Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

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 0. 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 0.

В,

[graphic]
[ocr errors]

Q.E.D.

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.

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

Pass plane AF through AB and EF, plane CF through

D

c CD and EF, and plane AD through AB and point D. Let planes AD and CF

F

E
intersect in DC".
Then
DC" is parallel to FE.

(466) Hence DC and DC" coincide,

(Ax. 11.) and

BA and DC are lI.

[graphic]

Q.E.D.

PROPOSITION V. THEOREM

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

[merged small][merged small][ocr errors]

Нур. . The sides of the angles ABC and A'B'C" are respectively parallel and lie in the same direction.

To prove

LABC= LA'B'C".

Proof.

Take AB = A'B' and BC= B'C".

Draw

Hence

AA', BB', CC", AC, and A'C".
AB' is a parallelogram.

(Why ?) BC" is a parallelogram.

(Why?)
AA' is equal and parallel to BB',
BB' is equal and parallel to CC".
AA' is equal and parallel to CC". (Ax. 1.) (467)

AACC is a parallelogram,

and

Therefore

Whence

and

AC = A'C".

.. A ABC= A A'B'C".

(Why?)

.. ZB= ZB'.

Q.E.D.

[blocks in formation]

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

M,

D

B в

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

[blocks in formation]

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

[blocks in formation]

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

Whence

CD is I 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.

Ex. 1035. A line joining the midpoints of two adjacent sides of a quadrilateral in space is equal and parallel to a line joining the midpoints of the other two sides.

R

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

[graphic][merged small][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][subsumed][merged small][merged small][merged small][ocr errors][merged small]

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 AM I BC intersecting PQ in DH.

Through A in tbe plane AM draw AF I 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.
Since
BC is I to plane AM,

(Con.) and

BC is to IK,
. . IK is I to plane AM.

(469) .. FA is I to IK.

(Why ?)

But

FA is I to DH.

(Con.)

.:. FA is I to plane PQ.

(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.

AL

C'

[blocks in formation]

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.

« ΠροηγούμενηΣυνέχεια »