*

angles will become equal. In this position C'P becomes perpendicular to AB 3(26). Again, if the line C'P revolve from the position in which the angles are equal, one angle will increase and the other diminish; hence there is only one position of the line on this side of AB in which the adjacent angles are equal. Therefore there can be one and only one perpendicular erected to AB at P, which shall lie on the same side of AB. Q. E. D.

123. COR. 1.-On the other side of the line a second perpendicular, and only one, can be drawn from the same point in the line.

124. COR. 2.-If one straight line meets another so as to make the angle on one side of it a right angle, the angle on the other side is also a right angle, and the first line is perpendicular to the second.

125. COR. 3.—If two lines intersect so as to make one of the four angles formed a right angle, the other three are right angles, and the lines are perpendicular to each other.

DEM. Thus, if CEB is a right angle, CEA, being equal to it, is also a right angle. Then, as AEC is a right angle, the adjacent angle AED is a right angle, since they are equal. Also, as CEB is a right angle, and BED equal to it, BED is a right angle. Hence CD being perpendicular to AB, AB is perpendicular to CD, as it meets CD so as to make the adjacent angles AEC and AED, or CEB and BED equal to each other (43).

A

ليا

E

B

D
FIG. 103.

PROPOSITION II.

126. Theorem.-When two straight lines intersect at right angles, if the portion of the plane of the lines on one side of either line be conceived as revolved on that line as an axis until it coincides with the portion of the plane on the other A side, the parts of the second line will coincide.

FIG. 104.

B

DEM.-Let the two lines AB and CD intersect at right angles at E; and let the portion of the plane of the lines on the side of CD on which B lies be conceived to revolve on the line CD as an axis,† until it falls in the

• When a preceding principle is referred to, it should be accurately quoted by the pupil.

+ As if the paper, which may represent the plane of the lines, were folded in the line CD. It is important that this process be clearly conceived, as it is to be made the basis of many subsequent demonstrations.

portion of the plane on the other side of CD. Then will EB fall in and coincide with AE.

For, the point E being in CD, does not change position in the revolution; and, as EB remains perpendicular to CD, it must coincide with EA after the revolution, or there would be two perpendiculars to CD on the same side and from the same point, E, which is impossible (122). Hence EB coincides with ÉA. Q. E. D.

PROPOSITION III.

127. Theorem.-From any point without a straight line, one perpendicular can be let fall upon that line, and only one.

A

DEM.-Let AB be any line, and P any point without the line; then one perpendicular, and only one, can be let fall from P upon AB.

DEC B
FIG. 105.

For, conceive any oblique line, as PC, drawn, making the angle PCB> PCA. Now, while the extremity P of this line remains fixed, conceive the line to revolve so as to make the greater angle PCB decrease, and the less angle PCA increase. At some position of the revolving line, as PD, the two angles which it makes with the line AB will adjacent angles are equal, the line, as PD, is perMoreover, there is only one position of the line in which these angles are equal; hence, only one perpendicular can be drawn from a given point to a given line. Q. E. D.

become equal. When these pendicular to AB (26, 43).

FIG. 106.

B

PROPOSITION IV.

128. Theorem.-From a point without a straight line, a perpendicular is the shortest distance to the line.

DEM.-Let AB be any straight line, P any point without it, PD a perpendicular, and PC any oblique line; then is PD<PC.

Produce PD, making DP' = PD, and draw P'C. Then let the portion of the plane of the lines above AB be revolved upon AB as an axis, until it coincides with the portion below AB. Since PP' and AB intersect at right angles, PD will fall in DP' (126); and, since PD = DP', P will fall at P', and PC P'C, since they coincide when applied. Finally, PP' being a straight line, is shorter than

=

PCP', which is a broken line, since a straight line is the shortest distance between two points. Hence PD, the half of PP', is less than PC, the half of the broken line PCP'. Q. E. D.

PROPOSITION V.

129. Theorem.-If a perpendicular be erected at the middle point of a straight line,

1st. Any point in the perpendicular is equally distant from the extremities of the line.

2d. Any point without the perpendicular is nearer the extremity of the line on its own side of the perpendicular.

DEM.-1st. Let PD be a perpendicular to AB at its middle point D. Then, O being any point in the perpendicular, OA = OB.

For, revolve the figure OBD upon OD as an axis until it falls in the plane on the other side of PD. Since ODB and ODA are right angles, DB will fall in DA (126); and, since DB DA, B will fall at A. Hence, OA and OB coincide, and OA = OB.

2d. O' being any point without the perpendicular on the same side as B, O'B <O'A.

P

FIG. 107.

For, drawing O'A and O'B, let O be the point at which O'A cuts the perpendicular. Draw OB. Now O'B<BO+00', since O'B is a straight and O'OB is a broken line. But, as OA=OB, we may substitute it in the inequality, and have O'B<OA + OO', which sum = O'A.

130. COR.-If each of two points in one line is equally distant from the extremities of another line, the former line is perpendicular to the latter at its middle point.

DEM.-Every point equally distant from the extremities of a straight line lies in a perpendicular to that line at its middle point, by the proposition. But, two points determine the position of a straight line. Hence, two points, each equally distant from the extremities of a straight line, determine the position of the perpendicular at the middle point of the line.

EXERCISES.

1. Prob.-To erect a perpendicular to a given line at a given point in the line.

A

FIG. 108.

SOLUTION.-[The process is given in (44), and should be repeated here exactly as given there, with the reasons for the solution, as follows.] A is one point in the line OA, which is equally distant Y from B and C, by construction, and O is another. Hence OA is perpendicular to BC at A, by (130).

2. Prob.-To bisect a given line.

A

Xm

Xn

FIG. 109.

FIG. 110.

The
The

SOLUTION.-[For the process see (39). student should first do it as he did then. reason why this process bisects AB is as follows.] Since m is one point equally distant from the extremities A and B, and n another, there are two points in mn each equally distant from the extremities of AB. Hence mn is perpendicular to AB at its middle point O, by (130). [The reason for the process in Fig. 20 is the same. Let the student give this method, and show how the corollary (130) applies.]

3. Prob.-From a point without a given line, to let fall a perpendicular upon the line.

SOLUTION.-[Repeat the process as in (45), and give the reason for it as follows.] O is one point equally distant from B and C, and D is another. Hence a line drawn from O to D is perpendicular to BC by (130).

4. Wishing to erect a line perpendicular to AB at its centre, I

P'

FIG. 111.

B

take a cord or chain somewhat longer than AB, and, fastening its ends at A and B, take hold of the middle of the cord or chain and carry it as far from AB as I can, first on one side and then on the other, sticking pins at the most remote points, as at P and P'. These points determine the

perpendicular sought. What is the principle involved?

5. Two boys are skating together on the ice, and both start from

the same point at the same time, one skating directly to the shore and the other obliquely. They both reach the shore at the same time. Which skates the faster? What principle is involved?

6. Several persons start at different times from the same point in a straight road that runs along a wood, and each travels directly away from the road. Will they come out at the same, or at different points on the opposite side of the wood? What principle is involved? What is the geometrical language for the colloquial phrase "Directly away from the road"?

7. If I go from A to B, Fig. 111, by first passing over AP, will I gain or lose in distance by going on a little farther in the direction of AP before I turn and go straight to B? What principle is in-volved? Would I gain or lose by stopping short of P on the line AP? Why?

131. Theorem.- When an oblique line meets another straight line forming two adjacent angles, the sum of these angles is two right angles.

DEM.-Let the oblique line CD meet the straight line AB forming the two adjacent angles CDB and CDA; then CDB + CDA equals two right angles.

C'

A

B

FIG. 112.

For suppose CD to revolve toward the position of the perpendicular C'D; the angle CDB will increase at the same rate that CDA diminishes; hence their sum will remain constant (i. e., the same). But, when CD becomes perpendicular, the sum of the adjacent angles formed with AB is two right angles by definitions (26, 43). Therefore CDB + CDA = two right angles. Q. E. D.

132. COR.-The sum of all the consecutive angles formed by any number of lines meeting a given line on the same side and at a given point is two right angles.