To prove AB, CD, and EF either meet in a point or are parallel.

Proof.

AB and EF lie in a plane.

Hence, they either meet or are parallel.

I. Suppose they meet (produced) in O. O being a point of AB lies in plane AD, and being a point of FE lies in plane CF.

Therefore, O must lie in the intersection of planes AD and CF, or in the line CD; that is, CD (produced) must pass through O.

B

C

II. Suppose AB and EF are parallel.

Then CD cannot meet AB, for otherwise FE would have to meet AB.

(Case I.)

Hence CD and AB, which lie in the same plane, must be parallel.

Similarly CD is parallel to EF.

Q.E.D.

467. COR. If two lines AB and CD are parallel to a third line EF, then AB and CD are parallel.

B

D

E

A

C

C'

PROPOSITION V. THEOREM

468. If two angles, not in the same plane, have th sides parallel and lying in the same direction, they equal.

C'

Нур.

The sides of the angles ABC and A'B'C' are respe

tively parallel and lie in the same direction.

Therefore

Whence

and

AA' is equal and parallel to CC'. (Ax. 1.) (467)

AA'C'C is a parallelogram,

PROPOSITION VI. THEOREM

469. If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to the plane.

Hyp. AB is to CD, and AB is to plane MN.

Proof. In the plane MN through B and D respectively draw two parallel lines BE and DF.

and CD is perpendicular to any line drawn through its foot.

Whence

CD is to plane MN.

Q.E.D.

Ex. 1034. If ABCD is a quadrilateral in space (i.e. A, B, C, and D do not lie in the same plane) and AB = BC, CD = DA, then plane angle A is equal to plane angle C.

PROPOSITION VII. PROBLEM

470. Through a given point, to draw a line perpendicular to a given plane.

Given. Plane PQ and point A.

Required. Through A, to draw a line I to PQ.

Construction. In PQ, draw any line BC.

Through A construct a plane AML BC intersecting PQ

in DH.

Through A in the plane AM draw AF DH.

AF is the required perpendicular.

Proof. Through the foot of AF (F in Fig. I and A in Fig. II) in plane PQ draw IK || to BC.

But

FA is to DH.

.. FA is to plane PQ.

(Con.)

(Why?) Q.E.F.

471. COR. 1. Through a given point, one and only one perpendicular can be drawn to a given plane.

For if there were two perpendiculars passing through the point, their plane would intersect the given plane in a straight line, perpendicular to both perpendiculars. Or, in a plane, there would be two perpendiculars to a given line, through a given point, which is impossible.

472. COR. 2. If two lines are perpendicular to the same plane, these lines are parallel.

But through D only one perpendicular can be drawn upon MN.

Hence DC and DC' coincide, and DC is || to AB.

Q.E.D.

Ex. 1036. The lines joining, in succession, the midpoints of the sides of a quadrilateral in space enclose a parallelogram.

Ex. 1037. The lines joining the midpoints of opposite sides of any quadrilateral in space bisect each other.