Proposition 3. Theorem.

500. If a straight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of those lines.

Hyp. Let OP be to PA, PB at the pt. P.

To prove OP is to the plane MN of these lines.

Proof. Join AB, and through P draw in MN any other st. line PC cutting AB in C.

M

B

N

Produce OP to O' making PO' = PO, and join O, O' to each of the

pts. A, B, C.

Since PA, PB are to 00' at its mid. pt., (Hyp.) (Cons.)

... OA = O'A, and OB = 0’B.

(66)

.'. ▲ OAB = ▲ O’AB, and .'. ≤OAC = ¿O'AC.

(108)

... OC

=

having two sides and the included equal each to each (104).
O'C, and... PC is 1 to 00' at its mid. pt. P. (67)
to any st. line in MN passing through its foot P.
to the plane MN.

.. OP is

.. OP is

(487)

Q. E. D.

501. COR. 1. At a given point in a plane, only one perpendicular to the plane can be erected.

For, if there could be two Is at the same pt. P, pass a plane through them whose intersection with the plane MN is AP; then these two Is would be both to the line AP at the same pt. P, which is impossible.

(51)

502. COR. 2. From a given point without a plane only one perpendicular can be drawn to the plane.

For, if OP, OA be two such Is, the ▲ OPA contains two rt. Zs, which is impossible.

(101)

Proposition 4. Theorem.

503. Conversely, all the perpendiculars to a straight line at the same point lie in a plane perpendicular to the

But in the plane OPC only one can be drawn to OP

at P,

(51)

... PC and PC' coincide, and PC lies in the plane APB.

Q. E. D. NOTE.-Hence a plane is determined by one point and the normal to the plane at that point.

504. COR. 1. At a given point in a straight line one plane, and only one, can be drawn perpendicular to the line.

505. Cor. 2. If a right angle be turned round one of its arms as an axis, the other arm will generate a plane.

506. COR. 3. Through a given point without a straight line one plane, and only one, can be drawn perpendicular to the line.

For, in the plane of OP and the pt. C, the CP can be drawn to OP; then the plane generated by turning PC round OP will be to OP; and it is clear that there is only one such plane.

Proposition 5. Theorem.

507. If from the foot of a perpendicular to a plane a straight line is drawn at right angles to any line in the plane, and its intersection with that line is joined to any point of the perpendicular, this last line will be perpendicular to the line in the plane.

Hyp. Let OP be a to the plane MN, PA a from P to any line BC in MN, and OA a line joining A with any pt. O in OP.

M.

N

Then since O and A are each equally distant from B and C,

.•. OA is ¦ to BC.

(67)

Q.E.D.

508. COR. 1. The line BC is perpendicular to the plane

of the triangle OPA.

For, it is to the st. lines AP, AO at pt. A.

(500)

COR. 2. The line PA measures the shortest distance between OP and BC.

EXERCISES.

1. Find the locus of points equally distant from two given points.

2. Given a straight line and any two points: find a point in the straight line equally distant from the two points.

Proposition 6. Theorem.

509. Two straight lines perpendicular to the same plane are parallel.

Hyp. Let AB, CD, be to the

plane MN at the pts. B, D.

To prove

M

AB || to CD.

... CD, AD, BD, are all in the same plane,

all the 1s to a st. line at the same pt. lie in the same plane (503). ... AB and CD lie in the same plane,

510. COR. 1. If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to that plane.

For, if AB is || to CD, and to the

plane MN, then a

A C

M

to MN at D will be ||

B D

N

to AB (509), and will coincide with CD (501). ... CD is to MN.

511. COR. 2. Two straight lines that are parallel to a

third straight line are parallel to each

other.

For, each of the lines AB, CD, is

to a plane MN that is to EF (510); ... AB and CD are || .

(509)

N

A CE

M

PARALLEL PLANES.

Proposition 7. Theorem.

512. If two straight lines are parallel, each of them is parallel to every plane passing through the other and not containing both lines.

Hyp. Let AB, CD be two || st. lines, A.

and MN any plane passing through

M

N

... if AB meets the plane MN, it must meet it in some

pt. of the intersection CD, of the two planes.

But AB is | to CD, and so cannot meet it.

... AB cannot meet the plane MN.

... AB is to MN.

(Hyp.)

Q.E.D.

513. COR. 1. A line parallel to the intersection of two planes is parallel to each of those planes.

514. COR. 2. Through any given straight line, a plane can be passed parallel to any other given straight line. For, through any pt. C of CE draw CD to AB; then the plane of DCE is || to AB.

(512)

515. COR. 3. Through a given point a plane can be passed parallel to any two given straight lines in space.

For, draw through the given pt. O, in the plane of the given line AB and O, the line A'B' || to AB, and in the plane of the given line CD and O, the line C'D' || to CD; then the plane of the lines A'B', C'D' is || to each of the lines AB and CD.

M

A

A.

B

-E

B

O

(512)