14. Same as above when the sides are 10, 4, and 7, and the perpendicular is let fall from the angle included by the sides 10 and 4. Draw the figure. Why is one of the segments nagative?

15. What is the area of a regular octagon inscribed in a circle whose radius is 1? What is its perimeter? What if the radius is 10? 2.612 6.12

2,61200-75544

OF THE

PROP. I. Of chords.

PROP. II. Of secants.

PROP. III. Of secants and tangents.

PROP. IV. How divide sides.

PROP. V. Of exterior angles.

PROP. VI. Length of in relation to other parts.
PROP. VII. All intersect at one point.

PROP. VIII. Of triangles.

PROP. XI. Rectification of circum- Sch. Signification and

ference whose radius

is 1.

importance of x.

PROP. XII. Circumference of any circleCor. Also π D.

= 2′′r.

PROP. XIII. Whose radius is 1.

Cor. Of sector.

PROP. XIV. Of any circle. Sch. Squaring the circle.

EXERCISES.

{

Prob. To divide a line in extreme and mean ratio.
Prob. To inscribe a regular decagon, etc.

CHAPTER II.

SOLID GEOMETRY.*

SECTION I

OF STRAIGHT LINES AND PLANES.

PERPENDICULAR AND OBLIQUE LINES.

380. Solid Geometry is that department of geometry in which the forms (or figures) treated are not limited to a single plane.

381. A Plane (or a Plane Surface) is a surface such that a straight line joining two points in it lies wholly in the surface.

ILL.-The surface of the blackboard is designed to be a plane. To ascertain whether it is truly so, take a ruler with a straight edge, and apply this edge in all directions upon it. If it always coincides, i. e., touches throughout its whole length, the surface is a plane. Is the surface of the stove-pipe a plane? Will a straight line coincide with it in any direction? Will it in every direction?

PROPOSITION I.

382. Theorem.-Three points not in the same straight line determine the position of a plane.

DEM.-Let A, B, and C be three points not in the same straight line; then one plane can be passed through them, and only one; i. e., they determine the position of a plane.

* In some respects, perhaps, "Geometry of Space" is preferable to this term; but, as neither is free from objections, and as this has the advantage of simplicity and long use, the author prefers to retain it.

A

For, pass a straight line through any two of these points, as A and B. Now, conceive any plane containing these two points; then will the line passing through them lie wholly in the plane (381). Conceive this plane to revolve on the line as an axis until the point C falls in the plane. Thus we have one plane passed through the three points. That there can be only one is evident, since when C falls in the plane, if it be revolved either way, C will not be in it. The same may be shown by first passing a plane through B and C, or A and C. There is, therefore, only one position of the plane in which it will contain the third point. Q. E. D.

FIG 255.

383. COR. 1.-Through one line, or two points, an infinite number of planes can be passed.

384. COR. 2.-Two intersecting lines determine the position of a plane.

DEM.-For, the point of intersection may be taken as one of the three points requisite to determine the position of a plane, and any other two points, one in each of the lines, as the other two requisite points. Now, the plane passing through these points contains the lines, for it contains two points in each line.

385. COR. 3.- Two parallel lines determine the position of a plane.

DEM.-For, pass a plane through one of the parallels, and conceive it revolved until it contains some point of the second parallel; then as the plane cannot be revolved either way from this position without leaving this point without it, it is the only plane containing the first parallel and this point in the second. But parallels lie in the same plane (66), whence the plane of the parallels must contain the first line, and the specified point in the second. Therefore, the plane containing the first line and a point in the second is the plane of the parallels, and is fixed in position.

386. Cor. 4.—The intersection of two planes is a straight line.

For two planes cannot have even three points, not in the same straight line, common, much less an indefinite number, which would be required if we conceived the intersection (that is, the common points) to be any other than a straight line.

387. A Perpendicular to a Plane is a line which is perpendicular to all lines of the plane passing through its foot. Conversely, the plane is perpendicular to the line.

PROPOSITION IL

388. Theorem.-A line which is perpendicular to two lines of a plane, at their intersection, is perpendicular to the plane.

DEM.-Let PD be perpendicular to AB and CF at D; then is it perpendicular to MN, the plane of the lines AB and CF.

M

C

P

FIG. 226.

Let OQ be any other line of the plane MN, passing through D. Draw FB intersecting the three lines AB, CF, and OQ. Produce PD to P', making P'D PD, and draw PF, PE, PB, P'F, P'E, and P'B. Then is PF P'F, and PB PB, since FD and BD are perpendicular to PP', and PD = P'D (284). Hence, the triangles PFB and P'FB are equal (292); and, if PFB N be revolved upon FB till P falls at P', PE will fall in P'E. Therefore OQ has E equally distant from P and P', and as D is also equidistant from the same points, OQ is perpendicular to PD at D (130). Now, as OQ is any line, PD is perpendicular to any line of the plane passing through its foot, and consequently per

pendicular to the plane (387). Q. E. D.

389. COR.-If one of two perpendiculars revolves about the other as an axis, its path is a plane perpendicular to the axis. Thus, if AB revolves about PP'as an axis, it describes the plane MN.

PROPOSITION III.

390. Theorem.-At any point in a plane one perpendicular can be erected to the plane, and only one.

M

Α

P

FIG. 227.

E

N

DEM.-Let it be required to show that one perpendicular, and only one, can be erected to the plane MN at D. Through D draw two lines of the plane, as AB and CE, at right angles to each other. CE being perpendicular to AB, let a line be conceived as starting from the position ED to revolve about AB as an axis. It will remain perpendicular to AB (389). Conceive it to have passed to P'D. Now, as it continues to revolve, P'DC diminishes continuously, and at the same rate as P'DE grows greater; hence, in only one, as PD, PDE will equal Therefore, PD is perpendicular to and is the only line that can be

one position of the revolving line, and in PDC, and PD will be perpendicular to CE. two lines of the plane, at their intersection, thus perpendicular, whence it is perpendicular to the plane (388), and is the only perpendicular. Q. E. D.

PROPOSITION IV.

391. Theorem.-If from any point in a perpendicular to a plane, oblique lines be drawn to the plane, those which pierce the plane at equal distances from the foot of the perpendicular are equal; and of those which pierce the plane at unequal distances from the foot of the perpendicular, those which pierce at the greater distances are the greater.

DEM.-Let PD be a perpendicular to the plane MN, and PE, PE', PE", and PE', be oblique lines piercing the plane at equil distances ED, E'D, E'D, and E'''D, from the foot of the perpendicular; then PE = PE' = PE" PE"". For each of the triangles PDE, PDE', etc., has two sides and the included angle equal to the corresponding parts in the other.

=

Again, let FD be longer than E'D. Then is PF> PE'. For, take ED E'D; then PE = PE', by the preceding part of the demonstration. But PF > PE by (139). Hence, PF > PE'. Q. E. D.

M

E"

E'

FIG. 258.

E

392. The Inclination of a line to a plane is measured by the angle which the line makes with a line of the plane passing through the point in which the line pierces the plane and the foot of a perpendicular to the plane from any point in the line.

Thus PFD is the inclination of PF to the plane MN.

393. COR. 1.-The angles which oblique lines drawn from a com-· mon point in a perpendicular to a plane, and piercing the plane at equal distances from the foot of the perpendicular, make with the perpendicular, are equal; and the inclinations of such lines to the plane are equal.

Thus the equality of the triangles, as shown in the demonstration, shows that EPD = E'PD = E’PD = E'''PD, and PED = PE'D = PE'D = PE'''D.

394. COR. 2.-Conversely, If the angles which oblique lines drawn from a point in a perpendicular to a plane, make with the perpendicular, are equal, the lines are equal, and pierce the plane at equal distances from the foot of the perpendicular.

=

DEM.-Thus, in the figure, let DPE' DPE'; then PE' = PE" and DE' DE". For, revolve DE'P about PD; DE' will continue in the plane MN, and when angle DPE' coincides with its equal DPE", PE' coincides with PE", and DE' with DE".