AP be perpendicular to every line of the plane which passes through P, and consequently, to the plane itself.

For, through P, draw in the plane MN, any line PQ; through any point of this line, as Q, draw the line BC, so that BQ shall be equal to Q0 (B. IV., Prob. V.); draw AB, AQ, and AC.

B

A

N

B

The base BC, of the triangle BPC, being bisected at Q, we have (B. IV., P. XIV.),

PC2 + PB2 = 2PQ2 + 2Q02.

In like manner, we have, from the triangle ABC,

Subtracting the first of these equations from the second, member from member, we have,

2

ᎪᏛ . PC+AB2 PB = 2AQ2 - 2PQ2.

But, from Proposition XI., C. 1, Book IV., we have,

AC2 - PC2 = AP2 and AB2 PB2 = AP2;

hence, by substitution,

AQ2 - PQ2; or, AP2 + PQ2 = AQ2.

The triangle APQ is, therefore, right-angled at P (B. IV., P. XIII., S.), and consequently, AP is perpendicular to PQ hence, AP is perpendicular to every line of the plane MN passing through P, and consequently, to the plane itself; which was to be proved.

Cor. 1. Only one perpendicular can be drawn to a plane

A

from a point without the plane.
For, suppose two perpendiculars,
as AP and AQ, could be
drawn from the point A to the
plane MN. Draw PQ; then
the triangle APQ would have
two right angles, APQ and
AQP; which is impossible (B. I., P. XXV., C. 3).

M

P

Cor. 2. Only one perpendicular can be drawn to a plane from a point of that plane. For, suppose that two perpendiculars could be drawn to the plane MN, from the point P. Pass a plane through the perpendiculars, and let PQ be its intersection with MN; then we should have two perpendiculars drawn to the same straight line from a point of that line; which is impossible (B. I., P. XIV., C.).

If from a point without a plane, a perpendicular be drawn to the plane, and oblique lines be drawn to different points of the plane :

1o. The perpendicular will be shorter than any oblique line: 2°. Oblique lines which meet the plane at equal distances from the foot of the perpendicular, will be equal:

3. Of two oblique lines which meet the plane at unequal distances from the foot of the perpendicular, the one which meets it at the greater distance will be the longer.

Let A be a point without the plane MN; let AP be perpendicular to the plane; let AC, AD, be any two oblique lines meeting the plane at equal distances from the foot of the perpendicular; and let AC and AE be any

two oblique lines meeting the plane at unequal distances from

the foot of the perpendicular:

1o. AP will be shorter than any oblique line AC.

For, draw PC; then will

AP be less than AC (B.

I, P. XV.); which was to be proved.

A

E

D

M

2o. AC and AD will be equal.

For, draw PD; then the right-angled triangles APC, APD, will have the side AP common, and the sides PC, PD, equal: hence, the triangles are equal in all their parts, and consequently, AC and AD will be equal; which was to be proved.

3o. AE will be greater than AC.

For, draw PE, and take PB equal to PC; draw AB: then will AE be greater than AB (B. I., P. XV.); but AB and AC are equal: hence, AE is greater than AC; which was to be proved.

Cor. The equal oblique lines AB, AC, AD, meet the plane MN in the circumference of a circle, whose centre is P, and whose radius is PB: hence, to draw a perpendicular to a given plane MN, from a point A, without that plane, find three points B, C, D, of the plane equally distant from A, and then find the centre P of the circle whose circumference passes through these points: then will - AP be the perpendicular required.

Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN. The equal oblique lines AB, AC, AD, are all equally inclined to the plane MN. The inclination of AE is less than the inclination of any shorter line AB.

If from the foot of a perpendicular to a plane, a line be drawn at right angles to any line of that plane, and the point of intersection be joined with any point of the per pendicular, the last line will be perpendicular to the line of the plane.

Let AP be perpendicular to the plane MN, P its foot, BC the given line, and A any point of the perpendicular; draw PD at right angles to BC, and join the point D with A then will AD be perpendicular to BC.

equal to PC, we have AB equal to AC (P. V.). The line AD has, therefore, two of its points A and D, each equally distant from B and C hence, it is perpendicular to BC (B. I., P. XVI., S.); which was to be proved.

Cor. 1. The line BC is perpendicular to the plane of the triangle APD; because it is perpendicular to AD and PD, at D (P. IV.).

Cor. 2. The shortest distance between AP and BC is measured on PD, perpendicular to both. For, draw BE between any other points of the lines: then will BE be greater than PB, and PB will be greater than PD: hence, PD is less than BE.

In

Scholium. The lines AP and BC, though not in the same plane, are considered perpendicular to each other. general, any two straight lines not in the same plane, are considered as making an angle with each other, which angle is equal to that formed by drawing through a given point, two lines respectively parallel to the given lines.

PROPOSITION VII. THEOREM.

If one of two parallels is perpendicular to a plane, the other one is also perpendicular to the same plare.

Let AP and ED be two parallels, and le AP be perpendicular to the plane MN: then will ED be also perpendicular to the plane MN.

For, pass a plane through the parallels; its intersection with MN will be PD; draw AD, and in the plane MN draw BC perpendicular to PD at D. Now, BD is perpendicular to the plane APDE (P. VI., C.);

M

E

N

the angle BDE is consequently a right angle; but the angle EDP is a right angle, because ED is parallel to AP (B. I., P. XX., C. 1): hence, ED is perpendicular to BD and PD, at their point of intersection, and consequently, to their plane MN (P. IV.); which was to be proved.

Cor. 1. If the lines AP and ED are perpendicular to the plane MN, they are parallel to each other. For, if not, draw through D a line parallel to PA; it will be perpendicular to the plane MN, from what has just been proved; we shall, therefore, have two perpendiculars to the the plane MN, at the same point; which is impossible (P. IV., C. 2).