Proposition 15. Theorem.

537. If two planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their intersection is perpendicular to the other.

Hyp. Let the planes PQ and MN

be to each other, and let AB be drawn in PQ to their intersection

...ABD is the plane of the rt. diedral / PCQN. (528)

Because

PQ is to MN,

...ABD is a rt..

(Hyp.)

(536)

.. AB is to QC and BD at their pt. of intersection.

... AB is to the plane MN.

(500) Q.E.D.

538. COR. 1. If two planes are perpendicular to each other, a straight line through any point of their intersection perpendicular to one of the planes will lie in the other.

See (501).

539. COR. 2. If two planes are perpendicular to each other, a straight line from any point of one plane perpendicular to the other will lie in the first plane.

See (502).

Proposition 16. Theorem.

540. If a straight line is perpendicular to a plane, every plane passed through the line is perpendicular to that plane.

Hyp. Let AB be to the plane MN, and PQ any plane passed through AB, intersecting MN in QC.

To prove plane PQL to plane MN. Proof. In the plane MN draw BD to QC at B.

Because AB is to MN, (Hyp.)

... AB is to QC and BD.

M

P

N

(487)

.. ABD is the plane of the diedral / PCQN. (528)

But

ZABD is a rt. 2.

... PQ is to MN.

(Proved above)

Q.E.D.

541. COR. A plane perpendicular to the edge of a diedral angle is perpendicular to its faces.

542. SCH. When three straight lines, AB, CB, DB, are perpendicular to one another at the same point, each line is perpendicular to the plane of the other two, and the three planes are perpendicular to one another.

EXERCISE.

If two planes which intersect contain two lines parallel to each other, the intersection of the planes will be parallel to the lines.

Proposition 17. Theorem.

543. Through a given straight line oblique to a plane, a plane can be passed perpendicular to the given plane, and but one.

Hyp. Let AB be the given st. line oblique to the plane MN.

To prove that one plane can be passed through AB to MN, and but one. Proof. Draw AD to MN from any pt. A of AB.

Through AB and AD pass a plane AC.

A

M

CL

D

N

Because the plane AC passes through the IAD, (Cons.)

... the plane AC is to the plane MN.

(540)

Also, because any plane passed through ABL to MN must contain the

AD,

(539)

... the plane AC is the only plane to MN that can be passed through AB.

(485)

Q.E.D.

544. COR. 1. The projection of a straight line on a plane is a straight line.

For, the 1s from all pts. of AB to MN lie in the plane ACL to MN (539), and therefore these Is all meet MN in the intersection CD of the two planes, which is a straight line. (495) 545. COR. 2. If a line intersects a plane, its projection passes through the point of intersection.

Proposition 18. Theorem.

546. If two intersecting planes are each perpendicular to a third plane, their intersection is perpendicular to the third plane.

Hyp. Let the planes PQ, RS, intersecting in AB, be to the third

Р.

R

N

(538)

Because this lies in each plane PQ and RS, it is their line of intersection AB.

.. AB is to the plane MN.

Q. E.D.

547. COR. 1. A plane perpendicular to each of two intersecting planes is perpendicular to their intersection.

548. COR. 2. If the planes PQ and RS include a right diedral angle, the three planes PQ, RS, MN, are perpendicular to one another; the intersection of any two of these planes is perpendicular to the third plane; and the three intersections are perpendicular to one another.

Compare (542).

EXERCISE.

Show that three planes in general intersect in one point: what are the exceptions?

Proposition 19. Theorem.

549. Every point in the plane bisecting a diedral angle is equally distant from the faces of the angle.

Hyp. Let AM be the plane which bisects the diedral / CABD, and let PE, PH be s from any pt. P in the plane AM to the planes CB and DB.

To prove

PE = PH.

Proof. Let O be the pt. where

the plane EPH cuts the line AB.

ய

B

M

H

DB.

PH is to DB,

(Cons.)

the plane EPH is to each of the planes CB and

.. the plane EPH is ... the s POE, POH

Zs CABM and DABM.

(540)

to their intersection AB. (547) are the plane s of the diedral

(528)

Because the diedral s CABM, DABM are equal, (Hyp.)

.. ≤ POE = ≤ POH.

.. rt. A POE = rt. ▲ POH,

having the hypotenuse and an acute =, each to each (106).

(536)

1. If three planes have a common line of intersection, the normals drawn to these planes from any point of that line are in one plane.

2. If from any point perpendiculars be drawn to the faces of a diedral angle, the angle between these perpendiculars is the supplement of the diedral angle between the planes in which the point is situated.