PROPOSITION XXVII. THEOREM.

If a straight line intersect two other straight lines which are in the same plane, and make the alternate angles equal to each other, those two lines are parallel. See Appendix to Book I.

Let the straight line EF, which intersects the two straight lines AB, CD, make the alternate angles AEF, EFD equal to each other; AB is parallel to CD.

For if AB and CD be not parallel, they are inclined to each other, and, if produced, will meet either toward B, D, or toward A, C. Let them be produced and meet toward B, D, in the point G; then GEF is a triangle, and its exterior angle AEF is greater than the interior and opposite angle EFG (Prop. 16.). But the angle AEF is also equal to EFG, which is impossible; therefore A AB and CD being produced do not meet toward B, D. In like manner it may be demonstrated that they do C not meet toward A, C. Therefore AB, CD are

F

F

B

D

not inclined to each other. Therefore AB is parallel to CD (14 Def.). Wherefore, if a straight line &c. Q. E. D.*

PROPOSITION XXVIII. THEOREM.

If a straight line intersect two other straight lines which are in the same plane, and make the exterior angle equal to the interior and opposite angle on the same side of the line, or make the two interior angles on the same side together equal to two right angles; the two straight lines are parallel.

Let the straight line EF, which intersects the two straight lines AB, CD, make the exterior angle EGB equal to the inte

* By reason of the false supposition, and construction of the figure, this demonstration is not intelligible to learners.

Because the angle EGB is equal to GHD, and EGB is also equal to the angle AGH (Prop. 15.), the angle AGH is equal to GHD; and they are alternate angles; therefore AB is parallel to CD (Prop. 27.).

Again, the angles BGH, GHD are together equal to two right angles (by Hyp.), and AGH, BGH, are together equal to two right angles (Prop. 13.); therefore the angles AGH, BGH, are together equal to the angles BGH, GHD. Take away the common angle BGH, then the remaining angles AGH, GHD are equal; and they are alternate angles; therefore AB is paral lel to CD. Wherefore, if a straight line &c. Q. E. D.

Cor. Straight lines which are perpendicular to the same straight line, and in the same plane with it, are parallel to one another.

If AB and CD be perp. to EF, the angles at E and F will be right angles (10 Def.), and therefore equal (11 Def. Cor.); wherefore AB is parallel to CD.

A

E

ED.

B

PROPOSITION XXIX. THEOREM.

C

F

If a straight line intersect two parallel straight lines it makes the alternate angles equal to each other; and the exterior angle equal to the interior and opposite angle on the same side of it; and likewise the two interior angles on the same side together equal to two right angles.

AXIOM 10. Two straight lines which intersect each other cannot be both parallel to the same straight line.*

Let the straight line EF intersect the parallel straight lines AB, CD; the alternate angles AGH, GHD are equal to each

*This axiom was omitted in its proper place by mistake.

other; and the exterior angle EGB is equal to the interior and opposite angle GHD) on the same side of EF; and the two interior angles BGH, GHD, on the same side, are together equal to two right angles.

E

If AGH be not equal to GHD, one of them must be greater than the other. Let AGH be the greater, and at the point G, in the straight line EF, make the angle KGH equal to GHD (Prop. 23.), and produce KG to L; KL will be parallel to CD (Prop. 27). But AB is also parallel to CD; therefore two straight lines AB, KL, which intersect each other, are parallel to the same line KCD, which is impossi- C. ble (10 Ax.), Therefore the angles AGH, GHD are not unequal.

Again, the angle EGB is equal to AGH (Prop. 15.); and AGH is proved to be equal to GHD;

G

-B

-D

H

therefore the angle EGB is likewise equal to GHD. Add to each of these the angle BGH, then the angles EGB, BGH are together equal to the angles BGH, GHD. But EGB, BGH are together equal to two right angles (Prop. 13); there fore also BGH, GHD are together equal to two right angles. Wherefore, if a straight line &c. Q. E. D.

Cor. If a straight line EF be perpendicular to one CD of two parallel straight lines AB, CD, it is also perpendicular to

the other AB.

For the angle AEF is equal to EFD, and the angle BEF is equal to A EFC. But the angles EFD and EFC are right angles, because EF is perp. to CD. Therefore AEF and BEF C are right angles, therefore EF is perpendicular to AB.

ED.

PROPOSITION XXX. THEOREM.

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

Ꮐ

Let the straight lines AB, CD be parallel to EF; AB is also parallel to CD.

G

B

Let the line GHK cut the straight lines AB, EF, CD. Because GHK cuts the parallel lines AB, EF, the angle AGH is equal to GHF (Prop. 29); A and because GK cuts the parallel lines EF, CD, the angle GHF is equal to GKD; there- E fore also the angle AGK is equal to GKD; and they are alternate angles; therefore AB is parallel to CD (Prop. 27). Wherefore straight lines &c. Q. E. D.

H

F

K

C

D

PROPOSITION XXXI. PROBLEM.

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 a straight

line through A parallel to BC. E

In BC take any point D, and

join AD at the point A, in the line AD, make the angle DAE

B

D

A

F

C

equal to ADC (Prop. 23); and produce the line EA to F. Then EF is parallel to BC.

Because AD meets the two straight lines BC, EF, and makes the alternate angles EAD, ADC equal to each other, EF is parallel to BC (Prop. 27). Therefore the line EAF is drawn. through the given point A parallel to the given line BC. Which was to be done.

PROPOSITION XXXII. THEOREM.*

1. If any side of a rectilineal triangle be produced, the exterior angle is equal to the two interior and opposite angles taken together.

Let ABC be a triangle, and let a side BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC together.

*Euclid's enunciation contains two distinct propositions, which are enunciated and demonstrated separately, as follows.

Through the point C draw CE parallel to the side AB (Prop. 31.). Because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (Prop. 29.); and because AB is parallel to

CE, and BD meets them, the exterior angle ECD is equal to the interior and opposite angle B. Hence the whole exterior angle ACD is equal to the two interior and opposite_angles A and B together. Therefore, if any side &c. Q. E. D.

Cor. The difference between the exterior angle and either of the interior and opposite angles is equal to the other interior and opposite angle.

2. The three angles of every rectilineal triangle are together equal to two right angles.

Let ABC be a triangle; the three angles A, B, C, are toge ther equal to two right angles.

Produce any side BC to D, then the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, together. To these equals add the angle ACB, then the two angles ACD, ACB are together equal to the three angles CBA, BAC, ACB. But the angles ACD, ACB are together equal to two right angles (Prop. 13.); therefore the three angles CBA, BAC, ACB are together equal to two right angles. Wherefore the three angles &c. Q. E.'D.

Cor. 1. If one angle in one triangle be equal to one angle in another, the other two angles of the former triangle are toge ther equal to the other two angles of the latter.

ED.

Cor. 2. If two angles of one triangle be together equal to two angles of another, the third angle of the former triangle is equal to the third angle of the latter (3 Ax.).

ED.

Cor. 3. If one angle in any triangle be either right or obtuse, then each of the other two angles is acute.

ED.