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.

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 cut off a part AD equal to AB.

Proof.

Join BD.

Then in the triangle ABD, because AB is equal to AD,

therefore the angle ABD is equal to the angle ADB.

I. 3.

I. 5.

But the exterior angle ADB of the triangle DCB is greater than the interior opposite angle DCB, that is, greater than the angle ACB.

I. 16.

Therefore also the angle ABD is greater than the angle ACB; still more then is the angle ABC greater than the angle

ACB.

Euclid enunciated Proposition 18 as follows:

Q.E.D.

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. If Euclid's enunciations of Props. 18 and 19 are adopted, it is important to remember that in each case the part of the triangle first named points out the hypothesis.]

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.

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; I. 5. but it is not.

Neither is AC less than AB;

Нур.

for then the angle ABC would be less than the angle ACB; I. 18.

but it is not.

Hyp.

That is, AC is neither equal to, nor less than AB.
Therefore AC is greater than AB. 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 suppositions must necessarily be true; and it consists in shewing that each of these suppositions is false 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 on Props. 18 and 19 see page 44.]

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

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 shall be greater than BA;

and CB, BA shall be greater than AC.

Construction. Produce BA to D, making AD equal to AC. I. 3.

Proof.

Join DC.

Then in the triangle ADC,

because AD is equal to AC,

Constr.

therefore the angle ACD is equal to the angle ADC. 1. 5. But the angle BCD is greater than its part the angle ACD; 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, 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 44.]

I. 19.

Q.E.D.

PROPOSITION 21. THEOREM.

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.

B

Let ABC be a triangle, and from B, C, the ends of the side BC, let the 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.

EXERCISES.

ON PROPOSITIONS 18 AND 19.

1. The hypotenuse is the greatest side of a right-angled triangle.

2. If two angles of a triangle are equal to one another, the sides also, which subtend the equal angles, are equal to one another. Prove this [i.e. Prop. 6] indirectly by using the result of Prop. 18.

3. BC, the base of an isosceles triangle ABC, is produced to any point D; shew that AD is greater than either of the equal sides.

4. If in a quadrilateral the greatest and least sides are opposite to one another, then each of the angles adjacent to the least side is greater than its opposite angle.

5. In a triangle ABC, if AC is not greater than AB, shew that any straight line drawn through the vertex A and terminated by the base BC, is less than AB.

6. ABC is a triangle, in which OB, OC bisect the angles ABC, ACB respectively: shew that, if AB is greater than AC, then OB is greater than OC.

ON PROPOSITION 20..

7. The difference of any two sides of a triangle is less than the third side.

8. In a quadrilateral, if two opposite sides which are not parallel are produced to meet one another; shew that the perimeter of the greater of the two triangles so formed is greater than the perimeter of the quadrilateral.

9. The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter.

10.

The perimeter of a quadrilateral is greater than the sum of its diagonals.

11.

Obtain a proof of Proposition 20 by bisecting an angle by a straight line which meets the opposite side.

ON PROPOSITION 21.

12. In Proposition 21 shew that the angle BDC is greater than the angle BAC by joining AD, and producing it towards the base.

13. The sum of the distances of any point within a triangle from its angular points is less than the perimeter of the triangle.