PROPOSITION 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 in each, or the sides opposite to them; then shall the other sides be equal, each to each, and also the third angle of the one equal 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, each to each, namely, ABC to DEF, and BCA to EFD; also one side equal to one side.

First, let those sides be equal which are adjacent to the angles that are equal in the two triangles, namely, BC to EF.

Then the other sides shall be equal, each to each, namely, 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 greater than the other.
If possible, let AB be greater than DE,
make BG equal to ED, (1. 3.) and join GC.
Then in the two triangles GBC, DEF,

because GB is equal to DE, and BC to EF, (hyp.) 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 the base GC is equal to the base DF, (1. 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 the angle ACB is, by the hypothesis, equal to the angle DFE; wherefore also the angle GCB is equal to the angle ACB; (ax. 1.) the less angle equal to the greater, which is impossible; therefore AB is not unequal to DE,

that is, AB is equal to DE.

Hence, in the triangles ABC, DEF;

because AB is equal to DE, and BC to EF, (hyp.) and the angle ABC is equal to the angle DEF; (hyp.) therefore the base AC is equal to the base DF, (1. 4.) and the third angle BAC to the third angle EDF.

Secondly, let the sides which are opposite to one of the equal angles in each triangle be equal to one another, namely, AB equal to DE.

Then in this case likewise the other sides shall be equal, AC to DF, and BC to EF, and also the third angle BAC to the third angle EDdf.

For if BC be not equal to EF,

one of them must be greater than the other.
If possible, let BC be greater than EF';
make BH equal to EF, (1. 3.) and join AH.
Then in the two triangles ABH, DEF,
because AB is equal to DE, and BH to EF,
and the angle ABH to the angle DEF; (hyp.)
therefore the base AH is equal to the base DF, (1. 4.)
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 the angle BHA is equal to the angle EFD;
but the angle EFD is equal to the angle BCA; (hyp.)
therefore the angle BHA is equal to the angle BCA, (ax. 1.)
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, BC is equal to EF

Hence, in the triangles ABC, DEF;

because AB is equal to DE, and BC to EF, (hyp.) and the included angle ABC is equal to the included angle DEF; (hyp.) therefore the base AC is equal to the base DF, (1. 4.) and the third angle BAC to the third angle EDF. Wherefore, if two triangles, &c. Q. E.D.

PROPOSITION XXVII. THEOREM.

If a straight line falling on two other straight lines, make the alternat angles equal to each other; 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 Then AB shall be parallel to CD.

For, if AB be not parallel to CD,

then AB and CD being produced will meet, either towards A and C, or towards B and D.

Let AB, CD be produced and meet, if possible, towards B and D, in the point G,

then GEF is a triangle.

And because a side GE of the triangle GEF is produced to A, therefore its exterior angle AEF is greater than the interior and opposite angle EFG; (1. 16.)

but the angle AEF is equal to the angle EFG; (hyp.)

therefore the angle AEF is greater than, and equal to, the angle EFG; which is impossible.

Therefore AB, CD being produced, do not meet towards B; D. In like manner, it may be demonstrated, that they do not meet when produced towards A, C.

But those straight lines in the same plane, which meet neither way, though produced ever so far, are parallel to one another; (def. 35.) therefore AB is parallel to CD.

Wherefore, if a straight line, &c. Q. E.D.

PROPOSITION 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 of the line EF; or make the two interior angles BGH, GHD on the same side together equal to two right angles.

Then AB shall be parallel to CD.

Because the angle EGB is equal to the angle GIID, (hyp.) and the angle EGB is equal to the angle AGH, (1. 15.) therefore the angle AGH is equal to the angle GHD; (ax. 1.) and they are alternate angles,

therefore AB is parallel to CD. (1. 27.)

Again, because the angles BGH, GHD are together equal to two right angles, (hyp.)

and that the angles AGH, BGH are also together equal to two right angles; (I. 13.)

therefore the angles AGH, BGH are equal to the angles BGH, GHD; (ax. 1.)

take away from these equals, the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; (ax. 3.)

and they are alternate angles;

therefore AB is parallel to CD. (1.27.)

Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION XXIX. THEOREM.

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.

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Let the straight line EF fall upon the parallel straight lines AB, CD. Then the alternate angles AGH, GHD shall be equal to one another the exterior angle EGB shall be equal to the interior and opposite angle GHD upon the same side of the line EF; and the two interior angles BGH, GHD upon the same side of EF shall be together equal to two right angles.

First. For, if the angle AGH be not equal to the alternate angle GHD, one of them must be greater than the other;

if possible, let AGH be greater than GHD,

then because the angle AGH is greater than the angle GHD, add to each of these unequals the angle BGH;

therefore the angles AGH, BGH are greater than the angles BGH, GHD; (ax. 4.)

but the angles AGH, BGH are equal to two right angles; (1.13.) therefore the angles BGH, GHD are less than two right angles;

but those straight lines, which with another straight line falling upon them, make the two interior angles on the same side less than two right angles, will meet together if continually produced; (ax. 12.) therefore the straight lines AB, CD, if produced far enough, will meet towards B, D;

but they never meet, since they are parallel by the hypothesis; therefore the angle AGH is not unequal to the angle GHD, that is, the angle AGH is equal to the alternate angle GHD. Secondly. Because the angle AGH is equal to the angle EGB, (1. 15.) and the angle AGH is equal to the angie GHD,

therefore the exterior angle EGB is equal to the interior and opposite angle GHD, on the same side of the line.

Thirdly. Because the angle EGB is equal to the angle GHD, add to each of them the angle BGH;

therefore the angles EGB, BGH are equal to the angles BGH, GHD; (ax. 2.)

but EGB, BGH are equal to two right angles; (1. 13.) therefore also the two interior angles BGH, GHD on the same side of the line are equal to two right angles. (ax. 1.)

Wherefore, if a straight line, &c. Q. E. D.

Straight lines which are parallel to the same straight line are parallel to each other.

Let the straight lines AB, CD, be each of them parallel to EF. Then shall AB be also parallel to CD.

Let the straight line GHK cut AB, EF, CD.

Then because GHK cuts the parallel straight lines AB, EF, in G, H:

therefore the angle AGH is equal to the alternate angle GHF. (1. 29.) Again, because GHK cuts the parallel straight lines EF, CD, in H, K;

therefore the exterior angle GHF is equal to the interior angle HKD; and it was shewn that the angle AGH is equal to the angle GHF; therefore the angle AGH is equal to the angle GKD; and these are alternate angles;

therefore AB is parallel to CD. (1. 27.) Wherefore, straight lines which are parallel, &c. Q.E.D.

To draw a straight line through a given point parallel to a given straight line.

Let A be the given point, and BC the given straight line.

It is required to draw, through the point A, a straight line parallel to the straight line BC.

In the line BC take any point D, and join AD;

at the point A in the straight line AD,

make the angle DAE equal to the angle ADC, (1. 23.) on the opposite side of AD;

and produce the straight line EA to F.

Then EF shall be parallel to BC.

Because the straight line AD meets the two straight lines EF, BC, and makes the alternate angles EAD, ADC, equal to one another, therefore EF is parallel to BC. (1. 27.)

Wherefore, through the given point A, has been drawn a straight line EAF parallel to the given straight line BC. Q.E.F.