Given

< ABC 4 ECA.

AC not
AB.

AC not <
AB.

Make

AD AC.

< BCD> 4 BDC.

Proposition 19.-Theorem,

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

Let ABC be a triangle, of which the angle ABC is greater than the angle BCA;

The side AC shall be greater than the side AB.

PROOF.-If AC be not greater than AB, it must either be equal to or less than AB.

It is not equal, for then the angle ABC would be equal to the angle BCA (I. 5); but it is not (Hyp.);

Therefore AC is not equal to AB.

Neither is AC less than AB, for then the angle ABC would be less than the angle BCA (I. 18); but it is not (Hyp.);

Therefore AC is not less than AB.

And it has been proved that AC is not equal to AB;
Therefore AC is greater than AB.

Therefore, the greater angle, &c. Q. E. D.

Proposition 20.-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. CONSTRUCTION.-Produce BA to the point D, making AD equal to AC (I. 3), and join DC.

PROOF.-Because DA is equal to AC, the angle ADC is equal to the angle ACD (I. 5).

But the angle BCD is greater than the angle ACD (Ax. 9);
Therefore the angle BCD is greater than the angle ADC
D (or BDC).

C

And because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater angle is subtended by the greater side;

than the side BC (I. 19).

Therefore the side DB is greater

But BD is equal to BA and AC;

Therefore BA, AC are greater than BC.

In the same manner it may be proved that AB, BC are greater than AC; and BC, CA greater than AB. Therefore any two sides, &c. Q. E.D.

Proposition 21.-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 be less than the other two sides of the triangle, but shall contain a greater angle.

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

BD, DC shall be less than the sides BA, AC;

But BD, DC shall contain an angle BDC greater than the angle BAC.

CONSTRUCTION.-Produce BD to E. PROOF.-1. Because two sides of a triangle are greater than the third side (I. 20), the two sides BA, AE, of the triangle BAE are greater than BE.

To each of these add EC.

Therefore the sides BA, AC, are B greater than BE, EC (Ax. 4).

A

D

E

Again, because the two sides CE, ED, of the triangle EC.

CED are greater than CD (I. 20),

To each of these add DB.

Therefore CE, EB are greater than CD, DB (Ax. 4).

But it has been shown that BA, AC are greater than BE, EC;

Much more then are BA, AC greater than BD, DC. PROOF. 2. Again, because the exterior angle of a triangle is greater than the interior and opposite angle (I. 16), therefore BDC, the exterior angle of the triangle CDE, is greater than CED or CEB.

For the same reason, CEB, the exterior angle of the triangle ABE, is greater than the angle BAE or BAC.

EBCD + DB.

Again

< BDC > < CEB, and

< CEB >

BAE.

.. 4 BDC BAC.

And it has been shown that the angle BDC is greater than CEB;

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

Therefore, if from the ends, &c.

Q. E. D.

DF, FG,

=

Proposition 22.-Problem.

To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these lines must be greater than the third (I. 20).

Let A, B, C be the three given straight lines, of which any two whatever are greater than the third-namely, A and B greater than C, A and C greater than B, and B and C greater than A;

It is required to make a triangle of which the sides shall be equal to A, B, and C, each to each.

CONSTRUCTION.-Take a straight line DE terminated at the point D, but unlimited towards E.

GH respectively A, D B,C.

G

/H

L

A

B

FD as radius.

and GH as radius.

FK = A

GK = C.

E

Make DF equal to A, FG equal to B, and GH equal to C (I. 3).

From the centre F, at the distance FD, describe the circle DKL (Post. 3).

From the centre G, at the distance GH, describe the circle HLK (Post. 3).

Join KF,

KG.

Then the triangle KFG shall have its sides equal to the three straight lines A, B, C.

PROOF.-Because the point F is the centre of the circle
DKL, FD is equal to FK (Def. 15).
But FD is equal to A (Const.);

Therefore FK is equal to A (Ax. 1).

Again, because the point G is the centre of the circle
IILK, GH is equal to GK (Def. 15).
But GH is equal to C (Const.);
Therefore GK is equal to C (Ax. 1),

And FG is equal to B (Const.);

Therefore the three straight lines KF, FG, GK are equal to the three A, B, C, each to each.

Therefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines A, B, C. Q. E. F.

Proposition 23.-Problem.

At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle.

It is required to take an angle at the point A, in the straight line AB, equal to the rectilineal angle DCE.

FG B

Make
A AFG so

G that

AF CD
FG DE
AGCE.

Then the angle FAG shall be equal to the angle DCE. PROOF.-Because DC, CE are equal to FA, AG, each to each, and the base DE equal to the base FG (Const.), The angle DCE is equal to the angle FAG (I. 8). Therefore, at the given point A, in the given straight line < DCE= AB, the angle FAG has been made equal to the given rectilineal angle DCE. Q. E. F.

Proposition 24.-Theorem,

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other.

Then

< FAG

Let ABC, DEF, be two triangles which have

The two sides AB, AC equal to the two DE, DF, each to each-namely, AB to DE,and AC to DF,

But the angle BAC greater than the angle EDF;
The base BC shall be greater than the base EF.
CONSTRUCTION.-Let the side DF of the triangle DEF be

greater than its side DE.

Then at the point D, in the straight line ED, make the angle EDG equal to the angle BAC (I. 23).

Make DG equal to AC

or DF (I. 3).

Join EG, GF.

PROOF.-Because AB is

equal to DE (Hyp.), and AC to DG (Const.), the two sides
BA, AC are equal to the two ED, DG, each to each;

And the angle BAC is equal to the angle EDG (Const.);
Therefore the base BC is equal to the base EG (I. 4).
And because DG is equal to DF (Const.), the angle DFG
is equal to the angle DGF (I. 5).

But the angle DGF is greater than the angle EGF (Ax. 9);
Therefore the angle DFG is greater than the angle EGF;
Much more then is the angle EFG greater than the angle
EGF.

And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle is subtended by the greater side,

Therefore the side EG is greater than the side EF (I. 19).
But EG was proved equal to BC;

Therefore BC is greater than EF.

Therefore, if two triangles, &c. Q. E. D.

Proposition 25.—Theorem.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other, the angle contained by the sides of that which has the greater base shall be greater than the angle contained by the sides equal to them of the other.