THEOREM VIII. (103.)

If a plane be perpendicular to one of two parallel lines, it will be perpendicular to the other; and if two straight lines are perpendicular to the same plane, they are parallel to one another.

First. Let AP be perpendicular to the plane HK, and DE be parallel to AP; DE will also be perpendicular to HK.

Let the plane APDE which includes the parallels AP and DE, intersect the plane HK in PD; and in the plane HK let the line BC be drawn perpendicular to PD, and join AD.

H

E

F

P

D

B

By cor. th. 102, BC is perpendicular to the plane APDE, and therefore the angle BDE is a right angle. Also, since AP, DE, are parallels, and APD is a right angle (def. 89), the angle EDP is a right angle. Hence ED is perpendicular to the two straight lines DP, DB, and hence is perpendicular to the plane HK in which they lie.

Second. Let AP, DE be perpendicular to the plane HK, they shall be parallel to one another.

For if this be denied, let some other line DF through D be supposed parallel to AP. Then DF will (by the former part of the proposition) be perpendicular to HK. But, by hypothesis, DE, also passing through D, is perpendicular to HK: and hence through the same point D there can be two perpendiculars drawn to the plane, HK, which, (th. 99,) is impossible. Hence AP, DE, are parallel.

THEOREM IX. (104.)

If each of two lines which intersect one another be parallel to a plane, the plane in which these lines are situated, will also be parallel to it.

LET the two straight lines AR, AC, be each of them parallel to the plane MN; then the plane HQ, in which AR, AC, are situated, will be parallel to MN.

M

N

For if HQ be not parallel to MN they will meet, and their intersection will be a straight line (th. 2). Let NP be their intersection. Then, since NP is in the same plane with AR and AC, which meet in A, it cannot be parallel to both of them, and therefore will cut one at least, as AR, in the point R. Now the point R is situated also in the plane MN, and hence AR meets MN: but AR is parallel (hyp.) to the plane MN, which is impossible. Whence the plane in which AR, AC, are situated, cannot meet the plane MN to which AR, AC, are parallel; that is, HQ is parallel to MN.

Cor. Hence, through any given line which is parallel to a plane, a second plane may always be made to pass, that shall be parallel to that plane. For through any point in the line another line parallel to the given plane may be drawn; and the plane of these two lines will be parallel to the other plane.

THEOREM X. (105.)

[See figure to theorem 1.]

If one plane meet another plane, it will make angles with that other plane, which are together equal to two right angles.

LET the plane ACBD meet the plane AEBF; these planes make with each other two angles whose sum is equal to two right angles.

For, through any point G, in the common section AB, draw CG, EF, perpendicular to AB. Then, the line CG makes with EF two angles together equal to two right angles. But these two angles are (def. 91) the angles of inclination of the two planes. Therefore, the two planes make angles with each other, which are together equal to two right angles.

Scholium.

In like manner it may be demonstrated, that planes which intersect have their vertical or opposite angles equal; also, that parallel planes cut by a third make their alternate angles equal; and so on, as in parallel lines; but with obvious limitations. See pp. 301, 302.

THEOREM XI. (106.)

If two lines be parallel to a third line, though not in the same plane with it; they will be parallel to each other.

LET the lines AB, CD, be each of them parallel to the third line EF, though not in the same plane with it; then will AB be parallel to CD.

For, from any point G in the line EF, let GH, GI, be drawn perpendicular to EF, in the planes EB, ED, of the parallels AB, EF, and EF, CD.

Then, since the line EF is perpendicular to the two lines GH, GI, it is perpendicular to the plane GHI of those lines

(th. 98). And because EF is perpendicular to the plane GHI, its parallel AB is also perpendicular to that plane (cor. th. 103). For the same reason, the line CD is perpendicular to the same plane GHI. Hence, because the two lines AB, CD, are perpendicular to the same plane, these two lines are parallel (th. 103).

THEOREM XII. (107.)

If two lines, that meet each other, be parallel to two other lines that meet each other, though not in the same plane with them; the angles contained by those lines will be equal.

LET the two lines AB, BC, be parallel to the two lines DE, EF; then will the angle ABC be equal to the angle DEF.

For, make the lines AB, BC, DE, EF, all equal to each other, and join AC, DF, AD, BE, CF.

Then, the lines AD, BE, joining the equal and parallel lines AB, DE, are equal and parallel (th. 24). For the same reason, CF, BE, are equal and parallel. Therefore

D

C

VOL. I.

A a

AD, CF, are equal and parallel (th. 106); and consequently also AC, DF (th. 24). Hence, the two triangles ABC, DEF, having all their sides equal, each to each, have their angles also equal, and consequently the angle ABC is equal to the angle DEF.

THEOREM XIII. (108.)

The sections made by a plane cutting two parallel planes are parallel to one

another.

LET the parallel planes HK, MN, be cut by the plane EG H in the lines EF, GD; then EF will be parallel to GD.

For if EF, GD, be not parallel, since they are in the same plane EG, they would, if produced, meet: but as EF is in the plane HK, and GD in MN, these planes would in that case also meet. But these planes cannot meet, since, by hypothesis, they are parallel: and hence EF, GD, cannot meet, or they are parallel.

E

M

D

G

Cor. 1. The parallels ED, FG, comprehended between parallel planes HK, MN, are equal.

Let the plane in which the parallels ED, FG, lie, cut the parallel planes in EF, GD. Then these are also parallel: hence EG is a parallelogram, and its sides FG, ED, therefore equal.

Cor. 2. Hence two parallel planes are every where equi-distant. For in this case FG, ED, are perpendicular to the two planes (th. 100), and hence are parallel (th. 101), and hence again (cor. 1, th. 108) are equal to one another.

THEOREM XIV. (109.)

Straight lines being cut by parallel planes are divided proportionally.

LET there be, for instance, two straight lines AB, CD, cut

in A, E, B, C, F, D, by the three parallel planes HK, MN, PQ then we shall have AE: EB:: CF: FD.

:

M

K

G

N

Draw AD meeting the plane MN in G, and join AC, EG, GF, BD; the intersections EG, BD, of the plane ABD with MN, PQ, are parallel (th. 108); and hence AE:EB::AG: P GD. In like manner, AC, GF, are parallel, and hence AG: GD: CF: FD. Hence AE EB:: CF: FD. In the same way the property is established if there be more lines or more planes, or both.

Scholium.

If three or more straight lines meeting a plane be divided proportionally, and in the same order, all the sets of corresponding points of section will lie in planes parallel to the first: but if there be only two lines, the planes through the corresponding points may or may not be parallel (p. 360). When the planes are not parallel, however, their sections with each other will either coincide or be parallel to each other.

THEOREM XV. (110.)

1. If through a line which is perpendicular to a plane another plane be made to pass, this plane will be perpendicular to the former.

2. If two planes be perpendicular to one another, and in one of them a line be drawn perpendicular to the common section, it will be perpendicular to the other plane.

3. If a plane be perpendicular to a plane, and if at a point in their intersection a perpendicular be erected to the former plane, it will lie wholly in the latier. 4. If each of two planes be perpendicular to a third plane, their intersection will be perpendicular to it also.

First. If the line AP be perpendicular to the plane MN, any plane through AP will be perpendicular to MN.

Let the planes AB, MN intersect in BC, and in the plane MN draw DE perpendicular to BP. Then the line AP, being perpendicular to the plane MN, will be perpendicular to each of the straight lines BC, DE: but the angle formed by the two perpendiculars PA, PD, at the common intersection

measures the angle of the two planes, (def. 91); and hence (def. 90), since the angle is right, the two planes are perpendicular to each other.

Second. If the plane AB is perpendicular to the plane MN, and in AB the line AP is drawn perpendicular to the common section PB of the planes MN, AB, it will be perpendicular to the plane MN.

For, in the plane MN draw PD perpendicular to PB; then, because the planes are perpendicular, the angle APD is a right one: therefore the line AP is perpendicular to the two straight lines PB, PD. Hence it is perpendicular to their plane MN.

Third. If the plane AB be perpendicular to the plane MN, and if at a point P of the common intersection a perpendicular be erected to the plane MN, that perpendicular will be in the plane AB.

For if not, then in the plane AB a perpendicular AP might be drawn to the common intersection PB, which at the same time would be perpendicular to the plane MN. Hence two perpendiculars may be drawn from the same point to the same plane, which is impossible (th. 109).

Fourth. If each of two planes be perpendicular to a third plane, their common intersection will be perpendicular to it also.

That is, if the planes AB, AD, are perpendicular to a third plane MN, their common intersection AP will be also perpendicular to MN.

For at the point P, let a perpendicular be drawn to the plane MN. That perpendicular must be at the same time in the plane AB and in the plane AD, and hence it is their common intersection AP.

Scholium.

The properties in this theorem are the foundation of the method of co-ordinates in space, and of the principles and practice of Descriptive Geometry.

THEOREM XVI. (111.)

If any prism be cut by a plane parallel to its base, the section will be equal and similar to the base.

LET AG be any prism, and IL a plane parallel to the base, AC; then will the plane IL be equal and similar to the base AC, or the two planes will have all their sides and all their angles equal.

G

E

M

P

L

I

D

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B

For, the two planes AC, IL, being parallel by hypothesis: and two parallel planes, cut by a third plane, having parallel sections (th. 108); therefore IK is parallel to AB, and KL to BC, and LM to CD, MP to DN, and IP to AN. But AI and BK are parallels (def. 95); consequently AK is a parallelogram; and the opposite sides BA, IK, are equal (th. 22). In like manner, it is shown that KL = BC, and LM = CD, MP = DN, and IP = AN, or the two planes AC, IL, are mutually equilateral. But these two planes having their corresponding sides parallel, have the angles contained by them also equal (th. 107), namely, the angle A = the angle I, the angle B = the angle K, the angle C = the angle L, and the angle D = the angle M. So that the two planes AC, IL, have all their corresponding sides and angles equal, or they are equal and similar.

THEOREM XVII. (112.)

If a cylinder be cut by a plane parallel to its base, the section will be a circle, equal to the base.

LET AF be a cylinder, and GHI any section parallel to the base ABC; then will GHI be a circle, equal to ABC.

For, let the planes KE, KF, pass through the axis of the cylinder MK, and meet the section GHI in the three points H, I, L; and join the points as in the figure.

Then, since KL, CI, are parallel (def. 102); and the plane KI, meeting the two parallel planes ABC, GHI, makes the two sections KC, LI, parallel (th. 108); the figure KLIC is therefore a parallelogram, and consequently has the opposite sides LI, KC, equal, where KC is a radius of the circular base.

In like manner it is shown that LH is equal to the radius KB; and that any other lines, drawn from the point L to the circumference of the section GHI, are all equal to radii of the base; consequently GHI is a circle, and equal to ABC.

Scholium.

Had the base been any other curve whatever, it may be shown in the same manner, that the section parallel to the base will be a figure equal and similar to the base.