BOOK I. and because ADB is the exterior angle of the triangle BDC,

b xvi. 1. • v. 1.

it is greater than the interior and opposite angle DCB;

с

but ADB is equal to ABD,

because the side AB is equal to the side AD;

therefore the angle ABD is likewise greater than the angle

АСВ.

Wherefore much more is the angle ABC greater than ACB.
Therefore the greater side, &c.
Q. E. D.

a v. 1.

PROP. XIX. THEOR.

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 is likewise greater than the side AB.

therefore AC is not equal to AB;

neither is it less; because then

xviii. 1. the angle ABC would be less than the angle ACB; but it is not;

therefore the side AC is not less than AB;

and it has been shown that it is not equal to AB;

therefore AC is greater than AB.

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

[The latter of these propositions is the converse of the BOOK L former: the former shows that if one side be greater than another, the angle opposite that side is greater than the angle opposite the other; the latter, that if one angle is greater than another, the side opposite that angle is greater than the side opposite the other.]

PROP. XX. THEOR.

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

Let ABC be a triangle;

side.

any two sides of it together are greater than the third side, viz. the sides BA, AC greater than the side BC;

and AB, BC greater than AC;

and BC, CA greater than AB.

Produce BA to the point D, and make AD equal to AC; and join DC.

Because DA is equal to AC,

B

the angle ADC is likewise equal to ACD;

A

but the angle BCD is greater than the angle ACD; therefore the angle BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB

is greater than its angle BDC,

с

and that the greater side is opposite to the greater angle ; therefore the side DB is greater than the side BC;

but DB is equal to BA and AC;

therefore the sides BA, AC are greater than BC.

In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB.

Therefore any two sides, &c. Q. E. D.

* iii 1.

b v. 1.

c xix. 1.

BOOK I.

PROP. XXI.

THEOR.

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 the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC,

to the point D within it;

BD and DC are less than the other two sides

BA, AC of the triangle,

but contain an angle BDC greater than the angle BAC.

Produce BD to E;

And because two sides of a triangle are greater than the third side,

the two sides BA, AE of the triangle ABE

therefore the sides CE, EB are greater than CD, DB;
but it has been shown that BA, AC

are greater than BE, EC,

much more then are BA, AC greater than BD, DC.

Again, because the exterior angle of a triangle

is greater than the interior and opposite angle,
the exterior angle BDC of the triangle CDE

is greater than CED;

for the same reason,

C

the exterior angle CEB of the triangle ABE is greater than

BAC;

and it has been demonstrated that the angle BDC

is greater than the angle CEB;

much more then is the angle BDC

greater than the angle BAC.

Therefore, if from the ends of, &c. Q. E. D.

PROP. XXII. PROB.

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

Let A, B, C be the three given straight lines,

of which any two whatever are greater than the third, viz.

A and B greater than C;

A and C greater than B;

and B and C than A.

It is required to make a triangle,

of which the sides shall be equal to A, B, C, each to each.

Take a straight line DE terminated at the point D,

the triangle KFG has its sides equal to the three straight

lines A, B, C.

Because the point F is the centre of the circle DKL,

BOOK I.

a xx. 1.

a iii. 1.

b 3 Post.

a 15 Def.

BOOK I. therefore FK is equal to A:

Again, because G is the centre of the circle LKH,

с

⚫ 15 Def. GH is equal to GK;

but GH is equal to C;

therefore also GK is equal to C;

and FG is equal to B;

therefore the three straight lines KF, FG, GK,

are equal to the three A, B, C:

And therefore the triangle KFG has its three sides

KF, FG, GK

equal to the three given straight lines, A, B, C.
Which was to be done.

[Take three straight lines which do not fulfil the condi

A

B

C

Another mode of drawing such a triangle would be as

tance equal to C, draw a circular arc cutting the other in K. Join DK and KF. If B and C should be together equal to A, the point K would lie on the line DF. If they should together be less than A, the circles would have no point in common.]