506. A plane is determined if it passes through:

I. Three points not in the same straight line.
II. A straight line and a point without that line.
III. Two intersecting straight lines.

IV. Two parallel straight lines.

I. GIVEN-three points, A, B, and C not in the same straight line. TO PROVE that one and only one plane can be passed through them.

Pass a plane MN through one of the points, turn it about this point until it contains one of the other points, and then turn it about these two points until it contains the third. No other plane will contain these points.

For, suppose PQ to be such a plane.

Take X any point in PQ. X will be proved to be also in MN. Draw the straight lines AB and AC.

These will be in both planes, since A, B, and C lie in both planes.

$504

Through X draw a straight line in PQ cutting AB and AC in D and E.

Since D and E lie in the plane MN, the straight line DEX lies wholly in MN.

$504

Hence X, a point in DE, lies in the plane MN.

Thus any point, that is, every point in the plane PQ lies also in the plane MN, and in like manner it can be proved that every point in MN lies in PQ.

Therefore the two planes coincide.

II. GIVEN the straight line AB and the point C without AB.

Q. E. D.

TO PROVE that one and only one plane can be passed through them.

The plane passed through C and any two points of AB will contain AB. $504

No other plane can be passed through AB and C, for then there would be two planes containing three points not in the same straight line, which is impossible.

Q. E. D.

III. GIVEN the straight lines AB and AC intersecting in A. TO PROVE that one and only one plane can be passed through them.

The plane passed through the three points A, B, and C will contain the straight lines AB and AC. $504

No other plane can be passed through AB and AC, for then there would be two planes containing three points not in the straight line, which is impossible.

Q. E. D.

IV. GIVEN-the parallel straight lines FG and KL.

TO PROVE that one and only one plane can be passed through

them.

These parallel lines lie in the same plane.

831

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

Q. E. D.

50%. Def. The intersection of two planes is the line common to both planes.

PROPOSITION II. THEOREM

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

GIVEN

TO PROVE

two intersecting planes, MB and EB.
their intersection is a straight line.

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.

§ 506 I

Q. E. D.

PERPENDICULAR AND OBLIQUE LINES AND PLANES

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

510. 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 perpendicular to the line.

511. 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.

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

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

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.

$78

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. Therefore every straight line in MN passing through G

$ 104