BOOK VI.

PLANES AND POLYEDRAL ANGLES.

DEFINITIONS.

1. A straight line is perpendicular to a plane, when it is perpendicular to every straight line of the pls e which passes through its foot: conversely, the plane is perpendicular to the line. The point at which the perp ndicular meets the plane, is called the foot of the perpendic ular.

2. A line is parallel to a plane, when it cannot meet that plane, to what distance soever both be produced, Conversely, the plane is parallel to the line.

3. Two planes are parallel to each other, when they cannot meet, to what distance soever both be produced.

4. The indefinite space included between two planes which intersect each other, is called a diedral angle: the planes are called the faces of the angle, and their line of common intersection, the edge of the angle.

A diedral angle is measured by the angle contained be tween two lines, one drawn in each face, and both perpen. dicular to the common intersection at the same point. This angle may be acute, obtuse, or a right angle. If it is a right angle, the two faces are perpendicular to each other.

5. A POLYEDRAL angle is the indefinite space included by several planes meeting at a common point. Each plane is called a face: the line in which any two faces intersect, is called an edge: and the common point of meeting of all the planes, is called the vertex of the polyedral angle.

Thus, the polyedral angle whose vertex is is bounded by the four faces, ASB, BSC, CSD, DSA. Three planes, at least, are necessary to form a polyedral angle.

A polyedral angle bounded by three planes, is called a triedral angle.

POSTULATES.

1. Let it be granted, that from a given point of a plane, a line may be drawn perpendicular to that plane.

2. Let it be granted, that from a given point without a plane, a perpendicular may be let fall on the plane.

PROPOSITION I. THEOREM.

A straight line cannot be partly in a plane, and partly out of it.

For, by the definition of a plane (B. I., D. 9), when a straight line has two points common with it, the line lics wholly in the plane.

Scholium. To discover whether a surface is plane, apply a straight line in different ways to that surface, and ascer tain if it coincides with the surface throughout its whole extent.

PROPOSITION II. THEOREM.

Two straight lines which intersect each other, lie in the same plane, and determine its position.

B

Let AB, AC, be two straight lines which intersect each other in A; a plane may be conceived in which the straight line AB is found; if this plane be turned round AB, until it pass through the point C, then the line AC, which has two of its points A and C, in this plane, lies wholly in it; hence, the position of the plane is determined by the single condition of contain ing the two straight lines AB, AC.

the parallels AB, CD. But the lines AE, EF, determine this plane; therefore, so do the parallels, AB, CD.

PROPOSITION III. THEOREM.

If two planes cut one another, their common section will be a straight line.

Let the two planes AB, CD, cut ɔne another, and let E and F be two points of their common section. Draw the straight line EF. This line lies wholly in the plane AB, and also, wholly in the plane CD (B. I., D. 9): therefore, it is in both planes at once. But

B

D

A

since a straight line and a point out of it cannot lie in two planes at the same time (P. II., c. 1), EF contains all the points common to both planes, and consequently, is their common intersection.

PROPOSITION IV. THEOREM.

If a straight line be perpendicular to two straight lines at their point of intersection, it will be perpendicular to the plane of those lines.

Let MN be the plane of the two lines BB, CC, and let AP be perpendicular to each of them at their point of intersection P; then will AP be perpendicular to every line of the plane passing through P, and consequently to the plane itself (D. 1).

For, through P draw in the plane MN, any straight line as PQ. Through any point of this line, as Q, draw BQC, so that BQ shall be equal to QC (B. IV., PROB. 5); draw AB, AQ, AC.

The base BC being divided into two equal parts at the point Q, the triangle BPC gives (B. IV., P. 14).

M

B

PC2+PB2=2PQ2+2QC2.

The triangle BAC in like manner gives,

AC2+AB2✡2AQ2+2QC2.

B

Taking the first of these equals from the second, and observing that the triangles APC, APB, being right-angled at P, give

AC-PCAP2, and AB-PBAP2,

we shall have,

AP2+AP2

2AQ'—2PQ'.

Therefore, by taking the halves of both, we have

AP2AQ2-PQ3, or AQ2=AP2+PQa ;

hence, the triangle APQ is right-angled at P; hence, AP is perpendicular to PQ.

Scholium. Thus, it is evident, not only that a straight line may be perpendicular to all the straight lines which pass through its foot, in a plane, but that it always must be so, whenever it is perpendicular to two straight lines drawn in the plane: hence, a line and plane may fulfil the conditions of the first definition.

Cor. 1. The perpendicular AP is shorter than any oblique line AQ; therefore, it measures the shortest distance from the point A to the plane MN.

Cor. 2. At a given point P, on a plane, it is impossible to erect more than one perpendicular to the plane; for, if there could be two perpendiculars at the same point P, draw through these two perpendiculars a plane, whose sec tion with the plane MN is PQ; then these two perpen

diculars would be both perpendicular to the line PQ, at the same point, which is impossible (B. I., P. 14, c.)

It is also impossible to let fall from a given point, out of a plane, two perpendiculars to that plane; for, if AP, AQ, be two such perpendiculars, the triangle APQ will have two right angles APQ, AQP, which is impossible (B. L., P. 25, c. 3).

PROPOSITION V. THEOREM.

If, from a point without a plane, a perpendicular be drawn to the plane, and oblique lines be drawn to its different points: 1st. The oblique lines which meet the plane at points equally distant from the foot of the perpendicular, are equal:

2d. Of two oblique lines which meet the plane at unequal dis tances, the one passing through the remote point is the longer.

Let AP be perpendicular to the plane MN; AB, AC, AD, oblique lines intercepting the equal distances PB, PC, PD, and AE a line intercepting the larger distance PE: then will AB=AC=AD; and AE will be greater than AD.

For, the angles APB, APC, APD, being right angles, and the distances

PB, PC, PD, equal to each other, N
the triangles APB, APC, APD,
have in each an equal angle con-
tained by two equal sides: there-
fore they are equal (B. I., P. 5);
hence, the hypothenuses, or the
oblique lines AB, AC, AD, are equal
to each other.

BE

M

Again, since the distance PE is greater than PD, or its equal PB, the oblique line AE is greater than AB, or its equal AD (B. 1, p. 15).

Cor. All the equal oblique lines, AB, AC, AD, &c., terminate in the circumference BCD, described from P, the foot of the perpendicular, as a centre; therefore, a point A being given out of a plane, the point P at which the per