.

But AC is not equal to AB,

for then the angle ABC would be equal to the angle ACB; 1.5. but it is not.

Neither is AC less than AB;

Hyp.

for then the angle ABC would be less than the angle ACB; 1.18. but it is not:

Therefore AC is neither equal to, nor less than AB.
That is, AC is greater than AB.

Нур.

Q.E. D.

NOTE. The mode of demonstration used in this Proposition is known as the Proof by Exhaustion. It is applicable to cases in which one of certain mutually exclusive suppositions must necessarily be true; and it consists in shewing the falsity of each of these suppositions in turn with one exception: hence the truth of the remaining supposition is inferred.

Euclid enunciated Proposition 19 as follows:

The greater angle of every triangle is subtended by the greater side, or, has the greater side opposite to it.

[For Exercises see page 38.]

PROPOSITION 20. THEOREM.

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

A

B

Let ABC be a triangle:

then shall any two of its sides be together greater than the third side:

namely, BA, AC, shall be greater than CB;

AC, CB greater than BA;

and CB, BA greater than AC.

Construction. Produce BA to the point D, making AD equal

Then in the triangle ADC,
because AD is equal to AC,

1. 3.

Constr.

therefore the angle ACD is equal to the angle ADC. I. 5. But the angle BCD is greater than the angle ACD; Ax. 9. therefore also the angle BCD is greater than the angle ADC, that is, than the angle BDC.

And in the triangle BCD,

because the angle BCD is greater than the angle BDC, Pr. therefore the side BD is greater than the side CB.

But BA and AC are together equal to BD;
therefore BA and AC are together greater than CB.
Similarly it may be shewn

that AC, CB are together greater than BA;
and CB, BA are together greater than AC.

[For Exercises see page 38.]

I. 19.

Q. E. D.

If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, then these straight lines shall be less than the other two sides of the triangle, but shall contain a greater angle.

A

E

B

Let ABC be a triangle, and from B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle:

then (i) BD and DC shall be together less than BA and AC; (ii) the angle BDC shall be greater than the angle BAC.

Construction. Produce BD to meet AC in E.

Proof. (i) In the triangle BAE, the two sides BA, AE are together greater than the third side BE:

to each of these add EC ;

I. 20.

then BA, AC are together greater than BE, EC. Ax. 4. Again, in the triangle DEC, the two sides DE, EC are together greater than DC:

to each of these add BD;

I. 20.

then BE, EC are together greater than BD, DC. But it has been shewn that BA, AC are together greater than BE, EC:

still more then are BA, AC greater than BD, DC. (ii) Again, the exterior angle BDC of the triangle DEC is greater than the interior opposite angle DEC; I. 16. and the exterior angle DEC of the triangle BAE is greater than the interior opposite angle BAE, that is, than the angle BAC;

I. 16.

still more then is the angle BDC greater than the angle BAC.

Q.E.D.