For, let M'N' be any other plane, A' any point without it, and

A'P' the perpendicular from A' to

MN, erected at P. There can be no other: for let PB be any other straight line drawn through P; let the plane determined by the two lines PA, PB, intersect the plane MN in the line PC; then, since APC is a right angle, BPC is not a right angle, and therefore BP is not perpendicular to the plane.

9. Scholium. By the distance of a point from a plane is meant the shortest distance; hence it is the perpendicular distance from the point to the plane.

PROPOSITION IV.-THEOREM.

10. Oblique lines drawn from a point to a plane, at equal distances from the perpendicular, are equal; and of two oblique lines unequally distant from the perpendicular the more remote is the greater.

1st. Let AB, AC be oblique lines from the point A to the plane MN, meeting the plane at the equal distances PB, PC, from the foot of the perpendicular AP; then, AB AC. For, the right triangles APB,

A

C

E

11. Corollary I. Conversely, equal oblique lines from a point to a plane meet the plane at equal distances from the perpendicular; and of two unequal oblique lines, the greater meets the plane at the greater distance from the perpendicular.

12. Corollary II. Equal straight lines from a point to a plane meet the plane in the circumference of a circle whose centre is the foot of the perpendicular from the point to the plane. Hence we derive a method of drawing a perpendicular from a given point A to a given plane MN: find any three points, B, C, E, in the plane, equidistant from A, and find the centre P of the circle passing through these points; the straight line AP will be the required perpendicular.

PROPOSITION V.-THEOREM.

13. If a straight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of those lines. Let AP be perpendicular to PB and PC, at their intersection P; then, AP is perpendicular to the plane MN which contains those lines.

For, let PD be any other straight line drawn through P in the plane MN. Draw any straight line BDC intersecting PB, PC PD, in B, C, D; produce AP to A' making PA' = PA, and join A and A' to each of the points B, C, D.

M

C

A'

=

CA';

Since BP is perpendicular to AA', at its middle point, we haye BA =BA', and for a like reason CA therefore, the triangles ABC, A'BC, are equal (I. 80). If, then, the triangle ABC is turned about its base BC until its plane coincides with that of the triangle A'BC, the vertex A will fall upon A'; and as the point D remains fixed, the line AD will coincide with A'D; therefore, D and P are each equally distant from the extremities of AA', and DP is perpendicular to AA' or AP(I. 41). Hence AP is perpendicular to any line PD, that is, to every line, passing through its foot in the plane MN, and is consequently perpendicular to the plane.

14. Corollary I. At a given point P of a straight line AP, a plane can be passed perpendicular to that line, and but one. For, two perpendiculars, PB, PC, being drawn to AP in any two different planes APB, APC, passed through AP, the plane of the lines PB, PC, will

be perpendicular to the line AP. Moreover, no other plane passed through P can be perpendicular to AP; for, any other plane not containing the point C would cut the oblique line AC in a point C' different from C, and we should have the angle APC' different from. APC, and therefore not a right angle.

M

-D

C

N

15. Corollary II. All the perpendiculars PB, PC, PD, etc., drawn to a line AP at the same point, lie in one plane perpendicular to AP. Hence, if an indefinite straight line PQ, perpendicular to AP, be made to revolve, always remaining perpendicular to AP, it is said to generate the plane MN perpendicular to AP; for the line PQ passes successively, during its revolution, through every point of this plane.

B

16. Corollary III. Through any point C without a given straight line AP, a plane can be passed perpendicular to AP, and but one. For, in the plane determined by the line AP and the point C, the perpendicular CP can be drawn to AP, and then the plane generated by the revolution of PC about AP as an axis will, by the preceding corollary, be perpendicular to AP; and it is evident that there can be but one such perpendicular plane.

PROPOSITION VI.-THEOREM.

17. If from the foot of a perpendicular to a plane a straight line is drawn at right angles to any line of the plane, and its intersection with that line is joined to any point of the perpendicular, this last line will be perpendicular to the line of the plane.

Let AP be perpendicular to the plane MN; from its foot P let PD be drawn at right angles to any line BC of the plane; then, A being any point in AP, the straight line AD is perpendicular to BC.

For, lay off DB = DC, and join PB, PC, AB, AC. Since DB DC, we have

M

B

N

(10). Therefore, A and D being each equally distant from B and C, the line AD is perpendicular to BC (I. 41)..

PARALLEL STRAIGHT LINES AND PLANES.

18. Definitions. A straight line is parallel to a plane when it cannot meet the plane though both be indefinitely produced.

In the same case, the plane is said to be parallel to the line.

Two planes are parallel when they do not meet, both being indefinite in extent.

PROPOSITION VII.-THEOREM.

19. If two straight lines are parallel, every plane passed through one of them is parallel to the other.

Let AB and CD be parallel lines, and MN any plane passed through CD; then, the line AB and the plane MN are parallel.

M

A

B

D

N

For, the parallels AB, CD, are in the same plane, ACDB, which intersects the plane MN in the line CD; and if AB could meet the plane MN, it could meet it only in some point of CD; but AB cannot meet CD, since it is parallel to it; therefore AB cannot meet the plane MN.

20. Corollary I. Through any given straight line HK, a plane can be passed parallel to any other given straight line AB.

For, in the plane determined by AB and any point H of HK, let HL be drawn parallel to AB; then, the plane MN, determined by HK and HL, is parallel to AB.

A

M

K

H

L

N

3

21. Corollary II. Through any given point O, a plane can be passed parallel to any two given straight lines AB, CD, in space.

For, in the plane determined by the

given point and the line AB let a Ob be drawn through O parallel to AB; and in the plane determined by the point 0 and the line CD, let cOd be drawn through O parallel to CD; then, the plane determined by the lines ab and cd is parallel to each of the lines AB and CD.

M

M

C

Ᏼ .

N

PROPOSITION VIII.-THEOREM.

22. If a straight line and a plane are parallel, the intersections of the plane with planes passed through the line are parallel to that line and to each other.

Let the line AB be parallel to the plane MN, and let CD, EF, etc., be the intersections of MN with planes passed through AB; then, these intersections are parallel to AB and to each other.

For, the line AB cannot meet CD, since

M

A

B

E

H

N

it cannot meet the plane in which CD lies; and since these lines are in the same plane, AD, and cannot meet, they are parallel. For the same reason, EF, GH, are parallel to AB.

Moreover, no two of these intersections, as CD, EF, can meet; for if they met, their point of meeting and the line AB would be at once in two different planes, AD and AF, which is impossible (4).

23. Corollary. If a straight line AB is parallel to a plane MN, a parallel CD to the line AB, drawn through any point C of the plane, lies in the plane.

For, the plane passed through the line AB and the point C intersects the plane MN in a parallel to AB, which must coincide with CD, since there cannot be two parallels to AB drawn through the same point C.

PROPOSITION IX.-THEOREM.

24. Planes perpendicular to the same straight line are parallel to each other.

The planes MN, PQ, perpendicular to the same straight line AB, cannot meet; for, if they met, we should have through a point of their intersection two planes perpendicular to the same straight line, which is impossible (16); therefore these planes are parallel.