THEOREM II.

Parallel lines intercepted between parallel planes are

equal.

Let FH and EG be any two parallel lines intercepted between two parallel planes, AB and CD. It is to be shown that FH is equal to EG.

Through the parallels FH and EG let the plane FG pass. Its intersections with the parallel

planes AB and CD will be parallel lines (Def. 7); that is, FE is parallel to HG. Hence, FEGH is a parallelogram, and FH is equal to EG (Theo. XX, Book I). Therefore, parallel lines, etc.

Cor. The perpendicular distance between two parallel planes is everywhere the same.

THEOREM III.

A straight line perpendicular to one of two parallel planes is perpendicular to the other.

Pass a plane through AO and AB. It will cut DP in some line, as HO, parallel to AB (Def. 7). But BA is perpendicular to AO (Def. 3), and, therefore, HO is perpendicular to AO. But HO is any line in the plane DP, therefore, AO is perpendicular to the plane DP (Def. 3).

THEOREM IV.

When an oblique line meets a plane, the angle which it makes with its own projection is the least angle it makes with any line in the plane.

From a point without a plane there can be only one straight line drawn perpendicular to the plane.

If there can be two, as AB and AC, draw the line BC, and we have the triangle ABC containing two right angles.

Ques.-What false assumption leads to this absurdity?

Cor. From the point A an infinite number of equal lines can be drawn to the plane

B

N

PN, and their points of intersection with the plane will form a circle.

Ques. 1.-In what case would this infinite number be reduced to one?

Ques. 2.-How can a perpendicular to a plane be drawn from a point without the plane?

Ques. 3.-A room is ten feet high. How, by the use of an inflexible rod eleven feet long, can a point on the floor be found directly beneath a given point on the ceiling (Def. 11, Sec. XIV, Cor.)?

THEOREM VI.

A line, perpendicular to two lines of a plane at their intersection, is perpendicular to the plane.

H

A

B

If PC is perpendicular to two lines, CA and CD, of the plane HO, at their intersection, then will it be perpendicular to any other line of the plane, as CB, drawn to its foot; and, therefore (Def. 3, Sec. XIV), be perpendicular to the plane.

Extend PC below the plane

till P'C = PC. Draw the line AD cutting CB in B, and draw PA, PB, PD, and P'A, P'B, P'D.

PB is equal to P'B.

For PA P'A (Book I, Theo. IX).

=

PD = P'D.

.. the triangles PAD and P'AD are equal (?).

And the triangles PAB and P'AB are equal (Book I, Theo. XII).

.. PB P'B, and, as PC P'C, the line CB is perpendicular to PP' (Book I, Theo. XIII, Cor. 5). And PC is perpendicular to the plane HO.

Ques. 1.-How many perpendiculars to the plane HO can be drawn from P?

Ques. 2.-How many parallels to the plane can be drawn through P?

THEOREM VII.

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

PC is perpendicular to the plane AB. Pass through PC the plane HO. Then will HO be perpendicular to AB.

From C draw CD, in the plane AB, perpendicular to the intersection ON; then will PCD be a right angle, and, as it measures the an

A

H

N

gle of inclination (Def. 5, Sec. XIV), the plane HO is perpendicular to the plane AB, and HO is any plane passed through PC.

THEOREM VIII.

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

The planes HS and MN are perpendicular to each other. AB is a straight line in HS perpendicular to OS, their intersection. BC is drawn in the plane MN perpendicular to OS

A

M

S

at B. Then, ABC is a right angle (?).

N

The line AB is perpendicular to BS and BC in the plane MN. It is therefore perpendicular to the plane (Theo. VI).

Ques. How can you draw a line upon the wall, Ques.-How which will be perpendicular to the floor?

THEOREM IX.

If two planes are each perpendicular to a third plane, their intersection is perpendicular to that plane.

The angle included by two perpendiculars drawn from any point within a diedral to its faces, is the supplement of the diedral angle.

T

S

Let A be a point within the diedral H-OT-S, and AB and AC the perpendiculars to its faces, HO and SO. Then will BAC be the supplement of the diedral angle; or, more strictly, the supplement of the angle which measures the diedral. Apply Theorems VII and IX, and prove that BDC is the diedral angle. Finish the demonstration.

Cor. 1.-Any two parallel lines lie in the same plane. Cor. 2. Any two lines forming an angle lie in the same plane.

Cor. 3.-Through a given line of a plane, only one plane can be passed perpendicular to the given plane.