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

Take in CD, CE, any points D, D E, and join DE; and make the

triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC, so that CD be equal to AF, CE to AG, and DE to FG; and because DC, CE are equal to FA, AG, each to each, and the base DE to the base FG; the angle DCE is equal to the angle FAG. Therefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE.

XXIV. 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.

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 the angle EDG equal to BAC; and make DG equal to AC or DF, and join EG, GF.

the

Because AB is equal to DE, and AC to DG, two sides BA, AC, are equal to the two ED, DG, each to each, and the angle

D

DGF; but DGF is greater

F

than EGF; therefore DFG is greater than EGF; and

much more is EFG greater than EGF; and because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater side is opposite to the greater angle; the side EG is therefore greater than the side EF; but EG is equal to BC; and therefore also BC is greater than EF. XXV.-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.

B

D

For if it be not greater it must either be equal to it, or less; but the angle BAC is not equal to EDF, because then the base BC would be equal to EF; but it is not; therefore the angle BAC, is not equal to EDF; neither is it less; because then the base BC would be less than EF; but it is not; therefore the angle BAC is not less than EDF; and it was shown that it was not equal to it; therefore the angle BAC is greater than EDF.

XXVI.-If two triangles have two angles of 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 DEF, EFD; viz., ABC to DEF, and BCA to EFD; also one side equal to one

B

E

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 BAC 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, and make BG equal to DE, and join GC; therefore, because BG is equal to DE, add BC to EF, the two sides GB, BC, are equal to the two, DE, EF, each to each; and the angle GBC is equal to DEF; therefore the base GC is equal to DF, and the triangle GBC to DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle GCB is equal to DFE; but DFE is, by the hypothesis, equal to BCA; wherefore also the angle BCG is equal to BCA, the less to the greater, which is impossible; therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF; therefore the two, AB, BC, are equal to the two DE, EF, each to each; and the angle ABC is equal to DEF; therefore the base AC is equal to 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,

B

HC E

viz., AB to DE; likewise in this case, the other sides shall be equal, AC to DF, and BC to EF; and also the 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, the two AB, BH, are equal to the two DE, EF, each to each; and they contain equal angles; therefore the base AH is equal to DF, and the triangle ABH to

DEF, and the other angles shall be equal, each to each, to which the equal sides are opposite; therefore the angle BHA is equal to the angle EFD; but EFD is equal to the angle BCA; therefore also 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; wherefore BC is not unequal to EF, that is, it is equal to it; and AB is equal to DE; therefore the two AB, BC, are equal to the two DE, EF, each to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. XXVII.—If a straight line falling upon two other

straight lines makes 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 AE
GEF is a triangle, and its exterior
angle AEF is greater than the in-
terior and opposite angle EFG; but

F

B

D

it is also equal to it, which is impossible; therefore AB and CD being produced do not meet towards B, D. In like manner it may be demonstrated that they do not meet towards H, C; but those straight lines which meet neither way, though produced ever so far, are parallel to one another. AB therefore is parallel to CD. XXVIII.—If a straight line, falling upon two other

straight lines, makes the exterior angle equal to the interior and opposite upon the same side of the line; or makes 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.

Because the angle EGB is equal to GHD, and the angle EGB equal to AGH, the angle AGH is equal to GHD, and they are the alternate

angles; therefore AB is parallel to CD. Again, because the angles BGH, GHD, A are equal to two right angles; and that AGH, BGH, are equal to two right angles; the angles AGH, BGH,

E

G

H

are equal to the two BGH, GHD: take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD. XXIX.-If a straight line fall upon two parallel straight

lines, it makes the alternate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles.

Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHD, are equal to one another; and the exterior angle EGB is equal to the interior and opposite, upon the same side, GHD; and the two interior angles BGH, GHD, upon the same side, are to- E gether equal to two right angles.

A

B

D

F

For, if AGH be not equal to GHD, one of them must be greater than the other; let AGH be the greater; and because the angle AGH, is greater than GHD, add to each of them the angle BGH; therefore the angles AGH, BGH, are greater than BGH, GHD; but the angies AGH, BGH, are equal to two right angles; therefore the angles BGH,