By definition these parallel lines lie in the same plane.

831

There cannot be two planes passed through them, for then we would have two planes containing three points F, G, and K, not in the same straight line, which is impossible.

Q. E. D.

527. Def.-The intersection of two planes is the line common to both planes.

PROPOSITION II. THEOREM

528. If two planes intersect, their intersection is a straight

If possible, suppose the intersection is not straight.

It would then contain three points not in the same straight line.

That is, the two planes would contain three points not in the same straight line, which is impossible. Therefore the intersection must be a straight line.

§ 526 I

Q. E. D.

PERPENDICULAR AND OBLIQUE LINES AND PLANES

529. Def.—If a straight line meet a plane, its point of meeting is called its foot.

530. Defs.-A straight line is perpendicular to a plane, if it is perpendicular to every straight line in the plane drawn through its foot. In the same case the plane is said to be perpendicular to the line.

PROPOSITION III. THEOREM

531. If two intersecting straight lines are perpendicular to a third at the same point, their plane is perpendicular to that straight line.

GIVEN-the two intersecting straight lines GC and GD perpendicular to the straight line BG at the point G.

TO PROVE that the plane MN passed through GC and GD is perpendicular to BG.

In the plane MN draw through G any straight line GH.

Let CD be any straight line cutting the lines GC, GH, and GD in C, H, and D.

Produce the line BG to A making GA equal to GB, and join A and B to C, H, and D.

Then, since GC is perpendicular to BA at its middle point,

And the homologous angles BCH and ACH are equal.

Hence the triangles BCH and ACH are equal.

$79

Therefore their homologous sides BH and AH are

equal.

Therefore GH is perpendicular to BA.

But GH is any straight line in MN passing through G.

S104

Therefore every straight line in MN passing through G is perpendicular to BA; that is, MN is perpendicular to BA.

$530

Q. E. D.

532. COR. I. At a given point in a straight line one and only one plane can be drawn perpendicular to that straight

line.

Hint.-Let AB be the straight line and G the point.

At G draw the straight lines GC and GD perpendicular to AB.
The plane of these lines will be perpendicular to AB. (Why?)

Only one such plane can be drawn.

For any other plane passing through G cannot contain both of the lines GC and GD. (Why?)

It must therefore cut one of the planes BGC and BGD, say BGC, in some line GC' other than GC and GD.

Since BGC' is not a right angle, this second plane is not perpendicular to AB. (Why?)

533. COR. II. Through a given point without a straight line one and only one plane can be passed perpendicular to that straight line.

Hint.-Use § 531 to draw one such plane. Any other plane cuts 40 either at O or at some other point, O'. § 532 shows that the first is not perpendicular. Show also that the second is not.

534. COR. III. All the perpendiculars to a given straight line at the same point lie in a plane perpendicular to that line at that point.

Hint.-Every pair of these perpendiculars, as OA and OB, determines a plane perpendicular at O. (Why ?)

And all the planes thus determined must coincide. (Why?) Hence, etc.

535. COR. IV. At a point in a plane one and only one perpendicular to the plane can be drawn.