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

3. GH is equal to GK;

but GH is equal to C; therefore also

4. GK is equal to C;

and FG is equal (Constr.) to B; therefore the three straight lines KF, FG, GK, are equal to the three A, B, C. And therefore

5. 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.

PROP. 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 the given point in it, and DCE the given rectilineal angle; it is required to make an angle at the given point in the given straight line AB, that shall be equal to the given rectilineal angle DCE.

Take in CD, CE, any points D, E, and join DE; and make (I. 22.) the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, CE, so that CD be equal to AF, CE to AG, and DE to FG; the angle FAG is equal to the angle DCE.

Because DC, CE, are equal to FA, AG, each to each, and the base DE to the base FG; (I. 8.)

The angle DCE is equal to the angle FAG.

Therefore, at the given point ▲ in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done.

PROP. 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, viz. AB equal to DE, and AC to

DF; but the angle BAC greater than the angle EDF; the base BC is also greater than the base EF.

A

D

Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make (I. 23.) the angle EDG equal to the angle BAC; and make DG equal (I. 3.) to AC or DF, and join EG, GF.

Because AB is equal to DE, and AC to DG, the two sides BA, AC, are equal to the two ED, DG, each to each, and the angle BAC is equal (Constr.) to the angle EDG; therefore (I. 4.)

1. The base BC is equal to the base EG.

And because DG is equal to DF, (1, 5.)

2.

The angle DFG is equal to the angle DGF; but the angle DGF is greater than the angle EGF, therefore The angle DFG is greater than EGF;

3.

and much more

4. The angle EFG is greater than the angle EGF.

And because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater (I. 19.) side is opposite to the greater angle; therefore

5. The side EG is greater than the side EF;

but EG is equal to BC; and therefore also 6. BC is greater than EF.

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

PROP. 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 the one greater than the base of the other; the angle also 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.

Let ABC, DEF, be two triangles which have the two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the base CB greater than the base EF; the angle BAC is likewise greater than the angle EDF.

For if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle EDF, because then the base BC would be equal (I. 4.) to EF; but it is not, therefore

1. The angle BAC is not equal to the angle EDF.

Neither is it less; because then the base BC would be less (I. 24.) than the base EF; but it is not, therefore

2.

The angle BAC is not less than the angle EDF; and it was shown that it is not equal to it; therefore

3. The angle BAC is greater than the angle EDF.

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

PROP. XXVI.- -THEOREM.

If two triangles have two angles of the one equal to two angles of the other, each to each; and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each; then shall the other sides be equal, each to each; and also the third angle of the one to the third angle of the other.

Let ABC, DEF, be two triangles which have the angles ABC, BCA, equal to the angles DEF, EFD, viz. ABC to DEF, and BCA to EFD, also one side equal to one side; and first let those sides be equal which are adjacent to the angles that are equal in the two triangles, viz. BC to EF; the other sides shall be equal, each to each, viz. AB to DE, and AC to DF; and the third angle BẮC to the third angle EDF.

For if AB be not equal to DE, one of them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join GC; therefore, because BG is equal to DE, and BC to EF,

1.

The two sides GB, BC, are equal to the two DE, EF,
each to each;

and the angle GBC is equal to the angle DEF; therefore (I. 4.)
2. The base GC is equal to the base DF, and the triangle

GBC to the triangle DEF,

and the other angles to the other angles, cach to each, to which the equal sides are opposite; therefore

3.

The angle GCB is equal to the angle DFE;

but DFE is, by the hypothesis, equal to the angle BCA; wherefore also The angle BCG is equal to the angle BCA,

4.

the less to the greater, which is impossible; therefore

5. AB is not unequal to DE,

that is, it is equal to it; and BC is equal (Hyp.) to EF; therefore The two AB, BC, are equal to the two DE, EF, each

6.

to each;

and the angle ABC is equal (Hyp.) to the angle DEF; therefore (I. 4.) 7. The base AC is equal to the base DF, and the third angle BAC to the third angle EDF.

Next, let the sides which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this case, the other sides shall be equal, AC to DF, and BC to EF; and also the third angle BAC to the third EDF.

For if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH; and because BH is equal to EF, and AB to DE,

1.

to each;

The two AB, BH, are equal to the two DE, EF, each

and they contain equal angles; therefore (I. 4.)

2. The base AH is equal to the base DF, and the triangle

ABH to the triangle DEF,

and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore

3.

The angle BHA is equal to the angle EFD;

but EFD is equal (Hyp.) to the angle BCA; therefore also

4. The angle BHA is equal to the angle BCA,

that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA, which is impossible (I. 16.); wherefore 5. BC is not unequal to EF,

that is, it is equal to it; and AB is equal (Hyp.) to DE; therefore 6. The two AB, BC, are equal to the two DE, EF, each

to each;

and they contain (Hyp.) equal angles; wherefore (I. 4.)

7. The base AC is equal to the base DF, and the third angle BAC to the third angle EDF.

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

PROP. XXVII.-THEOREM.

If a straight line, falling upon two other straight lines, make the alternate angles equal to one another, these two straight lines shall be parallel,

Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD, equal to one another; AB is parallel to CD.

For, if it be not parallel, AB and CD being produced, shall meet either towards B, D, or towards A, C; let them be produced and meet towards B, D, in the point G; therefore

1. GEF is a triangle, and its exterior angle AEF is greater (I. 16.) than the interior and opposite angle EFG;

but it is also equal to it, which is impossible; therefore

2.

AB and CD being produced do not meet towards B, D.

In like manner it may be demonstrated that

3.

AB and CD do not meet towards A, C;

but those straight lines which meet neither way, though produced ever so far, are parallel (Def. 35.) to one another; therefore

4. AB is parallel to CD.

Wherefore if a straight line, &c.

Q.E.D.

PROP. XXVIII.-THEOREM.

If a straight line falling upon two other straight lines make the exterior angle equal to the interior and opposite upon the same side of the line; or make the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another.

Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD upon the same side; or make the interior angles on the same side BGH, GHD, together equal to two right angles; AB is parallel to CD.

E