Definitions.

1. A straight line is perpendicular to a plane, when it is perpendicular to all the straight lines which pass through its foot in the plane. Conversely, the plane is perpendicular to the line.

The foot of the perpendicular is the point in which the perpendicular line meets the plane.

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

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

4. The angle or mutual inclination of two planes is the quantity, greater or less, by which they separate from each other; this angle is measured by the angle contained between two lines, one in each plane, and both perpendicular 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 planes are perpendicular to each other.

5. A solid angle is the angular space included between several planes which meet at the same point.

Thus, the solid angle S, is formed by the union of the planes ASB, BSC, CSD, DSA.

Three planes at least, are requisite to form a solid angle.

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PROPOSITION I. THEOREM.

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

For, by the definition of a plane, when a straight line has two points common with a plane, it lies wholly in that plane.

Scholium. To discover whether a surface is plane, it is necessary to apply a straight line in different ways to that surface, and ascertain if it touches 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.

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, B 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 containing the two straight lines AB, AC.

Cor. 1. A triangle ABC, or three points A, B, C, not in a straight line, determine the position of a plane.

If two planes cut each other, their common intersection will be a straight line.

C

Let the two planes AB, CD, cut each other. Draw the straight line EF, joining any two points E and F in A the common section of the two planes. This line will lie wholly in the plane AB, and also wholly in the plane CD (Book I. Def. 6.): therefore it will be in both planes at once, and consequently is their common intersection.

PROPOSITION IV. THEOREM.

E

B

F

P

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 them at their point of intersection P; then will AP be perpendicular M to every line of the plane passing through P, and consequently to the plane itself (Def. 1.).

B

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

N

The base BC being divided into two equal parts at the point Q, the triangle BPC will give (Book IV. Prop. XIV.), PC2+PB2=2PQ2+2QC2.

The triangle BAC will in like manner give,

AC2+AB2-2AQ2+2QC2.

B

A

D

AC2 PC-AP2, and AB2-PB2=AP2; we shall have

Taking the first equation from the second, and observing that the triangles APC, APB, which are both right angled at P, give

AP2+AP2=2AQ2—2PQ2.

Therefore, by taking the halves of both, we have
AP2=AQ2 PQ2, or AQ2=AP+PQ2;

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, when. ever it is perpendicular to two straight lines drawn in the plane: which proves the first Definition to be accurate.

Cor. 1. The perpendicular AP is shorter than any oblique line AQ; therefore it measures the true 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 that plane; for if there could be two perpendiculars at the same point P, draw through these two perpendiculars a plane, whose intersection with the plane MN is PQ; then these two perpendiculars would be perpendicular to the line PQ, at the same point, and in the same plane, which is impossible (Book I. Prop. XIV. Sch.).

It is also impossible to let fall from a given point out of a plane two perpendiculars to that plane; for let AP, AQ, be these two perpendiculars, then the triangle APQ would have two right angles APQ, AQP, which is impossible.

PROPOSITION V. THEOREM.

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

1st. Any two oblique lines equally distant from the perpendicular will be equal.

2d. Of any two oblique lines unequally distant from the perpendicular, the more distant will be the longer.

Let AP be perpendicular to the plane MN; AB, AC, AD, oblique lines equally distant from the perpendicular, and M AE a line more remote: then will AB AC=AD; and AE will be greater than AD.

E

N

For, the angles APB, APC, APD, being right angles, if we suppose the distances PB, PC, PD, to be equal to each other, the triangles APB, APC, APD, will have in each an equal angle contained by two equal sides; herefore they will be equal; hence the hypothenuses, or the oblique lines AB, AC, AD, will be equal to each other. In like

manner, if the distance PE is greater than PD or its equal PB, the oblique line AE will evidently be greater than AB, or its equal AD.

Cor. All the equal oblique lines, AB, AC, AD, &c. terminate in the circumference BCD, described from Pthe foot of the M perpendicular as a centre; therefore a point A being given out of a plane, the point P at which the perpendicular let fall from A would meet that plane, may be found by marking upon that plane three points B, C, D, equally distant from the point A, and then finding the centre of the circle which passes through these points; this centre will be P, the point sought.

N

P

Let AP be perpendicular to the plane NM, and PD perpendicular to BC; then will AD be also perpen- M dicular to BC.

B

Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN; which inclination is evidently equal with respect to all such lines AB, AC, AD, as are equally distant from the perpendicular; for all the triangles ABP, ACP, ADP, &c. are equal to each other.

E

PROPOSITION VI. THEOREM.

If from a point without a plane, a perpendicular be let fall on the plane, and from the foot of the perpendicular a perpendicular be drawn to any line of the plane, and from the point of intersection a line be drawn to the first point, this latter line will be perpendicular to the line of the plane.

P

C

D

Take DB DC, and draw PB, PC, AB, AC. Since DB=DC, the oblique line PB PC: and with regard to the perpendicular AP, since PB= PC, the oblique line AB-AC (Prop. V. Cor.); therefore the line AD has two of its points A and D equally distant from the extremities B and C; therefore AD is a perpendicular to BC, at its middle point D (Book I. Prop. XVI. Cor.).

E

B

N