Join the points A and D by the right line A D.

The AACD, which is thus formed, is an isosceles ▲. Therefore, the CAD is equal to the ▲ C D A. (Prop. V.)

The CDA is the exterior formed by the production of the side BD in the ▲ A B D. Therefore, the exterior CDA is greater than the interior and opposite DBA. (Prop. XVI.)

Now the

CA B is greater than the

It is therefore also greater than the

is equal to the CAD.

CAD. CDA, which

Much more then must the CAB be greater than the CBA, which was before shown to be less than the CDA.

This proposition may also be proved in the following

B

Produce CA, the shorter of the two sides, and cut off from the produced line a part, CE, equal to the line CB.

Join the points E and B.

The AEC B, which is thus formed, is an isosceles A, conse

CEB is equal to the

CBE.

Now, the CAB is the exterior formed by producing the side E A in the ▲ BEA. Consequently, the CAB is greater than the interior and opposite AE B.

It follows, therefore, that the CAB is also greater CBE, which is equal to the CEB.

than the

Much more is the

which is less than the

CAB greater than the CBA,
CBE.

PROPOSITION XIX.

If one angle of a triangle be greater than another, the side which is opposite the greater angle is greater than the side which is opposite to the less.*

The proof of this proposition is of the indirect kind For the proof of it we must know,—

1. That if two sides of a A are equal, the s opposite the equal sides are also equal. (Prop. V.)

2. That if one side of a ▲ is greater than another, opposite the greater side is greater than opposite the less. (Prop. XVIII.)

the

the

CBA

Let A B C be a A, with respect to which we know that the is greater than the CAB. It has to be proved that the side CA is longer than the side C B.†

Three alternatives only are possible, one of which must be true.

*This proposition is often expressed thus:-" The greater of every▲ is subtended by the greater side, or has the greater side opposite to it." The meaning of this is precisely the same as what is given above.

Let the beginner carefully distinguish between this proposition and the last. In the 18th proposition we were supposed to know something about the length of two sides of a▲, but nothing about the s, except what we could deduce from our knowledge respecting the sides. In the 19th proposition we are supposed to know something respecting two s of a ▲, but nothing respecting the sides, except what we can deduce from our knowledge respecting the s. Compare the 5th and 6th propositions.

CB is greater than the side CA, or C B is equal to CA, or CB is less than CA.

Now, it is impossible that C B should be equal to CA, for, if it were, the CAB would be equal to the CBA; whereas, we know that the CBA is

greater than the CA B.

It is also impossible that C A should be less than C B, for, if it were, the CBA would be less than the CAB; whereas, we know that the CBA is greater than the CA B. It follows, therefore, that the side CA is greater than the side C B.

PROPOSITION XX.

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

For the construction employed in this proposition, we must be able,—

1. To produce a given line to any length. (Post. II.)

2. From the greater of two given lines to cut off a part equal to the less. (Prop. III.)

3. To join two given points by a straight line. (Post. I.)

For the proof of the proposition we must know,— 1. That if equal quantities are added to the same quantity, the sums are equal.

2. That if two sides of a ▲ are equal, the s opposite those sides are also equal. (Prop. V.)

Let the beginner remember that the object of this proposition is not to convince him of the truth stated, but to show how it may be connected with and deduced from the fundamental axioms and definitions.

3. That if one of a ▲ be greater than another, the side opposite the greater is greater than the side opposite the less. (Prop. XIX.)

Let it be required to prove that the sides AC and CB in

the A A C B are together greater than the side A B.

Produce one of the two sides

A

D

C

B

whose sum is to be compared with the third, as A C, to any length, and from the produced part cut off a part, CD, equal to C B.

Join the points D and B by the right line D B.

In the ▲ D C B the sides DC and C B are equal. It follows, therefore (according to the fifth proposition), that the CDB is equal to the

CBD.

Now, the ABD is greater than the CBD. Therefore the ABD is also greater than the CD B, which is equal to the CBD.

Consequently, in the ▲ ABD, one Z, namely ABD, is greater than another, namely A D B.

Therefore (as was proved in the nineteenth proposition), the side A D is greater than the side A B.

Now, A D is made up of the parts A C and CD, of which C D is equal to C B.

Therefore, the sum of A C and C D (i.e., the line A D) is equal to the sum of A C and C B. (Ax. II.)

The line A D was shown to be greater than the line A B.

Therefore, the sum of the lines A C and C B is also greater than the line A B.

It would have done equally well if the side B C had been produced, and a part, CE, cut off equal to the side CA. We should then have had the BAE

greater than the CA E, and, consequently, greater

E

than the CE A, which is equal to CAE. It would have followed that the side B E in the AABE is longer than the side BA. Then, BC and CE, taken together, being equal to BC and CA taken together, it would have followed that the sum of B C and CA is greater than the line BA.

A

B

PROPOSITION XXI.

If from the extremities of one side of a triangle lines be drawn to a point within the triangle, these two lines will together be less than the other two sides of the triangle, but will contain a greater angle.

For the proof of this proposition we must know that,

1. Any two sides of a ▲ are together greater than the third side. (Prop. XX.)

2. If the same quantity be added to each of two unequal quantities, the sum of that quantity, and the greater of the two unequals, will be greater than the sum of the same quantity, and the lesser of the unequals. (Ax. IV.)

3. The exterior, formed by producing a side of a A, is greater than either of the two interior opposites. (Prop. XVI.)

Let ABC be a ▲; and suppose that lines A P and BP are drawn from the extremities of the side A B, to the point P, taken within the A. It has to be proved that the sum of the lines A P and PB is