Then the angle ABC shall be greater than the angle ACB.

A

D

B

Since the side AC is greater than the side AB, (hyp.)
make AD equal to AB, (1. 3.) and join BD.

Then, because AD is equal to AB, in the triangle ABD, therefore the angle ABD is equal to the angle ADB, (1. 5.) but because the side CD of the triangle BDC is produced to A, therefore the exterior angle ADB is greater than the interior and opposite angle DCB; (1. 16.)

but the angle ADB has been proved equal to the angle ABD, therefore the angle ABD is greater than the angle DCB; wherefore much more is the angle ABC greater than the angle ACB. Therefore the greater side, &c. Q. E.D.

PROPOSITION XIX. 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.

Then the side AC shall be greater than the side AB.

A

B

For, if AC be not greater than AB,
AC must either be equal to, or less than AB;
if AC were equal to AB,

then the angle ABC would be equal to the angle ACB; (1. 5.) but it is not equal; (hyp.)

therefore the side AC is not equal to AB.
Again, if AC were less than AB,

then the angle ABC would be less than the angle ACB; (1. 18.) but it is not less, (hyp.)

therefore the side AC is not less than AB; and AC has been shewn to be not equal to AB;

therefore AC is greater than AB. Wherefore the greater angle, &c.

PROPOSITION XX.

Q.E.D.

THEOREM.

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

Let ABC be a triangle.

Then any two sides of it together shall be greater than the third side, viz. the sides BĂ, AC greater than the side BC;

AB, BC greater than AC; and BC, CA greater than AB.

A

B

Produce the side BA to the point D, make AD equal to AC, (1. 3.) and join DC. Then 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 the angle ACD; (ax. 9.) therefore also the angle BCD is greater than the angle ADC. And because in the triangle DBC,

the angle BCD is greater than the angle BDC,

and that the greater angle is subtended by the greater side; (1. 19.)
therefore the side DB is greater than the side BC;
but DB is equal to BA and AC,

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

PROPOSITION XXI. THEOREM.

If from the ends of a 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 a point D within the triangle.

Then BD and DC shall be less than BA and AC the other two sides of the triangle,

but shall contain an angle BDC greater than the angle BAC.

Produce BD to meet the side AC in E.

Because two sides of a triangle are greater than the third side, (1. 20.) therefore the two sides BA, AE of the triangle ABE are greater than BE;

to each of these unequals add EC;

therefore the sides BA, AC are greater than BE, EC. (ax. 4.) Again, because the two sides CE, ED of the triangle CED are greater than DC; (1. 20.)

add DB to each of these unequals ;

therefore the sides CE, EB are greater than CD, DB. (ax. 4.) But it has been shewn that BA, AC are greater than BÈ, 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; (1. 16.)

therefore the exterior angle BDC of the triangle CDE is greater than the interior and opposite angle CED;

for the same reason, the exterior angle CED of the triangle ABE is greater than the interior and opposite angle BAC;

and it has been demonstrated,

that the angle BDC is greater than the angle CEB; much more therefore is the angle BDC greater than the angle BAC. Therefore, if from the ends of the side, &c. Q. E.D.

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

of which any two whatever are greater than the third, (1. 20.) namely, A and B greater than C;

A and C greater than B;

and B and C greater than 4.

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

Take & straight line DE terminated at the point D, but unlimited towards E,

make DF equal to A, FG equal to B, and GH equal to C; (1. 3.) from the center F, at the distance FD, describe the circle DKL; (post 3.)

from the center G, at the distance GH, describe the circle HLK; from K where the circles cut each other, draw KF, KG to the points F, G;

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

Because the point Fis the center of the circle DKL,
therefore FD is equal to FK; (def. 15.)

but FD is equal to the straight line 4;

therefore FK is equal to 4.

Again, because G is the center of the circle HKL;
therefore GH is equal to GK, (def. 15.)
but GH is equal to C;

therefore also GK is equal to C; (ax. 1.)
and FG is equal to B;

therefore the three straight lines KF, FG, GK, are respectively 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. Q.E.F.

PROPOSITION XXIII. 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, at the given point A in the given straight line AB, to make an angle that shall be equal to the given rectilineal angle DCE.

In CD, CE, take any points D, E, and join DE;

on AB, make the triangle AFG, the sides of which shall be equal o the three straight lines CD, DE, EC, so that AF be equal to CD, AG to CE, and FG to DE. (1. 22.)

Then the angle FAG shall be equal to the angle DCE.
Because FA, AG are equal to DC, CE, each to each,
and the base FG is equal to the base DE;

therefore the angle FAG is equal to the angle DCE. (1.8.) Wherefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Q.E.F.

PROPOSITION XXIV. 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.

Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, namely, AB equal to DE, and ACto DF; but the angle BAC greater than the angle EDF. Then the base BC shall be greater than the base EF.

Of the two sides DE, DF, let DE be not greater than DF, at the point D, in the line DE, and on the same side of it as DF, make the angle EDG equal to the angle BAC; (1. 23.) make DG equal to DF or AC, (1. 3.) and join EG, GF. Then, because DE is equal to AB, and DG to AC, the two sides DE, DG are equal to the two AB, AC, each to each and the angle EDG is equal to the angle BAC; therefore the base EG is equal to the base BC. (1. 4.) And because DG is equal to DF in the triangle DFG, therefore the angle DFG is equal to the angle DGF; (1. 5.) but the angle DGF is greater than the angle EGF; (ax. 9.) therefore the angle DFG is also greater than the angle EGF; much more therefore is the angle EFG greater than the angle EGF. And because in the triangle EFG, the angle EFG is greater than the angle EGF,

and that the greater angle is subtended by the greater side; (I. 19.) therefore the side EG is greater than the side EF; but EG was proved equal to BC; therefore BC is greater than EF. Wherefore, if two triangles, &c. Q. E.D.

PROPOSITION XXV. THEOREM.

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

Let ABC, DEF be two triangles which have the two sides AB, AC, equal to the two sides DE, DF, each to each, namely, AB equal to DE, and AC to DF; but the base BC greater than the base EF. Then the angle BAC shall be greater than the angle EDF.

For, if the angle BAC be not greater than the angle EDF,
it must either be equal to it, or less than it.

If the angle BAC were equal to the angle EDF,
then the base BC would be equal to the base EF; (1. 4.)
but it is not equal, (hyp.)

therefore the angle BAC is not equal to the angle EDF.
Again, if the angle BAC were less than the angle EDF,
then the base BC would be less than the base EF; (1. 24.)
but it is not less, (hyp.)

therefore the angle BAC is not less than the angle EDF; and it has been shewn, that the angle BAC'is not equal to the angle EDF; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c.

Q.E.D.