II. In the triangle ABC the angle BAC is bisected by the line AD. Prove that BA is greater than BD; and CA greater than CD. (Notice that

B

ADC is the exterior angle of the triangle A B D, and therefore (Euc. I. 16) greater than the interior opposite angle BAD.)

THEOREM (Euclid I. 19).

Repeat. The enunciation of Euc. I. 5, and of Euc. I. 18.

General Enunciation.

The greater angle of every triangle is subtended by the greater side.

NOTE 1.-An angle is said to be subtended by a side, when that side is opposite to it.

NOTE 2.- -This theorem is the converse of the preceding. The pupil is not to suppose that the converse of every theorem is true because the theorem itself is shown to be true. This is far from being the case.

NOTE 3.-This proposition is another example of the negative method of proof. This method is generally used by Euclid to prove the converse of a theorem. Particular Enunciation.

Given. --The triangle A B C having the angle A B C greater than the angle A CB.

Required. To prove that the side AC is greater than the side AB.

Proof.

B

Let us suppose that AC is not greater than A B.

Then one of two things must be the case: (a) A C must be equal to A B, or

(b) A C must be less than AB.

(a) Now A C cannot be equal to A B,

for, if it were, angle ABC would be equal to angle A CB (by Euc. I. 5).

But angle ABC is not equal to ACB (see Given). (d) .. A C is not equal to A B.

(b) A C cannot be less than AB,

for, if it were, angle ABC would be less than angle ACB (by Euc. I. 18).

But A B C is not less than ACB (see Given). ... A C is not less than A B.

(e)

A C is not equal to A B (d).

A C is not less than A B (e).

... AC must be greater than A B.

Q. E. D.

EXERCISE.-Given. Angle BAC greater than angle ABC. Prove that side B C is greater than side A C (in figure of above proposition).

THEOREM (Euclid I. 20).

Repeat. The enunciations of Euc. I. 5 and I. 19, and Axiom 9.

General Enunciation.

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

Particular Enunciation.

Given. The triangle ABC.

(a) BA, A C are greater than B C,
(b) AB, BC are greater than A C,
(c) A C, CB are greater than A B.

A

B

Construction.-Produce BA,

(d) Making A D equal to A C.
Join DC.

F

C

Proof.

ADC is an isosceles triangle, having AD equal to AC (d).

... By Euc. I. 5 (e) angle A DC is equal to angle A CD,

But, angle B CD is greater than angle A CD. .. By (e), angle B C D is greater than angle ADC.

Again, by Euc. I. 19, the greater angle of every triangle is subtended by the greater side.

(f) ... side B D is greater than side B C.
But side BD is made up of B A, A D,
and AD is equal to A C (d).

... sides BA, A C are greater than B C.

Q. E. D.

EXERCISES.-I. By producing CB prove that A B,

B C are greater than A C.

II. By producing another side prove that A C, CB are greater than AB.

THEOREM (Euclid I. 21).

Repeat. The enunciations of Euc. I. 20, and

I. 16.

General Enunciation.

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.

Given. (a) The triangle ABC,

(b) The two straight lines BD, CD, drawn from the ends of the side B C to the point D within the triangle ABC.

A

B

Construction.

Produce BD to meet A C in E (see next page).

Proof (of a).

By Euc. I. 20. Any two sides of a triangle are greater than the third side.