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.

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.

C

B

D

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.

line, as Q, draw BQC, so that BQ shall be equal to QC (Book

IV. Prob. V.); draw AB, AQ, AC.

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.

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

AC2-PC2-AP2, and AB2-PB2=AP2;

we shall have

AP2+AP2=2AQ2-2PQ2.

Therefore, by taking the halves of both, we have

AP2=AQ2-PQ2, or AQ=AP+PQ2;

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

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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 AE a line more remote: then will AB AC=AD; and AE will be greater than AD.

For, the angles APB, APC, APD, being right angles, if we suppose the distances PB, PC,

M

E

N

PD, to be equal to each other, the triangles APB, APC, APD, will have in each an equal angle contained by two equal sides; therefore they will be equal; hence the hypothenuses, or the oblique lines AB, AC, AD, will be equal to each other. In like

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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 P the foot of the M perpendicular as a centre; therefore a point A being given out of a plane, the point Pat which the perpendicular let fall from A would meet that plane, may be found by marking upon

E

B

N

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.

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.

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.

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

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

P

B

E

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.).

Cor. It is evident likewise, that BC is perpendicular to the plane APD, since BC is at once perpendicular to the two straight lines AD, PD.

Scholium. The two lines AE, BC, afford an instance of two lines which do not meet, because they are not situated in the same plane. The shortest distance between these lines is the straight line PD, which is at once perpendicular to the line AP and to the line BC. The distance PD is the shortest distance between them, because if we join any other two points, such as A and B, we shall have AB>AD, AD>PD; therefore AB>PD.

The two lines AE, CB, though not situated in the same plane, are conceived as forming a right angle with each other, because AE and the line drawn through one of its points parallel to BC would make with each other a right angle. In the same manner, the line AB and the line PD, which represent any two straight lines not situated in the same plane, are supposed to form with each other the same angle, which would be formed by AB and a straight line parallel to PD drawn through one of the points of AB.

PROPOSITION VII. THEOREM.

If one of two parallel lines be perpendicular to a plane, the other will also be perpendicular to the same plane.

Let the lines ED, AP, be parallel; if AP is perpendicular to the plane NM, M then will ED be also perpendicular to it.

Through the parallels AP, DE, pass a plane; its intersection with the plane MN will be PD; in the plane MN

draw BC perpendicular to PD, and draw AD.

By the Corollary of the preceding Theorem, BC is perpendicular to the plane APDE; therefore the angle BDE is a right angle; but the angle EDP is also a right angle, since AP is perpendicular to PD, and DE parallel to AP (Book I. Prop. XX. Cor. 1.); therefore the line DE is perpendicular to the two straight lines DP, DB; consequently it is perpendicular to their plane MN (Prop. IV.).