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, 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 (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, BC, are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF; the base therefore 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, namely, 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.

HC E

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 the base DF, and the triangle ABH to the triangle 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 (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 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.

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.

F

E

B

D

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 A the point G; therefore 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 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 meet neither way, though produced ever so far, are parallel to one another (Def. 38). AB therefore is parallel to CD.

PROPOSITION XXVIII. THEOREM.

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.

E

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.

G

B

Η

D

Because the angle EGB is equal to the angle GHD, and the angle EGB equal to the angle AGH (1. 15), the angle AGH is equal to the angle GHD; and

they are the alternate angles; therefore AB is parallel to CD (I. 27). Again, because the angles BGH, GHD, are equal (by Hyp.) to two right angles; and that AGH, BGH, are also equal to two right angles (I. 13); the angles AGH, BGH, are equal to the angles 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.

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.

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 together equal to two right angles.

K

C

E

H

For if AGH be not equal to GHD, let KG be drawn making the angle KGH equal to GHD, and produce KG to L; then KL will be parallel to CD (I. 27); but AB is also parallel to CD; therefore two straight lines Aare drawn through the same point G, parallel to CD, and yet not coinciding with one another, which is impossible (Ax. 11). The angles AGH, GHD, therefore are not unequal; that is, they are equal to one another. Now, the angle EGB is equal to AGH (I. 15); and AGH is proved to be equal to GHD; therefore EGB is likewise equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH, are equal to the angles BGH, GHD; but EGB, BGH, are equal to two right angles (I. 13); therefore also BGH, GHD, are equal to two right angles.

COR.-If two lines KL and CD make, with EF, the two angles KGH, GHC, together less than two right

angles, KG and CH will meet on the side of EF on which the two angles are that are less than two right angles.

For, if not, KL and CD are either parallel, or they meet on the other side of EF; but they are not parallel; for the angles KGH, GHC, would then be equal to two right angles. Neither do they meet toward the points L and D ; for the angles LGH, GHD, would then be two angles of a triangle, and less than two right angles; but this is impossible; for the four angles KGH, HGL, CHG, GHD, are together equal to four right angles, of which the two KGH, CHG, are by supposition less than two right angles; therefore the other two, HGL, GHD, are greater than two right angles. Therefore, since KL and CD are not parallel, and do not meet towards L and D, they will meet if produced towards K and C.

PROPOSITION XXX. THEOREM.

Straight lines which are parallel to the same straight line are parallel to one another.

Let AB, CD, be each of them parallel to EF, AB is also parallel to CD.

Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal to the angle GHF (I. 29).

Again, because the straight line GK cuts the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD; and it was shown that the angle AGK is equal to the angle GHF; therefore

also AGK is equal to GKD, and they are alternate angles; therefore AB is parallel to CD (I. 27).

PROPOSITION XXXI. PROBLEM.

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

E

Let A be the given point, and BC the given straight line; it is required to draw a straight line throught the point A, parallel to the straight line BC.

B

In BC take any point D, and join AD; and at the point A, in the straight line AD make (I. 23) the angle DAE equal to the angle ADC; and produce the straight line EA to F. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADČ, equal to one another, EF is parallel to BC (I. 27). Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC.

PROPOSITION XXXII. THEOREM.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.

Let ABC be a triangle, and let one of its sides BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle; namely,

ABC, BCA, CAB, are together equal to two right angles.

Through the point C draw CE pa

rallel to the straight line AB (1.31); and because AB is parallel to CE and AC meets them, the alternate angles BAC, ACE, are equal (I. 29). Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shown to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these angles add the angle ACB, and the angles ACD, ACB, are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB, are equal to two right angles (I. 13); therefore also the angles CBA, BAC, ACB, are equal to two right angles.

COR. I.-All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

For any rectilineal figure ABCDE can

E

C