Of Planes.

to be drawn through A, that shall be perpendicular to AP.

Now, every line drawn through A, and perpendicular to AP, will be a line of this last plane (Def. 1): hence, this last plane will contain the lines AB, AD,

B

P

and consequently, a line which is perpendicular to two lines at the point of intersection, will be perpendicular to the plane of those lines.

THEOREM V.

If two straight lines are perpendicular to the same plane they will be parallel to each other.

Let the two lines AB, CD, be perpendicular to the plane EF: then will they be parallel to each other.

For, join B and D, the points. in which the lines meet the plane EF.

E

B

Then, because the lines AB, CD, are perpendicular to the plane EF, they will be perpendicular to the line BD (Def. 1). Now, if BA and DC are not parallel, they will meet at some point as : then, the triangle OBD would have two right angles, which is impossible (Bk. I. Th. xvii. Cor. 4).

Cor. If two lines are parallel, and one of them is perpendicular to a plane, the other will also be perpendicular to the same plane.

Of Planes.

THEOREM VI.

If two planes intersect each other at right angles, and a line be drawn in one plane perpendicular to the common intersection, this line will be perpendicular to the other plane.

Let the plane FE be perpendicular to MN, and AP be drawn in the plane FE, and perpendicular to the common intersection DE: then will AP be perpendicular to the plane MN.

For, in the plane MN draw CP perpendicular to the common

M

F

B

N

intersection DE. Then, because the planes MN and FE are perpendicular to each other, the angle APC, which measures their inclination, will be a right angle (Def. 6). Therefore, the line AP is perpendicular to the two straight lines PC and PD; hence, it is perpendicular to their plane MN (Th. iv).

THEOREM VII.

If one plane intersect another plane, the sum of the angles on the same side will be equal to two right angles.

the lines EG and DEC, one in each plane, and both perpendicular to the common intersection at E. Then, the line GE makes, with the line DEC, two angles, which together are

Of Planes.

equal to two right angles (Bk I. Th. ii): but these angles measure the inclination of the planes; therefore, the sum of the angles on the same side, which two planes make with each other, is equal to two right angles.

Cor. In like manner it may be demonstrated, that planes which intersect each other have their vertical or opposite angles equal.

THEOREM VIII.

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

Let the planes MN and PQ be perpendicular to the line AB: 0 then will they be parallel.

For, if they can meet any where, let O be one of their their common points, and draw OB, in the plane PQ, and OA,

in the plane MN.

M

D

N

C

Now, since AB is perpendicular to both planes, it will be perpendicular to OB and OA (Def. 1): hence, the triangle OAB will have two right angles, which is impossible (Bk. I. Th. xvii. Cor. 4); therefore, the planes can have no point, as O, in common, and consequently, they are parallel (Def. 4).

THEOREM IX.

If a plane cuts two parallel planes, the lines of intersection will be parallel.

meet each other, and, consequently, could not be parallel which would be contrary to the supposition.

THEOREM X.

If two lines are parallel to a third line, though not in the same plane with it, they will be parallel to each other.

At any point, G, in the line EF, let GI and GH be drawn in the planes FC, BE, and each perpendicular to FE at G. Then, since the line EF is perpendicular to the lines GH GI, it will be perpendicular to the plane HGI (Th. iv). And since FE is perpendicular to the plane HGI, its parallels AB and DC will also be perpendicular to the same plane (Th. v). Hence, since the two lines AB, CD, are both perpendicular to the plane HGI, they will be parallel to each other.

Of Planes.

THEOREM XI.

If two angles, not situated in the same plane, have their sides parallel and lying in the same direction, the angles will be equal.

Let the angles ACE and BDF

have the sides AC parallel to BD, and CE to DF: then will the angle ACE be equal to the angle BDF.

For, make AC equal to BD, and CE equal to DF, and join AB, CD, and EF; also, draw AE, BF.

Now since AC is equal and parallel to BD, the figure AD will be a parallelogram (Bk. I. Th. xxv); therefore, AB is equal and parallel to CD.

B

D

G

Again, since CE is equal and parallel to DF, CF will be a parallelogram, and EF will be equal and parallel to CD. Then, since AB and EF are both parallel to CD, they will be parallel to each other (Th. x); and since they are each equal to CD, they will be equal to each other. Hence, the figure BAEF is a parallelogram (Bk. I. Th. xxv), and consequently, AE is equal to BF. Hence, the two triangles ACE and BDF have the three sides of the one equal to the three sides of the other, each to each, and therefore the angle ACE is equal to the angle BDF (Bk. I. Th. viii).

THEOREM XII.

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