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 to observe if it touches the surface throughout its whole extent.

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 B 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 containing the two straight lines AB, AC. Therefore, two

straight lines, &c.

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

Cor. 2. Hence also two parallels

AB, CD, determine the position of A
a plane; for, drawing the secant
EF, the plane of the two straight
lines AE, EF, is that of the paral-
lels AB, CD.

E

B

C

F

D

PROPOSITION III. THEOREM.

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

Let the two planes AB, CD, cut

each other, and let E and F be two A

points of their common intersec

tion. Draw the straight line EF.

E

D

(Def. 8. 1;) and therefore must be their common intersection, Hence, If two planes cut each other, &c.

PROPOSITION IV. THEOREM.

If a straight line AP is perpendicular to two other straight lines PB, PC, which cross each other at its foot in the plane MN, it will be perpendicular to any straight line PQ drawn through its foot in the same plane, and thus it will be perpendicular to the plane MN.

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

AC-PC2 AP2, and AB'-PB2=AP2;

we shall have

AP2+AP2-2AQ2—2PQ2.

Therefore, by taking the halves of both, we have

AP2=AQ3-PQ2, or AQ2=AP2+PQ2;

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

If a straight line is perpendicular to two other straight lines at the point of their intersection, it is also perpendicular to the plane in which those lines are.

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; which proves our 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 PQ.

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 those two perpendiculars would be perpendicular to the line PQ, at the same point, and in the same plane, which is impossible. (14. 1. 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. (28. 1. Cor. 3.)

Oblique lines equally distant from the perpendicular drawn from the same point without a plane as the perpendicular, are equal; and, of two oblique lines unequally distant from the perpendicular, the more distant is the longer.

fore 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. Hence, Oblique lines equally distant, &c.

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 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.

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.

Let AP be a perpendicular to the plane MN, and BC a line situated in that plane; if from P the foot of the perpendicular, PD is drawn at right-angles to BC, and AD is joined, AD will be perpendicular to BC.

Take BD DC, and join PB, PC, AB, AC; since DB=DC, the oblique line PB=PC:(15.1:) M and with regard to the perpendicular AP, since PB-PC, the oblique line AB=AC; (5.6;)

P

B

therefore ABC is an isosceles

triangle; consequently the line AD, drawn from its vertex to the

E

middle of its base BC, is perpendicular to BC. (11. 1. Sch.) Hence,

If from a point without a plane, a perpendicular be let fall on the plane, and from the foot of this perpendicular a straight line be drawn at right-angles to any line in that plane, and from the point of intersection a line be drawn to the first point, this last line will be perpendicular to the line in the plane.

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.