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

For in the plane MN, draw CD through the point B perpendicular to EF. Then, because the planes AE and MN are perpen-. dicular, the angle ABD 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 perpendicN ular to their plane MN (Pr. 4). Therefore,

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 (Pr. 4, 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 AB be perpendicular to the plane MN.

For, from the point B, erect a perpendicular to the plane MN. Then, by the N 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, etc.

Two straight lines which are perpendicular to the same plane· are parallel to each other.

C

A

Let the two straight 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 (Pr. 6); M also, the -line CD will lie in this plane, because it is perpendicular to MN (Pr. 7, Cor.).

E

B

D

Now, because AB and CD are both perpendicular to the plane MN, 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 (B. I., Pr. 20). Therefore, two straight lines, etc.

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 (Pr. 6) 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 (Pr. 7) it is perpendicular to the plane MN.

Cor. 2. Two straight lines parallel to the same straight line 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 plane MN.

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 the plane MN (Def. 2). Therefore, if a

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

M

P

N

Let the planes MN, PQ be perpendicular to the line AB; then will they be parallel to each other.

For, if they are not parallel, they will meet if produced. Let them be produced and meet in C. Join AC, BC.

Now the line AB, 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 (B. I., Pr. 16). Hence the planes MN, PQ can not meet when produced; that is, they are parallel to each other. Therefore two planes, etc.

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

PQ would also meet, which is impossible, because they are parallel. Hence the lines AB, CD are parallel. Therefore, if two parallel planes, etc.

If two planes are parallel, a straight line which is perpendicular to one of them is also perpendicular to the other.

Let the two planes MN, PQ be parallel, and let the straight line AB be perpendicular to the plane MN; AB will also be per- MA pendicular to the plane PQ.

Р

C

B

Through the point B draw any line BD in the plane PQ, and through the lines AB, BD suppose a plane to pass intersecting the plane MN in AC. The two lines AC, BD will be parallel (Pr. 12).. But the line AB, being perpendicular to the plane MN, is perpendicular to the straight line AC, which meets it in that plane; it must, therefore, be perpendicular to its parallel BD (B. I., Pr. 23, Cor. 1). But BD is any line drawn through B in the plane PQ; and, since AB is perpendicular to any line drawn through its foot in the plane PQ, it must be perpendicular to the plane PQ (Def. 1). Therefore, if two planes, etc.

Parallel straight lines included between two parallel planes are equal.

Let AB, CD be two parallel straight lines included between two parallel planes MN, PQ; then will AB be equal to CD.

M

Through the two parallel lines AB, CD, suppose a plane ABDC to pass, intersecting the parallel planes in AC and BD. The Р lines AC, BD will be parallel to each other

B

D

(Pr. 12). But AB is, by supposition, parallel to CD; therefore the figure ABDC is a parallelogram, and, consequently, AB is equal to CD (B. I., Pr. 30). Therefore parallel straight lines, etc. .Cor. Hence two parallel planes are every where equally distant; for if AB, CD are perpendicular to the plane MN, they will be perpendicular to the parallel plane PQ (Pr. 13), and, being both perpendicular to the same plane, they will be parallel to each other (Pr. 9), and consequently equal.

G

PROPOSITION XV. THEOREM.

If two angles not in the same plane have their sides parallel to each other and similarly situated, these angles will be equal, and their planes will be parallel.

M

B

H

D

E

G

C

Let the two angles ABC, DEF, lying in different planes MN, PQ, have their sides parallel each to each and similarly situated; then will the angle ABC be equal to the angle DEF, and the plane MN be parallel to the plane PQ.

Take AB equal to DE, and BC equal to EF, and join AD, BE, CF, AC, DF. Then, because AB is equal and parallel to DE, the figure ABED is a parallelogram (B. I., Pr.

32), and AD is equal and parallel to BE.

For the same reason, CF is equal and parallel to BE. Consequently, AD and CF, being each of them equal and parallel to BE, are parallel to each other (Pr. 9, Cor. 2), and also equal; therefore AC is also equal and parallel to DF (B. I., Pr. 32). Hence the triangles ABC, DEF are mutually equilateral, and the angle ABC is equal to the angle DEF (B. I., Pr. 15).

Also, the plane ABC is parallel to the plane DEF. For, if they are not parallel, suppose a plane to pass through A parallel to DEF, and let it meet the straight lines BE, CF in the points G and H. Then the three lines AD, GE, HF will be equal (Pr. 14). But the three lines AD, BE, CF have already been proved to be equal; hence BE is equal to GE, and CF is equal to HF, which is absurd; consequently, the plane ABC must be parallel to the plane DEF. Therefore, if two angles, etc.

Cor. 1. If two parallel planes MN, PQ are met by two other planes ABED, BCFE, the angles formed by the intersections of the parallel planes will be equal. For the section AB is parallel to the section DE (Pr. 12), and BC is parallel to EF; therefore, by the Proposition, the angle ABC is equal to the angle DEF.

Cor. 2. If three straight lines AD, BE, CF, not situated in the same plane, are equal and parallel, the triangles ABC, DEF, formed by joining the extremities of these lines, will be equal, and their planes will be parallel.

For, since AD is equal and parallel to BE, the figure ABED is a parallelogram; hence the side AB is equal and parallel to DE.