allel to AB by the proposition. But, by Cor. 1, there can be but one line through G parallel to AB. Hence the perpendicular to FE at G coincides with, or is, the parallel CD.

PROPOSITION II.

145. Theorem.-Two straight lines which are parallel to a third, are parallel to each other.

DEM.-Let AB and CD be each parallel to EF; then are they parallel to each other.

For draw H❘ perpendicular to EF; then will it be perpendicular to CD because CD is parallel to EF. For a like reason HI is perpendicular to AB. Hence CD and AB are both perpendicular to HI, and consequently parallel. Q. E. D.

Α

H

K

L

M

B

FIG. 122.

146. DEFINITIONS.-When two lines are cut by a third line the angles formed are named as follows:

Exterior Angles are those without the two lines, as 1, 2, 7, and 8.

Interior Angles are those within the two lines, as 3, 4, 5, and 6.

Alternate Exterior Angles are those without the two lines and on different sides of the secant line, but not adjacent, as 2 and 7, 1 and 8. Alternate Interior Angles are those within the two lines and on different sides of the secant line but not adjacent, as 3 and 6, 4 and 5.

34

56

78

FIG. 123.

Corresponding Angles are one without and one within the two lines, and on the same side of the secant line but not adjacent, as 2 and 6, 4 and 8, 1 and 5, 3 and 7.

PROPOSITION III.

147. Theorem.-If two lines are cut by a third line, making the sum of the interior angles on the same side of the secant line equal to two right angles, the two lines are parallel.

DEM.-Let AB and CD be met by the line EF, making EGD + FKB = two right angles; then are AB and CD parallel.

72

H G

P

ELEMENTARY PLANE GEOMETRY.

A

K

E

FIG. 124.

PK falls in PG.* Since PK

B

For, through P, the middle of GK, draw Hi perpendicular to AB. Since HPG and KPI are vertical angles, they are equal by (134). Also, since CKB and CGK are both supplements of DGK, the former by hypothesis, and the latter by (133), GKB = CGK. Now, conceive the portion of the figure below P, while remaining in the same plane (the plane of the paper), to revolve upon P (as a pivot) from right to left till PG, K will fall at G. Again, since KPI = GPH PI will take the direction PH, and I will fall in PH, or PH produced; and, since PKI - PGH, KI will take the direction CH, and I will fall somewhere in GC. Hence, as I falls in both PH and GC, it must fall at their intersection H; and KIP coincides with, and is equal to PHG. But KIP is a right angle by construction; hence GHP is a right angle. Therefore, AB and CD are both perpendicular to HI, and consequently parallel by (142). Q. E. D.

=

148. COR. 1.-If two lines are cut by a third, making the sum of the two exterior angles on the same side of the secant line equal to two right angles, the two lines are parallel.

DEM.-For, if FGD + EKB = two right angles, EKB is the supplement of FGD; so also is KGD (131, 133). Hence EKB = KGD. Again, for like reasons, FGD and GKB are both supplements of EKB and therefore equal. Hence, when FGD + EKB: two right angles, GKB + KGD = two right angles, and the lines are parallel by the proposition. The same is true for FGC and AKE. [Let the student prove it.]

149. COR. 2.-If two lines are cut by a third, making either two alternate interior, or either two alternate exterior, or any two corresponding angles, equal to each other, the lines are parallel.

DEM.-If CGK = GKB, KGD + GKB = two right angles, since CGK + KGD two right angles. Hence the lines are parallel by the proposition. So also if KGD AKG, or FGD = AKE, or CGF = EKB, or FCD GKB, or CGF = AKC, the two lines are parallel. [Let the student show the truth in each case.]

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The accompanying figures will aid the student in getting this conception. Fig. 125 represents the position of the lines after the revolution has gone about half a right angle, and Fig. 126 when the revolution is almost completed.

OF PARALLEL LINES.

73

PROPOSITION IV.

150. Theorem.-If two parallel lines are cut by a third line, the sum of the interior angles on the same side of the secant line is equal to two right angles.

DEM.-Let the parallels AB and CD be cut by EF, then is DGK + GKB = two right angles.

For, if DGK is not the supplement of GKB, let LM be drawn through G so as to make MCK that supplement. Then, by the preceding proposition, LM is parallel to AB; and we have two parallels to AB through the point G, which is impossible (143). Hence, as no line but a parallel can make this interior angle the supplement of the other, the parallel makes it SO. Q. E. D.

L

C

A

FIG. 127.

F

D

M

B

[Let the student demonstrate this proposition as the preceding was demonstrated. In this case CD and AB are parallel by hypothesis, and HI being drawn perpendicular to one is perpendicular to the other also. When K falls at G, KI falls on CG, since from a point without a line only one perpendicular can be drawn to that line.]

151. Cor. 1.—If two parallel lines are cut by a third line, the sum of either two exterior angles on the same side of the secant line is equal to two right angles.

DEM.-FGD + EKB = two right angles. For FGD and GKB are both supplements of DGK (133, 150), and therefore equal to each other. For like reasons, EKB KGD. Therefore, FGD + EKB = GKB + DGK = two right angles, by the proposition.

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152. COR. 2.-If two parallel lines are cut by a third line, either two alternate interior, or either two alternate exterior, or either two corresponding angles, are equal to each other.

For each is the supplement

DEM.-If CD and AB are parallel, CGK GKB. of KGD, the former by (131), the latter by (150). [Let the student show in like manner that AKG KGD, FGD = AKE, CGF = EKB, FGD GKB, and CGF= AKG.]

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153. COR. 3.-Of the eight angles formed when one line cuts two parallels, the four acute angles are equal each to each, and the four obtuse angles; or, in case any one angle is a right angle, all the others are right angles.

154. SCH.-The last two propositions and their corollaries are the converse of each other; i. e., the hypotheses or data and the conclusions or things proved are exchanged. Thus, in PROP. III., the hypothesis is, that The sum of the two interior angles on the same side of the secant line is equal to two right angles; and the conclusion is, that The two lines are parallel. Now, in PROP. IV., the hypothesis is, that The two lines are parallel; and the conclusion is, that The sum of the two interior angles on the same side of the secant line is two right angles.* [A clear conception of this scholium will save the student from confounding these propositions.]

+

PROPOSITION V.

155. Theorem.-If two straight lines are cut by a third line making the sum of the interior angles on one side of the secant line less than two right angles, the two lines will meet on this side of the secant line, if sufficiently produced.

156. Theorem.-Two parallels are everywhere equally distant from each other.

DEM.-Let E and F be any two points in the line CD, and EG and FH perpendiculars measuring the distances between the parallels CD and AB at these points; then is EG = FH.

For, let P be the middle point between E and F, and PO a perpendicular at

The learner may think that, if a proposition is true, its converse is necessarily true; and hence, that when a proposition has been proved, its converse may be assumed as also proved Now this is by no means always the case. Although in a great variety of mathematical prop! ositions, it happens that the proposition and its converse are both true, we never assume one from having proved the other; and we shall occasionally find a proposition whose converse is not true.

C

E

P

F

D

this point. Revolve the portion of the figure on the right of PO, upon PO as an axis, until it falls upon the plane of the paper at the left. Then, since FPO and EPO are right angles, PD will fall in PC; and, as PF= PE, F will fall on E. As F and E are right angles, FH will take the direction EG, and H will lie in EG or EG produced. Also, as POH and POG are right angles, OB will fall in OA, and H falling at the same time in EC and OA is at their intersection G. Hence FH coincides with and is equal to EG. Q. E. D.

A

B

H

FIG. 129.

EXERCISES.

1. Prob.-Through a given point to draw a line parallel to a given line, by the principle contained in PROP. I. of this section.

SUG's.-Draw a straight line on the blackboard. Designate with a dot some point without the line. To draw a line through the designated point and parallel to the given line, is the problem. Let fall a perpendicular upon the line from the point. Then through the given point draw a line perpendicular to this perpendicular. The latter line will be parallel to the given line. (By what proposition?)

2. Prob.-Through a given point to draw a parallel to a given line by PROP. III.

SUG's.-Through the given point draw an oblique line cutting the given line. Then draw a line through the given point making an angle with the oblique line equal to the supplement of the angle which is included between the oblique line and the given line, and on the same side of the former. [Of course the student will be required to do the work on the blackboard, guessing at nothing.]

3. Prob.-Through a given

point to draw a line parallel to a given line, upon the principle that the alternate angles made by a secant line are equal (152).

4. A bevel is an instrument much used by carpenters, and consists of a main limb AB, in which a tongue CD is placed, so as to open and shut like the blade of a knife. This tongue turns on the pivot o, which is a screw, and can be tightened so as to hold the tongue firmly at

m

B

A

FIG. 130.

any angle with the limb. The tongue can also be adjusted so as