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.

Given two triangles, ABC and DEF, which have the two sides AB and AC equal to the two DE and DF, each to eachnamely, AB equal to DE,

and AC to DF; but the angle
BAC greater than the angle EDF;
to prove that
the base BC is also

greater than the base EF.

(Const.) Of the two sides DE, DF, let DE be the side which is not greater than the other,

and at the

B

point D, in the straight line DE,

make (I. 23) the angle EDG, equal

to the angle BAC;

A

H

G

F

and make DG equal to AC or DF (I. 3),

and join EG and GF.

(Dem.) Because DE is not greater than DF or DG, therefore the angle DGE is not greater than DEG; DHG is greater than DEG (I. 16);

but angle

therefore DHG is greater than DGE, and hence the side DG or DF (I. 19) is greater therefore the line EG must lie between EF and

than DH;

ED, or the point F must be below EG.

Because AB is equal to DE, and AC to DG, the two sides BA and AC are equal to the two ED and DG, each to each, and the angle BAC is equal to the angle EDG; therefore the base BC is equal to the base EG (I. 4); and because DG is equal to DF, the angle DFG is equal to the angle DGF (I. 5);

and

and

but the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than EGF (Ax. 14); still more is the angle EFG greater than the angle EGF; because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater angle has the greater side opposite to it (I. 19), the side EG is therefore greater than the side EF; but EG is equal to BC; also (Ax. 15) BC is greater than EF.

Otherwise :

Given two triangles ABC and DBC, to DB, and BC common, angle DBC; to prove that

(Const.) Draw BE (I. 9)

and therefore

which have AB equal but the angle ABC greater than the AC is greater than CD. bisecting the angle ABD,

which

will therefore fall in the greater of the two angles ABC, and cut

AC in E; join ED.

(Dem.) Because AB is equal to DB, and
BE common to the two triangles ABE,
DBE;
the two sides AB and BE are equal

to the two DB and BE, and the contained
angle ABE is equal to the contained angle
ᎠᏴᎬ, therefore (I. 4) the base AE is
equal to DE; to each of these equals add
EC, and AC is equal to DE and EC,
but DE and EC are greater than DC (I. 20);
greater than DC.

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.

Given two triangles ABC and DEF, which have the two sides AB and AC equal to the two sides DE and DF, each to each- namely, AB equal to DE, and AC to DF; but have the base CB greater than the base EF; to prove that the angle BAC is likewise greater than the angle EDF.

B

(Dem.) For, if the angle BAC be not greater than EDF, the angle BAC must either be equal to EDF, or less than it; but the angle BAC is not equal to the angle EDF, because, then, the base BC would be equal to EF (I. 4), which it is not; therefore the angle BAC is not equal to the angle EDF; neither is it less; base BC would be less than the base EF (I. 24), therefore the angle BAC is not less than the and it was shewn that it is not equal to it; angle BAC is greater than the angle EDF.

because then the which it is not; angle EDF; therefore the

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-namely, either the sides adjacent to the equal angles, or the sides opposite to the equal angles in each; then shall the other sides be equal, each to each; and the triangles shall be equal; and also the third angle of the one to the third angle of the other.

Or, if two angles and a side in one triangle be respectively

equal to two angles and a corresponding side in another triangle, the two triangles shall be equal in every respect.

Given two triangles ABC and DEF, which have the angles ABC and BCA of the one equal to the angles DEF and EFD of the other- namely, ABC to

DEF,

and BCA to EFD;

one side equal to one side;

also

and

first let those sides be equal which are adjacent to the angles that are equal

in the two triangles

A

namely, BC

and

one of them must of the two, (Dem.) therefore the two sides GB

DF; and the third angle BAC to the third angle EDF. (Const.) For, if AB be not equal to DE, be the greater. Let AB be the greater make BG equal to DE, and join GC; because BG is equal to DE, and BC to EF, and BC are equal to the two DE and EF, each to each; and the angle GBC is equal to the angle DEF; therefore the base GC is equal to the base DF (I. 4), and the triangle GBC to the triangle 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 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, 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 and BC are equal to the two DE and EF, each to each; the angle ABC is equal to the angle DEF; therefore the base AC

and

is equal to the base DF, and the triangle ABC to the triangle DEF, and the third angle BAC to the third angle EDF.

Next, let the sides which are opposite to equal angles in each triangle be given equal to one another;

namely, AB to DE; to prove that likewise in this case, the other sides shall be equal,

DF,

and BC to EF;

AC to and also the third angle BAC to the third EDF.

(Const.) For, if BC be not equal to let BC be the greater of and make BH equal to EF,

EF,

them,

D

AA

B

H C

and join AH;

and because BH is equal to EF, and AB to DE,

(Dem.)

the two sides AB and they

and BH are equal to the two DE and EF, each to each; contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles are equal, each to each, to which the equal sides

C

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 (I. 16); wherefore BC is not unequal to EF; that is, it is equal to it; and AB is equal to DE; therefore the two sides AB and BC are equal to the two DE and EF, each to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the triangle ABC to the triangle DEF, the third angle BAC to the third angle EDF.

and

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.

Given that the straight line EF, which falls upon the two straight lines AB and CD, makes the alternate angles AEF and EFD equal to one another; to prove that AB is parallel

to CD.

(Const.) For, if AB and CD be not parallel,

being pro

duced, they shall meet either towards B, D, or towards A, C; let them be produced, and meet towards B, D, in the point G; (Dem.) therefore GEF is a triangle,

and its exterior angle AEF is greater (I. 16) than the interior and opposite angle EFG; but it was given equal to it, which is impossible;

C

F

E

B

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 A, C; but those straight lines which are in the same plane, and being produced ever so far both ways do not meet, are parallel to one another (Def. 34). Therefore AB is parallel to CD.

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.

Given that the straight line EF, which falls upon the two straight lines AB and CD, makes the exterior angle EGB equal to the interior and opposite angle GHD upon the same side; or makes the interior angles on the same side BGH and GHD

together equal to two right angles; to prove that

parallel to CD.

(Dem.) Because the angle EGB is equal

to the angle GHD,

and the angle EGB

equal to the angle AGH (I. 15),

the

angle AGH is equal to the angle GHD; and they are alternate angles;

therefore AB is parallel to CD

(I. 27). Again, because the angles BGH

right angles (I. 13);

AB is

and GHD are together equal (by Hyp.) to two right angles; and that AGH and BGH are also together equal to two the angles AGH and BGH are equal to the angles BGH and GHD. Take away the common angle BGH, and there remains the angle AGH equal to the angle GHD; and they are alternate angles; therefore (I. 27) AB is parallel to CD.

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.

Given the straight line EF falling upon the parallel straight lines AB, CD; to prove that 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; the two interior angles BGH and GHD, upon the same side, are together equal to two right angles.

and

(Const.) For if AGH be not equal to GHD, let KG be drawn making the angle KGH equal to GHD, and produce KG to

L. (Dem.) Then KL will be parallel to CD (I. 27);
AB is also parallel to CD;

there

Now, the angle EGB is equal to AGH (I. 15);

was proved to be equal to GHD;

but

therefore EGB is likewise equal to GHD; add to each of these the angle BGH; therefore the angles EGB and BGH are equal to the angles BGH and GHD; but EGB and BGH are equal to two right angles (I. 13); therefore also BGH and GHD are equal to two right

angles.