DEM. Thus ADC + CDC1 + CDC""' + C'''DC" + C"DC' + C'DC + CDB = ADC' + C'DB, which sum is two right angles by the proposition. Or, in general terms, the angles thus formed can always be united into two groups, constituting respectively the two adjacent angles formed by one line meeting

another.

133. DEF.-Two angles whose sum is two right angles, are called Supplemental Angles. Hence, the Supplement of an angle is what remains after subtracting it from two right angles.

PROPOSITION II.

134. Theorem.-When any two straight lines intersect, the opposite or vertical angles are equal to each other, and the sum of the four angles formed is four right angles.

DEM.-Let AB and CE intersect at D; then CDA = the opposite angle BDE, ADE = the opposite or vertical angle CDB, and ADC + CDB + BDE + EDA = four right angles. For, since CD meets AB, ADC is the supplement of CDB (131, 133). Also, since BD meets CE, BDE is the B supplement of CDB. Hence ADC BDE. In a similar manner ADE can be proved equal to CDB. [The student should give the proof.]

FIG. 114.

E

=

Again, since ADC + CDB = two right angles, and BDE + EDA = two right angles, by adding the corresponding members together, we have ADC + CDB + BDE + EDA = four right angles.

135. COR.-The sum of all the consecutive angles formed by any number of lines meeting at a common point is four right angles.

DEM. The truth of this corollary is rendered apparent by drawing a line through the common vertex, and observing that the sum of all the angles on each side thereof is two right angles; whence the sum of all the angles on both sides, which is the same as the sum of all the consecutive angles formed by the line, is four right angles. [Let the student put letters on the figure, and demonstrate by means of it.]

PROPOSITION III.

136. Theorem.-If two supplemental angles are so situated as to be adjacent to each other, the two sides not common will fall in the same straight line.

DEM.--Let the sum of the two angles BOA and CO'D be two right angles. Prolong CO', forming the angle DO'E. Then is DO'E supplemental to CO'D (131, 133), and hence equal to BOA, which is supplemental to CO'D by hypothesis. Now, if AOB be placed adjacent to CO'D, the vertex O being at O', and the side OA falling in O'D, OB will fall in O'E, since BOA = DO'E. Hence, when the angles are so situated, OB becomes the prolongation of CO'. Q. E. D.

B

O'

FIG. 116.

PROPOSITION IV.

137. Theorem.—If from a point without a straight line a perpendicular be drawn, oblique lines from the same point cutting the line at equal distances from the foot of the perpendicular are equal to each other; the angles which they form with the perpendicular are equal to each other; and the angles which they form with the line are equal to each other.

P

DEM-Let AB be any straight line, P any point without it, PD a perpendicular, and PC and PE oblique lines cutting AB at C and E, so that DC=DE; then PC=PE, angle CPD = angle DPE, and angle PCD = angle PED.

=

Revolve the figure PDE upon PD as an axis, until it falls in the plane on the other side of PD. Since AB is perpendicular to PD, DB will fall in DA; and, since DE will fall at C. Now, as P remains stationary, the triangles PDE and PDC coincide. Hence, PC = PE, angle CPD = angle DPE, and angle PCD = angle PED. Q. E. D.

DC, E

AC

F

FIG. 117.

QUERY.-How does the equality of PE and PC follow from (129) ?

E B

PROPOSITION V.

138. Theorem.-If from a point without a line a perpendicular be drawn to the line, and also from the same point two oblique

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.

A/C

F

P

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

=

Produce PD, making DP' = PD, and draw P'F and P'C, producing the latter until it meets PF in H. Revolve the figure FPD upon AB as an axis, until it falls in the plane on the opposite side of AB. Since PP' is perpendicular to AB, PD will fall in DP'; and, since PD DP', P will fall at P'. Then 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 PCPPF 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 draw PC. Then since PC = PE, if it is found less than PF, as in the demonstration, 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 supple

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 C, two times F, in the last figure, are added in order, how will AM and CN lie with reference to each other? Why?

Ans. They will coincide.

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)?

IF

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?

Rei.

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.

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 G 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 C where FE intersects CD, a perpendicular be drawn to FE, it is par

*Nous regarderons cette proposition comme ÉVIDENTE. P.-F. COMPAGNON. CHAUVENET.

So also.