lines making equal angles with the perpendicular, the oblique lines are equal to each other, cut the line at equal distances from the foot of the perpendicular, and make equal angles with it.*

DEM.-PD being a perpendicular to AB, and angle CPD equal to angle DPE, PC equals PE, CD equals DE, and PCD equals PED.

P

A/C

FIG. 118.

E B

Revolve the figure PDE upon PD as an axis, till it falls in the plane of PDC. Since angle EPD angle CPD, PE will take the direction PC, and E will fall somewhere in the indefinite line PF. But, since PDE and PDC are right angles, DE will fall in DA (126) and E will fall somewhere in the indefinite line DA. Now, as E falls at the same time in PF and DA, it must fall at their intersection

C. Hence, PE coincides with PC, and DE with DC. Therefore PE = = DC, and angle PED PCD. Q. E. D.

=

PC, DE

PROPOSITION VI.

139. Theorem.—If from a point without a line a perpendicular

DE

FIG. 119.

be let fall on the line, and two oblique lines be drawn, the oblique line which cuts off the greater distance from the foot of the perpendicular is the greater.

DEM.-Let AB be any straight line, P any point without it, and PC and PF two oblique lines of which PF cuts off the greater distance from the foot of the perpendicular; that is DF > DC. Then is PF > PC.

=

=

Revolve the figure FPD upon AB as an axis, until it falls in the plane on the opposite side of AB. Let P' be the point at which P falls; and revolve the figure FPD back to its original position. Draw P'D, P'F, and P'C producing the latter till it meets PF in H. Then P'D PD, P'C PC, and P'F PF. Now the broken line PCP'< than the broken line PHP', since the straight line PC the broken line PHC. For a like reason the broken line PHP' < PFP', since HP < HFP'. Hence PCP' < PFP', and PC the half of PCP'< PF the half of PFP'. Q E. D.

SCH.-If the two oblique lines to be compared lie on different sides of the perpendicular, as PF and PE, DF being greater than DE, lay off DC DE, and PE, if it is found less than PF, as in the demon

draw PC. Then since PC

stration, PE is less than PF.

=

* This proposition is the converse of the last. The significance of this statement will be more fully developed farther on (154).

140. Cor. 1.—From a given point without a line, there can not be two equal oblique lines drawn to the line on the same side of a perpendicular from the point to the line.

141. COR. 2.-Two equal oblique lines drawn from the same point in a perpendicular to a given line, cut off equal distances on that line from the foot of the perpendicular.

DEM.—For, if the distances cut off were unequal, the lines would be unequal.

EXERCISES.

1. Having an angle given, how can you construct its supplement? Draw any angle on the blackboard, and then construct its supplement.

CDE

FIG. 120.

2. The several angles in the figure are such parts of a rignt angle as are indicated by the fractions placed in them. If these angles are added together by bringing the vertices together and causing the adjacent sides of the angles to coincide, how will MA and CN lie? Construct seven consecutive angles of these several magnitudes. How do the two sides not common lie? Why?

3. If two times A, B, two times D, three times E, three times C, three times G, two times F, in the last figure, are added in order, how will AM and GN lie with reference to each other? Why?

Ans. They will coincide.

1 4. If you place the vertices of any two equal angles together so that two of the sides shall extend in opposite directions and form one and the same straight line, the other two sides lying on opposite sides thereof, how will the latter sides lie? By what principle?

5. Upon what principle in this section may the common method of erecting a perpendicular at the middle of a straight line (39, 44) be explained? Upon what the method of letting fall a perpendicular upon a straight line from a point without (45)?

6. A and B start at the same time, from the same point in a certain road; A travels directly to a point in another road at right angles to the first, and at ten miles from their intersection, and B travels directly toward a second point in the second road, which point is seven miles from the intersection. Both reach their destination at the same time. Which travels the faster? What principle is involved?

SECTION III.

OF PARALLELS.

PROPOSITION I.

142. Theorem.-Two straight lines lying in the same plane and perpendicular to a third line are parallel to each other.

A

H

'E

B

DEM.-Let AB and CD be two straight lines lying in the same plane and each perpendicular to FE; then are they parallel.

For if AB and CD are not parallel, they will meet at some point if sufficiently produced (66). But, if they could meet, we should have two straight lines from one point (their point of meeting), perpendicular to the same straight line, which is impossible (127). Therefore, as the lines lie in the same plane and cannot meet how far soever they be produced, they are parallel. Q. E. D.

FIG. 121.

143. COR. 1.-Through the same point one parallel can always be drawn to a given line, and only one.

DEM.-Let AB be the given line, and G the given point, there can be one and only one perpendicular through G to AB (127.) Let this be FE. Now through one and only one perpendicular can be drawn to FE. Let this be CD. Then is CD parallel to AB by the proposition. That there is only one such parallel, we shall assume as axiomatic.*

144. COR. 2.-If a straight line is perpendicular to one of two parallels, it is perpendicular to the other also.

DEM.-If FE is perpendicular to AB it is perpendicular to CD. For, if through G where FE intersects CD, a perpendicular be drawn to FE, it is par

Nous regarderons cette proposition comme ÉVIDENTE. P.-F. COMPAGNON. So ale

CHAUVENET.

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 HI 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.

IH

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.

12

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.

A

E

K

HG

P

F

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 GKB and CCK are both supplements of DGK, the former by hypothesis, and the latter by (133), CKB = 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 and I will fall in PH, or PH produced; and, since PGH, KI will take the direction CH, and I will fall somewhere in CC. 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 CHP is a right angle. Therefore, AB and CD are both perpendic ular to HI, and consequently parallel by (142). Q. E. D.

FIG. 124.

PK falls in PC.* Since PK

Pl will take the direction PH,
PKI:

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 must = KGD, since FGD + KGD = two right angles. Also, if FGD + EKB = two right angles, FGD must GKB, since GKB + EKB = two right angles. 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 either 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 FGD = GKB, or CGF = AKG, the two lines are parallel. [Let the student show the truth in each case.]

Fig. 125

The accompanying figures will aid the student in getting this conception. 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.