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Cor. 1. The perpendicular AB is shorter than any oblique ine AD; it therefore measures the true distance of the point A from the plane MN.

Cor. 2. Through a given point B in a plane, only one perpendicular can be drawn to this plane. For, if there could be two perpendiculars, suppose a plane to pass through them, whose intersection with the plane MN is BG; then these two perpendiculars would both be at right angles to the line BG, at the same point and in the same plane, which is impossible (Prop. XVI., Cor., B. I.).

It is also impossible, from a given point without a plane, to let fall two perpendiculars upon the plane. For, suppose AB, AG to be two such perpendiculars; then the triangle ABG will have two right angles, which is impossible (Prop. XXVII., Cor. 3, B. I.).

PROPOSITION V. THEOREM.

Oblique lines drawn from a point to a plane, at equal distances from the perpendicular, are equal; and of two oblique lines unequally distant from the perpendicular, the more remote is the longer.

Let the straight line AB be drawn perpendicular to the plane MN; and let AC, AD, AE be oblique lines drawn from the point A, equally distant from the perpendicular; also, let AF be more remote from the perpendicular than AE; then will the lines AC, AD, AE all be equal to each other, and AF be longer than AE.

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For, since the angles ABC, ABD, ABE are right angles and BC, BD, BE are equal, the triangles ABC, ABD, ÅBE have two sides and the included angle equal; therefore the third sides AC, AD, AE are equal to each other.

So, also, since the distance BF is greater than BE, it is plain that the oblique line AF is longer than AE (Prop. XVII., B. I.).

Cor. All the equal oblique lines AC, AD, AE, &c., term.nate in the circumference CDE, which is described from B, the foot of the perpendicular, as a center.

If, then, it is required to draw a straight line perpendicular to the plane MN, from a point A without it, take three points in the plane C, D, E, equally distant from A, and find B the

center of the circle which passes througn these points. Join AB, and it will be the perpendicular required.

Scholium. The angle AEB is called the inclination of the line AE to the plane MN. All the lines AC, AD, AE, &c., which are equally distant from the perpendicular, have the same inclination to the plane; because all the angles ACB, ADB, AEB, &c., are equal.

PROPOSITION VI. THEOREM.

If a straight line is perpendicular to a plane, every plane which passes through that line, is perpendicular to the firstmentioned plane.

Let the straight line AB be perpendicular to the plane MN; then will every plane which passes through AB be perpendicular to the plane MN.

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Suppose any plane, as AE, to pass through AB, and let EF be the common section of the planes AE, MN. In the plane MN, through the point B, draw CD perpendicular to the common section EF. Then, since the line AB is perpendicular to the plane MN, it must be perpendicular to each of the two straight lines CD, EF (Def. 1). But the angle ABD, formed by the two perpendiculars BA, BD, to the common_section EF, measures the angle of the two planes AE, MN (Def. 4); and since this is a right angle, the two planes must be perpendicular to each other. Therefore, if a straight line, &c.

Scholium. When three straight lines, as AB, CD, EF, are perpendicular to each other, each of these lines is perpendicular to the plane of the other two, and the three planes are perpendicular to each other.

PROPOSITION VII THEOREM.

If two planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their common section, will be perpendicular to the other plane.

Let the plane AE be perpendicular to the plane MN, and let the line AB be drawn in the plane AE perpendicular to the common section EF; then will AB be perpendicular to the plane MN.

For in the plane MN, draw CD through the point B perpendicular to EF. Then, because the planes AE and MN are perpendicular, the angle ABD M is a right angle. Hence the line AB is perpendicular to the two straight lines CD, EF at their point of intersection; it is consequently perpendicular to their plane MN (Prop. IV.). Therefore, if two planes, &c.

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Cor. If the plane AE is perpendicular to the plane MN, and if from any point B, in their common section, we erect a perpendicular to the plane MN, this perpendicular will be in the plane AE. For if not, then we may draw from the same point, a straight line AB in the plane AE perpendicular to EF, and this line, according to the Proposition, will be perpendicular to the plane MN. Therefore there would be two perpendiculars to the plane MN, drawn from the same point, which is impossible (Prop. IV., Cor. 2).

PROPOSITION VIII. THEOREM.

If two planes, which cut one another, are each of them perpendicular to a third plane, their common section is perpendicular to the same plane.

Let the two planes AE, AD be each of them perpendicular to a third plane MN, and let AB be the common section of the first two planes; then will M, AB be perpendicular to the plane MN.

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For, from the point B, erect a perpendicular to the plane MN. Then, by the Corollary of the last Proposition, this line must be situated both in the plane AD and in the plane AE; hence it is their common section AB. Therefore, if two planes, &c.

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PROPOSITION IX. THEOREM.

Two straight lines which are perpendicular to the same plane, ure rallel to each other.

Let the two stight lines AB, CD be each of them perpendicular to the same plane MN; then will AB be parallel to CD

In the plane MN, draw the straight line BD joining the points B and D. Through the lines AB, BD pass the plane EF; it will be perpendicular to the plane MN (Prop. VI.); also, the line CD will lie in this plane, because it is perpendicular to MN (Prop. VII., Cor.).

Now, because AB and CD are

both perpendicular to the plane MN,

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they are perpendicular to the line BD in that plane; and since AB, CD are both perpendicular to the same line BD, and lie in the same plane, they are parallel to each other (Prop. XX., B. I.). Therefore, two straight lines, &c.

Cor. 1. If one of two parallel lines be perpendicular to a plane, the other will be perpendicular to the same plane. If AB is perpendicular to the plane MN, then (Prop. VI.) the plane EF will be perpendicular to MN. Also, AB is perpendicular to BD; and if CD is parallel to AB, it will be perpendicular to BD, and therefore (Prop. VII.) it is perpendicular to the plane MN.

Cor. 2. Two straight lines, parallel to a third, are parallel to each other. For, suppose a plane to be drawn perpendicular to any one of them; then the other two, being parallel to the first, will be perpendicular to the same plane, by the preceding Corollary; hence, by the Proposition, they will be parallel to each other.

The three straight lines are supposed not to be in the same plane; for in this case the Proposition has been already demonstrated

PROPOSITION X. THEOREM.

If a straight line, without a given plane, be parallel to a straight line in the plane, it will be parallel to the plane.

Let the straight line AB be parallel to the straight line CD, in the plane MN; then will it be parallel to the M. plane MN.

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Through the parallels AB, CD suppose a plane ABDC to pass. If the line AB can meet the plane MN, it must meet it in some point of the line CD, which is the common intersection of the two planes. But AB can not meet CD, since they are parallel; hence it can not meet the plane MN; that is, AB is parallel to the plane MN (Def. 2). Therefore, if a straight line, &c.

PROPOSITION XI, THEOREM.

Two planes, which are perpendicular to the same straight line, are parallel to each other.

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which is perpendicular to the plane MN, is perpendicular to the line AC drawn through its foot in that plane. For the same reason AB is perpendicular to BC. Therefore CA and

CB are two perpendiculars let fall from the same point C upon the same straight line AB, which is impossible (Prop. XVI., B. I.). Hence the planes MN, PQ can not meet when produced; that is, they are parallel to each other. Therefore, two planes, &c.

PROPOSITION XII. THEOREM.

If two parallel planes are cut by a third plane, their common sections are parallel.

Let the parallel planes MN, PQ be M cut by the plane ABDC; and let their common sections with it be AB, CD; then will AB be parallel to CD.

For the two lines AB, CD are in the same plane, viz., in the plane ABDC which cuts the planes MN, PQ; and if these lines were not parallel, they would meet when produced; therefore

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the planes MN, PQ would also meet, which is impossible, be cause they are parallel. Hence the lines AB, CD are paral lel. Therefore, if two paralel planes, &c.

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